Complex Analysis through Examples and Exercises

Kluwer Text in the Mathematical Sciences

VOLUME21

A Graduate-Level Book Series

The titfes published in this series are listed at the end 0/ this vofume.

Complex Analysis through Examples and Exercises

by

Endre Pap Institute of Mathematics, University of Novi Sad, Novi Sad, Yugoslavia

SPRINGER-SCIENCE+BUSINESS MEDIA, B.V.

A c.I.P. Catalogue record for this book is available from the Library of Congress.

ISBN 978-90-481-5253-7 ISBN 978-94-017-1106-7 (eBook) DOI 10.1007/978-94-017-1106-7

Printed an acid-free paper

AII Rights Reserved

© 1999 Springer Science+Business Media Dordrecht Originally published by Kluwer Academic Publishers in 1999 N o part of the material protected by this copyright notice may be reproduced or utilized in any form or by any means, electronic or mechanical, inc1uding photocopying, recording or by any information storage and retrieval system, without written permission from the copyright owner.

Contents

Contents v

Preface IX

1 The Complex Numbers 1

1.1 Algebraic Properties 1

1.1.1 Preliminaries 1

1.1.2 Examples and Exercises 2

1.2 The Topology of the Complex Plane 32

1.2.1 Preliminaries ...... 32


2 Sequences and series 37

2.1 Sequences ..... 37



2.2 Series ......... 44



3 Complex functions 53

3.1 General Properties 53



vi CONTENTS

3.2 Special Functions .. 64



3.3 Multi-valued functions 68



4 Conformal mappings 73

4.1 Basics ....... 73



4.2 Special mappings .. 74



5 The Integral 103

5.1 Basics ••• 0 •• 103



6 The Analytic functions 129

6.1 The Power Series Representation 129

6.1.1 Preliminaries ...... 129


6.2 Composite Examples ...... 162

7 Isolated Singularities 171

7.1 Singularities . . . . 171



7.2 Laurent series .... 177


CONTENTS vii

7.2.2 Examples and Exercises ..................... 179

8 Residues

8.1 Residue Theorem

8.1.1 Preliminaries

8.1.2 Examples and Exercises

8.2 Composite Examples ..... .

9 Analytic continuation

9.1 Continuation ...

9.1.1 Preliminaries


9.2 Composite Examples . . . . . .

10 Integral transforms

10.1 Analytic Functions Defined by Integrals.

10.1.1 Preliminaries ..... .


10.2 Composite Examples ..... .

11 Miscellaneous Examples

Bibliography

List of Symbols

Index

191

· 191

· 191

· 193

.206

227

.227

.227

.228

.233

255

.255

.255

.256

.268

313

333

335

336

Preface

The book Complex Analysis through Examples and Exercises has come out from the lectures and exercises that the author held mostly for mathematician and physists . The book is an attempt to present the rat her involved subject of complex analysis through an active approach by the reader. Thus this book is a complex combination of theory and examples.

Complex analysis is involved in all branches of mathematics. It often happens that the complex analysis is the shortest path for solving a problem in real circumstances. We are using the (Cauchy) integral approach and the (Weierstrass) power se ries approach .

In the theory of complex analysis, on the hand one has an interplay of several mathematical disciplines, while on the other various methods, tools, and approaches. In view of that, the exposition of new notions and methods in our book is taken step by step. A minimal amount of expository theory is included at the beinning of each section, the Preliminaries, with maximum effort placed on weil selected examples and exercises capturing the essence of the material. Actually, I have divided the problems into two classes called Examples and Exercises (some of them often also contain proofs of the statements from the Preliminaries). The examples contain complete solutions and serve as a model for solving similar problems given in the exercises. The readers are left to find the solution in the exercisesj the answers, and, occasionally, some hints, are still given. Special sections contain so called Composite Examples which consist of combinations of different types of examples explaining, altogether, some problems completely and giving to the reader an opportunity to check his entire previously accepted knowledge.

The necessary prerequisites are a standard undergraduate course on real functions of real variables. I have tried to make the book self-contained as much as possible. For that reason, I have also included in the Preliminaries and Examples some of the mathematical tools mentioned.

The book is prepared for undergraduate and graduate students in matheniatics, physics, technology, economics, and everybody with an interest in complex analysis.

We have used for some calculations and drawings the mathematical software

ix

x PREFACE

packages Mathematica and Scientific Work Place v2.5.

I am grateful to Academician Bogoljub Stankovic for a long period of collaboration on the subject of the book, to Prof. Arpad TakaCi for his numerous remarks and advice about the text, and to Ivana Stajner for reading some part of the text. I would like to express my thanks to MarCicev Merima for typing the majority of the manuscript. It is my pleasure to thank the Institute of Mathematics in Novi Sad for working conditions and financial support. I would like to thank Kluwer Academic Publishers, especially Dr. Paul Roos and Ms. Angelique Hempel for their encouragement and patience.

Novi Sad, June 1998 ENDRE PAP

Chapter 1

The Complex N umbers

1.1 Algebraic Properties

1.1.1 Preliminaries

The field of complex numbers Cis the set of all ordered pairs (a, b) where a and b are real numbers and where addition and multiplication are defined by:

(a, b) + (c, d) = (a + c, b + d)

(a, b)(c, d) = (ac - bd, bc + ad).

We will write a for the complex number (a,O). In fact, the mapping a 1-+ (a,O) defines a field isomorphism of IR into C, hence we may consider IR as a subset of C. If we put z = (0,1), then (a, b) = a + bz. For z = a + zb we put Re z = a and Imz = b. Real numbers are associated with points on the x-axis and called the real axis . Purely imaginary numbers are associated with points on the y-axis and called the imaginary axis .

Note that z2 = -1, so the equation z2+ 1 = 0 has a root in C. If z = x+zy (x, y E IR), then we define

Izl = Jx2 + y2

to be the absolute value of z and z = x - zy is the conjugate of z. We have Izl 2 = zz and the triangle inequality

Iz + wl s:; Izl + Iwl (z,w E C).

By the definition of complex numbers, each z in C can be identified with a unique

point (Rez,Imz) in the plane IR2 •

1

E. Pap, Complex Analysis through Examples and Exercises© Springer Science+Business Media Dordrecht 1999

2 CHAPTER 1. THE COMPLEX NUMBERS

The argument of z i- 0, denoted by arg z, is the angle 0 (modulo 271") between the vector from the origin to z and positive x-axis, i.e., which satisfy

Rez cosO = Tz! and . 0 Imz

sm = Tz!.

The point z = x+zy =I- 0 has polar coordinates (r, 0) : x = r cos 0, y = r sin O. Clearly r = Izl and 0 is the angle between the positive real axis and the line segment from ° to z. Notice that 0 plus any multiple of 271" can be substituted for 0 in the above equations. The angle 0 is called the argument of z and is denoted by 0 = argz.

Let Zl = rl ( cos Ol + z sin Ol and Z2 = r2 ( cos O2 + i sin ( 2 ) then

In particular, if z = r( cos 0 + z sin 0), then

zn = rn(cos(nO) + zsin(nO». (1.1 )

As a special case of (1.1) we obtain DeMoivre's formula:

(cos 0 + z sin Ot = cos nO + z sin nO. (1.2)

The n-th root of z = r( cos 0 + z sin 0) are

o + 2h . 0 + 2h Zk = y'r(cos + zsm )

n n

for k = 0,1, ... , n - 1.

The complex number z = r( cos O+z sin 0) also has the exponential representation

z = r exp( zO).

For more explanations see the chapter on power series.


Example 1.1 Find the real numbers p and q such that the complex numbers

1 z = p + zq, W = P + z- be equal.

q

1.1. ALGEBRAIC PROPERTIES 3

Solution. We have that z = w is equivalent with Rez = Rew and Imz =

Immw. Therefore p = p and q = !, pEllt, q2 = 1, i.e., ql = 1, q2 = -1 and pEllt. q

Example 1.2 For z = 1 + z find w such that the real parts of the following numbers are equal to zero a)z + Wj b) Z· Wj c) fuj d) ~.

Solution. Let z = 1 + z = (1,1) and w = x + zy = (x, y). Then we have

a) Re(z + w) = ° {=} 1 + x = 0.

Hence x = -1 and y E llt is arbitrary.

b) Re (z . w) = ° {=} Re (x - y + z( x + y)) = 0.

Hence x - y = 0. Therefore w is given by w = x + zx = x(l + z) = x· z, for arbitrary xE llt.

c) We have fu = 0. Hence

z 1 + z x - zy x + y + z( x - y) - = -- . -- = ---'--:--'--::--~ W X + zy x - zy x 2 + y2

Therefore

Re(-=-) =o{=} ~+Y2 =O:::}x=-y,xi=0. w x +y

Finally we have w = x(l - z) for x E llt and x i= 0.

d) From Re (~) = ° it follows that for every x E llt :

w x + zy 1 - z x + y + z(x - y) -=--.--= z l+z 1-z 2

Hence W = x(l - z) for every x E llt.

Example 1.3 Prove that

Hint. It is easy to prove the case n = 2 and then use mathematical induction to prove the general case.

Example 1.4 Find for z = 1 + 2z the following numbers

a) zn j b) l/zj c) l/zn j d) Z2 + 2z + 5 + z.


Solution. a) We have

zn (1 + 2zt E (~)(2z)k [.!!.) [n-l) IJ-l)k(n)22k+Z 't(-l)k( n )22k+1,

k=O 2k k=O 2k + 1

where [xl is the greatest integer part of x.

b) We have 1 1 1 - 2z 1 - 2z 1 2 - = -- . -- = -- = - - - z. z 1 + 2z 1 - 2z 5 5 5

e) Sinee ~ = (~)n, we have by b) zn Z

1 (1 2)n 1 n (1 + 2z)n = 5" - 5" z = 5n (1 - 2z) .

Applying the same proeeclure as in a) we obtain (using that the imaginary part of z is -2) :

cl) Putting z = 1 + 2z in Z2 + 2z + 5 + z, we obtain 4 + 9z.

Example 1.5 Find the positions of the following points in the complex plane:

a + za, a - za, -a + za, -a - za fOT a E lR?

Solution. Using the trigonometrie representation (p,O) wc obtain

a + za V2 V2

V2aT + zv'2aT

(v'2' a ~) '4 '

a - za v'2 V2 V2a . - - zV2a-2 2

(v'2a -~) , 4'

-a + za (V2a, 3:),

-a - za (V2a, 3:).

Hence the four given points are the corners of the square in the circle with the center at origin and the radius J21a1-

Example 1.6 Which subsets of the complex plain correspond to the complex num-bers with the following properties:

a) Rez = Imz; b) Rez< 1; c) -1:::;Rez:::;l;

d) Imz 2 0; e) Izl :::; 2; f) 1 < Izl < 3;

g) Izl > 2; h) -7r < argz < 7r; i) ~ < argz < f?

Solution.

a) Rez = Imz {::::=:} x = y, where z = x + iy.

The desired sub set consists of the points of the straight line y = x (Figure 1.1).

x

Figure 1.1 Rez = Imz

b) Rez < 1 {::::=:} x < 1 and y is an arbitrary real number, where z = x + iy. The desired subset is the half plane left from the straight li ne x = 1 (without the points

of this straight line), Figure 1.2.

x

Figure 1.2 Rez < 1

c) -1 :::; Re z :::; 1 means -1 :::; x :::; 1 and y is an arbitrary real number. The desired subset is the strip between straight lines x = -1 and x = 1, Figure 1.3.

:..1 x

Figure 1.3 -1 :::; Rez :::; 1

d) Im x ~ 0 means y ~ 0 and x is an arbitrary real number. The desired subset

is the upper half plane with respect to the x- axis, Figure 1.4.

Figure 1.4 Imz ~ °

e) The eondition JzJ ~ 2, using the trigonometrie representation of the complex number z = (p, ()), reduced on JzJ = p ~ 2 and () is an arbitrary angle from [0, 27l'J.

The desired sub set is the disc with center at (0,0) and radius r = 2, Figure 1.5.

2 X

Figure 1.5 JzJ ~ 2

f) The ease 1 < JzJ < 3 reduces in a sirnilar way as in e) on 1 < JzJ = P < 3, where () is an arbitrary angle from the interval [0,27l'J. The desired sub set is the

annulus between the circles Izl = 1 and Izl = 3 without these circles, Figure 1.6.

Figure 1.6 1 < Izl < 3

g) For the case Izl > 2 we have JzJ = p > 2, () E [0, 27r]. Therefore the desired subset is the whole complex plane without the disc Jzl ~ 2, Figure 1.7.

Figure 1.7 Izl > 2

h) The condition -7r < arg z < 7r implies that p is arbitrary and -7r < () < 7r.

Therefore the desired subset is the whole complex plane without the negative part

of the real axis, Figure 1.8.

x

Figure 1.8 -'Fr < arg z < 'Fr

i) The condition 'Fr /6 < arg z < 'Fr /4 implies that p is arbitrary and e E ('Fr /6, 'Fr /4). The desired sub set are the points between half straight lines y = x . tan ~ and y = x . tan i without these half straight lines, Figure 1.9.

Figure 1.9 'Fr /6 < arg z < 'Fr /4

Example 1.7 Find the conditions for the real and imaginary part of a complex number z so that z belongs to the triangle with vertices 0,1 + zY3, 1 + z/ Y3, Figure 1.10.


Solution. We denote the vertices as:

Zl = 0, Z2 = 1 + zV3, Z3 = 1 + z/V3.

y

O=z 1

Figure 1.10

If we denote z = x + zy, then Rez = x and Imz = y and

Y [1C' 1C'] - = tan (), () E -, - . x 6 3

x

Therefore Im z = Re z . tan (). Since the function tan () is monotone increasing we have

1 y'3Rez ::::; Imz ::::; V3Rez.

Together with the condition 0 ::::; Re z ::::; 1 we have completely described the points in the given triangle with vertices Zt, Z2 and Z3.

Example 1.8 Prove that tor every z E C

Izl ::::; IRezl + IImzl ::::; -12 ·14

Solution. The left part of the inequality follows from

Izl = v'Re 2z + Im 2z::::; y'(IRezl + IImzl)2 = IRezl + IImzl

1.1. ALGEBRAIC PROPERTIES

The right part of the inequality follows from the following obvious inequality

(IRezl-IImzr ~ O.

Then IRezl 2 + IImzl 2 - 2lRezl·IImzl ~ 0,

Izl2 ~ 2IRezl·IIm z l·

Adding Izl 2 to the both sides of the last inequality we obtain

21z12 ~ 2lRezl'IImzl + Iz12,

21z12 ~ (IRezl + IImzlt Hence v'2lzl ~ IRezl + IImzl·

11

Example 1.9 Find the complex numbers which are the corners 01 the triangle with equal sides with vertices on unit circle and whose one vertex is on the negative part the real axis.

Solution. Let us put Zj = pj(cos8j + zsin8j),j = 1,2,3, for the soughtafter points.

/ /

Figure 1.11

We have for alt three vertices Pj = 1 and

Zl : 81 = 7r /3, z2: 82 = 7r, Z3: 83 = -7r /3.

Therefore:

Zl = (1, 7r/3) = 1/2 + zV3/2; Z2 = (1, 7r) = -1; Z3 = (1, -7r/3) = 1/2 - zV3/2,

Figure 1.11.


Exercise 1.10 Find the real and imaginary parts, modulus, argument and the eom-plex conjugate for the following numbers:

a) 51; b) 7r' , e) 1/2 + 1/2; d) 1 + l/VJ;

e) 1 1) (3 + 13)(1 + 1VJ); g) C +11VJf; h) 1 + 1

1 + 1VJ; 1 - 1

Answers.

Rez Imz Izl argz z

a) 0 5 5 7r/2 -51

b) 7r 0 7r 0 e) 1/2 1/2 ../2/2 7r/4 1/2 - 1/2

d) 1 1/v3 2../3/3 7r/6 1-1/../3

e) 1/4 -V3/ 4 1/2 57r/3 1/4 + 1V3/4

1) 3-3V3 3+3V3 V72 arctan( -2 - V3) 3 - 3V3 - (-3 + 3V3)1

g) -1/23 0 1/23 7r -1/23

h) 0 1 1 7r/2 -1

Example 1.11 Find t and 0 so that the eomplex numbers

z = t + tf) and w = t(cosO + lsinO)

would be equal.

Solution. The equality z = w implies

Izl = Iwl and tan(argz) = tan(argw).

The first condition Izl = Iwl implies t2 = t2 + 02 , i.e., 02 = O. Putting 0 = 0 in z and w we obtain z = t and w = t for t an arbitrary real number.

Example 1.12 Let C* be the set of alt eomplex numbers different from zero.

a) Prove that the set T of all eomplex numbers with modulus 1 is a multiplicative subgroup of the group (C*,.).

b) The multiplieative group C* is isomorphie with jR+ X T.

Solution. a) We have T = {zllzl = I} C C. Tunder the multiplication by the equality

IZI . z21 = IZII·lz21· The associativity follows by the associativity of multiplication in C. The neutral element is 1. The inverse of z = (a, b) E T is

C2 : b2 ' a2 -: b2) E T.

b) It is easy to check that an isomorphism is given by

J(z) = (lzl,cosO+zsinO),

where 0 = argz. nopagebreak c) We define an equivalence relation'" on R by rl '" r2 <===} rl - r2 = 2br, k is an integer. Let i be the corresponding quotient set. Prove that R+ x T isomorphie with R+ xi. The group c· is isomorphie with R+ x i (check). Therefore by b) and transitivity of the isomorphisms of groups it follows c).

Example 1.13 Find the sum of complex numbers which are the vertices of a npolygon in circle with radius r with the center in (0,0) for n = 4,6, ... , 2p, p E F:!, Figure 1.12.

Solution. Using the geometrical interpretation of the addition we see that the surn is O.

x

Figure 1.12 hexagon

Namely, adding the pairs of complex numbers on main diagonals we always


obtain 0 and then the total sum is 0 for n = 4,6, ... , 2p.

Second method. Using the Euler representation of complex number Z = re"" we obtain

Example 1.14 Prove that the condition Jor Jour points Zl,Z2,Z3 and Z4 be consequent vertices oJ a parallelogram is the Jollowing

Solution. The equality Zl - Z2 + Z3 - Z4 = 0 implies

Exercise 1.15 For three given complex numbers Zl, Z2 and Z3, find Z4 such that the corresponding points in the complex plane will be the corners oJ a parallelogram ..

Rint. Complex number Z4 can be obtained by Example 1.14:

Example 1.16 Starting with a complex number Z i- 0, find where the complex numbers 2z,3z, ... ,nz are?

Solution. For Z = p( cos () + z sin ()) we have

nz = np( cos () + z sin ()),

the complex numbers nx, for n = 1,2, ... are on the half straight line y = tan () . x with modulus np.

Example 1.17 Find the length oJ a side oJ a pentagon and the length oJ its diagonal iJ it is inside the unit circle.

Solution. We have Zl = 1. Then

27r . 27r Z2 = cos - + z sm -

5 5 and

47r . 47r Z3 = cos - + z sm -.

5 5


The length of the side Zl - Z2 is

I 2~ . 2~1 1 - cos 5" - z sm 5"

2~ . 2~ (1 - cos 5" )2 + (sm 5")2 =

= )2(1 _ cos 2; ) . ~

2sm 5,

where we have used the identity 2 sin2 ~ = 1 - cos t.

_-+-+-__ ~I--___ ~Z ' ____ )

1 X

Figure 1.13

15

We obtain analogously IZ2 - z31 diagonals.

2 sin 2;, Figure 1.13. Find the length of

Exercise 1.18 Find where the points Z are for a fixed Zo :

a) Iz-zol=l; b) Iz + zol = 1; ~

c) arg (z . zo) = 4'

Answers. a) The set {zllz - zol = I} is the circle with center Zo and radius 1, Figure 1.14.

y

o x

Figure 1.14 {zllz - zol = I}

b) {zllz + zol = I} is the circle with center -Zo and radius 1, Figure 15.

Figure 1.15 {zllz + zol = I}

c) {z 1 arg (z . zo) = !}, the corresponding complex numbers are on the half

straight line y = tan (! - O)x, where argzo = 00 , This follows by arg(z· zo) = 7r

00 + 0 = 4'

Exercise 1.19 Find where the points in the complex plane are if

a) Iz - 11 ::; 2; b) Iz + zl > 1; c) Iz - zol < r;

d) 7r

e) 7r 7r

f) larg (z - zo)1 < 0; arg (z - 1) = -; 4 < arg (z + z) ::; 4; 4

g) Re (zo . z) = 0, for Zo E li; h) III < r; i) Re(zz) = !; j)

z - = z. z


Answers. a) The closed disc with center at (1,0) and radius r = 2, Figure 1.16.

--~~~~~~r--' -1 X

Figure 1.16 Iz - 11 :s: 2

b) Wh oIe complex plane without the disc with center -2 and radius r = 1, Figure 1.17.

Figure 1.17 Iz + 21> 1

c) The disc with center Zo, and radius r (without the circle Iz - zol = r), Figure

1.18.

Figure 1.18 Iz - zol < r

cl) The straight half-li ne y = x-I, y 20, Figure 1.19.

y /

/ ./"---

// /// //

/' /./ ./ /./

---------<-/'-/-O~-ij--/~--·--·-·-x -.

Figure 1.19 y = x-I, Y 2 0

e) The region between the straight half-lines Y = x-I, y = ffx -1 ancl Y 2 1 inclucling the last straight half-line, Figure 1.20.

r -I

Figure 1.20

f) For Zo = a + zb the points z :

-0 < arg (z - zo) < 0

are in the region between straight half-lines

y - b = (x - a) tan 0 and y - b = (x - a) tan( -0), x ~ a

and y ~ b (y ::; b for the second), 0 E [0, i-l, and x ~ a and y ~ b (y ::; b for the

second), Figure 1.21.

~Y ,

Figure 1.21

g) The points z are on the imaginary axis fOT Zo #- O. FOT Zo = 0 the points z are

arbitrary complex numbers (Examine the case Zo E C; Y = kx, k = ~~:O).

h) From I!I < r we obtain Izl > ;, the points z are outside of the cirele Izl = ~.

i) The points z are on the straight li ne y = -i. j) Using the Euler representation z = pe'tp we obtain from ~ = z that

e'tp . 2,tp -- = z, z.e., e = l. e-'tp

Therefore the points z = pe'1r/4 satisfy the condition

z -= = z, y = x. z

Example 1.20 Prave for u, v E C the relation

and give a geometrie interpretation.

Solution. Since u . u = lul 2 we can easily obtain the desired equality. Narnely,

(u + v)· (u + v) + (u - v)· (u - v) = 2u· u + 2v· v,

where we have used :u:E""V = u ± v. The geornetrical interpretation: the surn of squares of diagonals of a square is

equal to the surn of the squares of its sides.

Exercise 1.21 Let t be a real parameter ZI = (PI, 81 ) and Z2 = (P2' 82 ) fixed complex numbers. Which curves are given in the complex plain by the following relations?

a) Z = (2, t), 0::; t < 271";

b) Z = (t,~), t ~ 0;

c) Z = ZI + cost + zsint, 0::; t < 271";

d) Z = ZI + t ( cos T + z sin~), t ~ 0;

e) Z = ZI + Z2(COS t + zsin t), 0::; t < 271";

J) Z = ZI + t(Z2 - zt), 0 ::; t ::; 1;

g) Z = ZI (cos t + z sin t) + Z2( cos t - z sin t), 0 ::; t < 271";

h) Z = t + (cos t + i sin t), t > O.

Answers.

a) The circle with the center 0 and radius r = 2.

b) The half straight line y = V3x, y ~ o. c) The circle with the center Z and radius r = 1.

d) The straight half-line y = x, y ~ 0 translated for PIon the half straight line

y = tan ()I . X, Y ~ 0, Figure 1.22.

o Figure 1.22

e) The circle with the center ZI and radius r = P2'

f) The line segment [zt, Z2], Figure 1.23.

f"Y r:/Z' I / .' ,/ .

2 - 2 2 /" /" ..... 0 1 /"/" I

0" I Figure 1.23 [zt, Z2]

g) The circle with parametrie equations

x = PI COS(()I + t) + P2 COS(()2 - t)

Y = PI sin(()I + t) + P2 sin(()2 - t), 0::; t < 211",

where ZI = PI( COS ()I + z sin ()I), Z2 = P2( tos ()2 + z sin ()2)'

In particular, for ZI = Z2 we obtain

x

x = 2PI COS ()I COS t and y = 2PI . sin ()I cos t, 0::; t < 211".

21


Eliminating the parameter t we obtain apart of the straight line

y = tan 01 . x, - 2Pl cos 01 :::; X :::; 2Pl cos 01.

h) The curve is given by the following parametrie equations

x = t + cos t, Y = sin t, t > 0,

Figure 1.24.

y

Example 1.22 Find

a) 0;


Figure 1.24

b) 0=1; c) ~; d) N.

{ji = y!(cosi+zsini)

for k = 0, 1, 2.

b) We have

=

"Fr - + 2k7r "Fr + 2k7r cos 2 3 + Z sin 2" 3 '

5 ~( 77r . 77r) v2 cos 4 + ZS1ll 5

10~( 1f + 2k7r . 1f + 2k7r) v2 cos + zsm , 5 5

x


for k = 0,1,2,3,4.

c) We have

lJlO(cos arctan(-1/3) + zsinarctan(-1/3))

417fi10 ( arctan(-1/3)+2k1r . arctan(-1/3)+2k1r) v IU COS 2 + z sm 2 .

Second method:

23

Starting from ~ = x + zy, find x and y taking equal the real parts and then the imaginary parts.

d) We have

4/""7 ~ . (7r k7r) . (7r k7r) V -1 = cos 7r + z sm 7r = cos "4 + 2 + z sm "4 + 4 '

for k = 0,1,2,3.

Write down in all examples all cases in the form a + zb, a, b E ~.

Example 1.23 Solve the following equations in C :

a) x8 - 16 = 0; b) x 3 + 1 = 0; c) x6 + z + 1 = O.

Solution.

a) The zeroes of the equation x 8 - 16 = 0 are the values of m,

m = y'16( cos 0 + z sin 0) = h( cos k: + z sin k47r),

for k = 0,1,2, ... , 7. Hence

Zl = V2, Z2 = 1 + Z, Z3 = V2z, Z4 = -1 + z,

Zs = -V2, Z6 = -1 - Z, Z7 = -V2z, Z8 = 1 - z.

b) The zeroes of the equation x3 + 1 = 0 are the values of A,

r-I {j . 7r + 2k1r . 7r + 2k1r -1= Cos7r+zsm7r = cos 3 +zsm 3 ' fork=0,1,2.

Hence

Zl = 1, 1 v'3

Z3 = -+z-. 2 2


c) The zeroes of the equation x6 + z + 1 = 0 are values of ~ -1 - z ,

571" 571" +2h -+2h

~ (cos T 6 + z sin 4 6 ) ,

for k = 0,1, ... ,5.

Example 1.24 Using the equality cos t = sin t = y.} find:

a) cos TI;; b) sin f6-. Solution.

a) Starting from the equality

(cosi+ zsin i)2 =cosz+zsinz,

we obtain . 2 . z z d sm z = sm 2 cos 2 an

2 Z • 2 Z cosz = cos 2 - sm 2'

Since cos i + zsin i E {z Ilzl = I},

we have cos2 t + sin2 ~ = 1. Putting this in the second identity we obtain

cos ~ = J cos z + 1 . 2 2

Applying the last formula two times on cos ~ we obtain

b) In this case we start from the equality

( z . Z)3 . cos 3" + z sm 3" = cos z + z sm z.

Example 1.25 Prove for x =I- 0 :

n cos R±J_ x sin nx a) ~ cos kx = ~ 2"" . L.J • X ,

k=l sm '2" b)

n. sin~x. sin~ E smkx = . x . k=l sm '2"


Solution. a) and b): We consider the sum

n n n

S = E cos kx + tE sin kx = E(cos kx + tsinkx), k=l k=l k=l

Then n 1 mx

S '" ,kx 'x - e =L...Je =e , k=l 1 - e'X

where we have used the Euler formula cos kx + tsin kx = e,kx. Using the formula

e1X _ e- tX

Slnx = ----2t

we can transform S in the following way:

S

25

If we apply the Euler formula on e(n+1},x/2 and compare with the starting sum we obtain

n n n + 1 sin nx + 1 sin nx E cos kx + z E sin kx = cos --x . T + z sin _n __ x . T· k=l k=l 2 sm "2" 2 sm "2"

Putting equal the real parts and then the imaginary parts we obtain the desired equalities.

Remark. Find in an analogous way the more general equalities for the sums

n n

E cos(a + kx); E sin(a + kx). k=O k=O

Example 1.26 Find the position 0] the vertices 0] the triangle with equal sides i] the two vertices are -1 and 2 + t, Figure 1.25.


JfA.y I

/ -~

I 2 I

'" I 'C

Figure 1.25

x

Solution. First we shall show that for any triangle with equal sides we have

where Zl, Z2, Z3 are the vertices of the triangle with equal sides. We have (see Figure 1.25)

Z2 - Zl = e 1r·/3(z3 - zt} and Zl - Z3 = e 1r·/3(z2 - Z3).

Dividing these two equalities we obtain

Since Zl and Z2 are known we can find Z3 by the last equality (we obtain two solutions).

Exercise 1.27 Find the following sums:

n

a) L cos(2k - l)x; k=l

Answers.

a) sin 2nx -2-'-; smx

b)

b)

n n L sin(2k - l)x; c) L (_l)k-l sin kx.

k=l

sin2 nx smx

k=l

sin n..tl cos T c)

cos~

1.1. ALGEBRA1C PROPERT1ES 27

Exercise 1.28 Find the following sums for real eonstants m and n 1= 2k7r

T T

a) L cos(m + kn); k;O

b) L sin(m + kn). k;O

Answers.

sin~n rn a) . n cos(m+-2 );

sm2

sin!±!.n rn b) . 2n sin(m + -).

sm 2 2

Exercise 1.29 Solve the equation z = zn-l (n is a natural number).

Exercise 1.30 Let m and n be integers. Prove for z 1= 0 :

a) that (y'Z)m has ~ different values where (m, n) is the greatest eommon ~n,mJ

divisor of the numbers m and n.

b) That the sets of values of (y'Z)m and yrzm are equal, i.e.,

(y'Z)m = ;:;zm if and only if(n,m) = 1, i.e., n and m has no non-trivial eommon divisors.

Exercise 1.31 Prove the identity

11 - ZlZ212 - IZI - z212 = (1 - IZlI2) . (1 - IZ212).

Exercise 1.32 Prove the inequality

Exercise 1.33 Find the vertices of regular n- polygon if its center is at Z = 0 and one vertex is known.

Answer . The vertices are

Zk = Zl· e'2k1n, k = 0,1, ... ,n -1,

where Zl is the given vertex.

Example 1.34 Prove:

a) 1f Zl + Z2 + Z3 = 0 and IZII = IZ21 = IZ31 = 1 then the points are vertices of a triangle with equal sides whieh is in the unit eirele.

b) 1f Zl + Z2 + Z3 + Z4 = 0 and IZII = IZ21 = IZ31 = IZ41 = 1, then the points Zl, z2, Z3, Z4 are either vertiees of a triangle with equal sides or they are equal in pairs.


Solution. a) The length of si des of the triangle with vertices Zb Z2 and Z3 are

IZ3 - zll, IZ2 - zt\, IZ3 - z21·

By the given conditions: Zl + Z2 + Z3 = 0 and IZil = 1, i = 1,2,3, we have

IZ3 - zll2 = 12z1 + z21 2 = (2z1 + z2)(2Z1 + Z2) = 5 + 2(ZlZ2 + ZlZ2),

and analogously

IZ3 - z21 2 = 12z2 + zll2 = 5 + 2(ZlZ2 + ZlZ2)'

Therefore IZ3 - zll = IZ3 - z21·

We can prove in a quite analogous way that IZ2 - zll = IZ3 - z21. Therefore the tri angle with vertices Zb Z2, Z3 is with equal sides.

Example 1.35 Let the points Zb Z2, ... , Zn be on the same side with respect 0/ a straight line wh ich cross (0,0). Prove that the points

1 1 1 , , ... ,

Zl Z2 Zn

have the same property and that Zl + Z2 + ... + Zn # 0, and that

1 1 1 -+-+ ... +- #0. Zl Z2 Zn

Solution. We can suppose without weakening the generality that the straight line from this example is just the imaginary axis and that all points are right from it (in the opposite it is enough to multiply Zk by some convenient number eil"). Then it is obvious that Re Zk > 0 and Re lk > 0 for all k, which implies the desired properties.

Exercise 1.36 Solve the equation (1 + ~) 3 = z.

Example 1.37 Prove that the field 0/ complex numbers is the smallest field which contains the field 0/ real numbers and the solution 0/ the equation x2 + 1 = O.

Solution. We will prove the desired result by reductio ad absurdum. Suppose the contrary, that there exists a field T which contains IR, the solution of the equation x 2 + 1 = 0 and is smaller than the field C, C :J T and C # T. Then there exists Zo = a + zb, z E C, such that a, b E IR and Zo (j. T. On the other side, since T contains the solution of the equation x 2 + 1 = 0 we have z E T. Therefore by T :J IR and the fact that T is a field we have x + yz E T for every x, y E IR. Hence Zo ETaIso, which is a contradiction.


Example 1.38 Prove that there does not exist a total order in the field of complex numbers whieh is compatible with the opemtions in this field and which extend the usual order of reals.

Solution. Suppose that there exists a total order ~ in the field C. We shall compare z = z and z = O. Suppose that z ~ O. Then Z2 ~ 0, -1 ~ O. Contradiction. Suppose now that z ~ O. Then 0 = z - z ~ i. Multiplying both sides by -z (-z ~ 0), we obtain (-z)(-z) ~ 0, i.e., -1 ~ O. Since we have obtained contradiction in both cases, we condude that such a total order ~ can not exist.

Example 1.39 Let p be the eommutative ring of all polynomials with real eoeffieients endowed with the usual addition and multiplieation.

Let J be a ideal ofthe elements oftheform (1+x2 )Q(x), where Q is a polynomial, in the ring P. We define in P an equivalenee relation'" in the following way:

Prove that

a) the set of all polynomials of first order with respeet to + is isomorphie with the set PI J of all equivalenee classesj

b) PIJ is a fieldj

c) the field PI J is isomorphie with the field of alt eomplex numbers C.

Solution. a) Each polynomial P from P can be written in the polynomial form

P(x) = (x 2 + 1)Q(x) + ax + b (x E lR)

for some a, b E lR. Therefore an equivalence dass from PI J has the form J + ax + b for some a, b E lR. The addition + in PI J is defined by

(J + (ax + b)) + (J + (ex + d)) = J + (a + e)x + b + d. (1.3)

This implies the isomorphism between (PI J, +) and (PI, +), where pI is the set of all polynomials of the first order.

b) By (1.3) the operation + in PI J is an inner operation. The neutral element is J + 0 and the inverse element of J + ax + b is the element J - ax - b. The multiplication in PI J is given by

(J + (ax + b)) . (J + (ex + d)) = J + ae(x2 + 1) + x(be + ad) + bd + ae,

i.e., (J + (ax + b)) . (J + (ex + d)) = J + x(be + ad) + bd - ae. (1.4)


The unit element is J + 1. The inverse element of J + ax + b for a2 + b2 =I 0 is J + a' x + b', where a' and 1/ are the unique solutions of the system of the equations

b . a' + a . b' = 0

b . b' - a . a' = 1.

c) Comparing the usual operations in C :

(az + b) + (cz + d) = (ae)z + b + d

(az + b) . (cz + d) = (bc + ad)z + bd - ae

with (1.3) and (1.4), respectively, follows the isomorphism between (C, +,.) and (PfJ, +, .). .

Example 1.40 Prove the isomorphisms:

a) the group (C \ {O},·) with the group of matrices of the following form

[_~ !], a, b E IR and a2 + b2 =I 0

with the matrix multiplication;

b) The group of quaternions

3

K \ {O} = {wo + w1i + w2j + w3kl Wi E IR,Ew~ =I 0, i = 0,1,2,3}, ;=0

where

ij = -ji = k, jk = -kj = i, ki=-ik=j;

1· k = k·1 = k, 1 . i = i . 1 = i, 1· j = j . 1 = j, with multiplication and the group of real matrices

:~ -:: :: 1 W3 Wo -Wl

-W2 Wl Wo

with matrix multiplications.

c) The group of quaternions (K \ {O},·) with the group of matrices of the second order

(u,v E C \ {O})

with matrix multiplication.

Rint. For c). U se a) and b) in the deeomposition of the following matrix

where u = Wo + iWI and v = W2 + iW3.

z-c Example 1.41 Let Icl < 1 and w = --_-.

1 - cz

What are the following sets in the z-complex plane:

Al = {zllwl < I}, A2 = {zllwl = I}, and A3 = {zllwl > I}?

31

Solution. For set AI, the eondition Iwl < 1 is equivalent with the inequality

Iz - cl 2 < 11 - czl\ i.e., (1 -lcI 2)(lzI 2 - 1) < O.

Henee Al = {zllzl < I}. In a quite analogous way we obtain

A 2 = {zllzl = I} and A 3 = {zllzl > I}.

Exercise 1.42 Prove for arbitrary Zl, Z2, Z3 E C that

1 Zl Zl

1 Z2 Z2 1 Z3 Z3

is areal number.

Exercise 1.43 Prove the inequality

Iz -11 ~ Ilzl-11 + Izllargzl (z E C).

Rint. Use the geometrie interpretation of the complex number.

Exercise 1.44 Which curves are given by the following sets

a) {zllz - 111z + 11 = const},

b) {ziliz - 111 = const}? z+l

Exercise 1.45 Prove the equality

(n - 2) E lakl 2 + I E akr E Elak +ail 2

l::;k< i::;n

1.2 The Topology of the Complex Plane

1.2.1 Preliminaries

We denote by D(zo, r) the open disc centered at Zo with radius r,

D(zo,r) = {z Ilz - zol < r}.

We call D(zo, r) also a neighborhood of zoo We denote by C(zo, r) the circle centered at Zo with radius r,

C(zo,r) = {z Ilz - zol = r}.

The function d : IC -+ 1m. given by

is ametrie on IC, i.e., d(zt, Z2) 2:: 0; d(ZI' Z2) = 0 if and only if ZI = Z2; d(Zt,Z2) = d(Z2,ZI) d(ZI,Z2):::; d(Zt,Z3)+d(Z3,Z2)' A set 0 C IC is open if for every z E 0 there exists c: > 0 such that D(z, c:) C O. A set Ais closed if its (set) complement AC is open. A point w is an accumulation point for a set A C IC if in every disc D (w, r) there are infinite elements from the set A. The boundary of a set AC IC, denoted by BA, is the set of complex numbers whose every neighborhoods have a nonempty intersection with both A and AC. A set A is bounded if A C D(O, M) for some M > O. A set A is compact if it is closed and bounded. A set A is connected if there do not exist two disjoint open sets 0 1 and O2 whose union contains A while neither 0 1 nor O2 alone contains A. We denote by [ZIZ2] the line segment with endpoints ZI and Z2' A set Ais polygonally connected if any two points Zo and Zn of A can be connected by a polygonal li ne [zn, ZI] U ... U [Zn-I, Zn] contained in A. A polygonally connected set is connected. A set A is a region if it is open and connected. Let

S = {(a, b, e) I a2 + b2 + (e -1/2? = 1/4}. (1.5)

The stereographie projection of S \ {(O, 0,1)} on IC ( of S on IC U {oo}) is a 1-1 correspondence obtained taking that the plain e = 0 coincides with the complex plane IC, and that a and b axes are the x and y axes, respectively, and we associate to (a, b, e) E S the eomplex number where the ray from (0,0,1) intersects c. We have

x a= .

x2 + y2 + l' b= Y ; x 2 + y2 + 1

1.2. THE TOPOLOGY OF THE COMPLEX PLANE 33


Exercise 1.46 Prove that the function d : !C -+ Im. given by

is ametrie on !C.

Exercise 1.47 Prove that the lunction d: !C -+ Im. given by

is ametrie on !C, where Zi = Xi + ZYi, i = 1,2.

Exercise 1.48 Which 01 the lollowing sets is open

a) {zllzl<3};

b) {zl1 < Rez < 3};

c) {zIRez<3}U{3}?

Answers. a) Yes. h) Yes. c) Not.

Example 1.49 Give an example lor a connected set which is not polygonally connected.

Solution. The set of points

is connected hut not polygonally connected, since the set A contains no straight line segments.

Exercise 1.50 Prove that a connected subset 01 Im. is an interval.

Hint. Prove that a suhset A of Im. is an interval if and only if for every two points rl and r2 in Im. with rl < r2 we have trI, r2] C A.

Exercise 1.51 Which 01 the lollowing subsets 01!C are connected:

a) {zllz-11<1}U{zllz-31<1};


b) {-~lnEN}U(0,11?

Example 1.52 Prove that a region is polygonally connected.

Solution. Let A be a region in IC and Zo E A. We denote by 0 1 the set of points of A which are polygonally connected by Zo in A and by O2 the set of points in A which are not. Both sets 0 1 and O2 are open, e.g., 0 1 is open since every point z can be connected by every point in D (z, c:). Since A = 0 1 U O2 and A is connected it follows that O2 is empty (we have Zo E Od. Now we have that every point in A can be polygonally connected by zoo Therefore every pair of points in A can be polygonally connected through the point zoo

Exercise 1.53 Prove that a polygonally connected subset of IC is connected.

Exercise 1.54 Prove that for stereographie projection the following formulas hold

a x=--;

1-e b

y=--. 1-e

Hint. Use that the points (O,O,l),(a,b,e) and (x,y,O) are collinear.

Exercise 1.55 For the points 0 and 1 + z in IC give the corresponding points of S by stereographie projeetions.

Exercise 1.56 Let S be given by (1.5) and C be a circle on S, i.e., the intersection of S with a plane of the form Aa+ Bb+ Ce = D. Prove that ifC. is the stereographie projection of C on IC then

a) for (0,0, 1) E C the projection C. is a line;

b) for (0, 0, 1) rf. C the projection C. is a circle.

Exercise 1.57 Prove that the function d : IC U {oo} -t ~ given for ZI, Z2 E IC by

and by 2

d(z, 00) = (1 + IzI2)1/2

for z E IC is ametrie on IC U {oo }.

1.2. THE TOPOLOGY OF THE COMPLEX PLANE 35

Exercise 1.58 Prove that the function d : C -t lR given by

is ametrie on C, where Zi = Xi + zy;, i = 1,2.

lu-vi Example 1.59 Prove that by d( u, v) = u, v E C is given a -/1 + lu12 -/1 + Iv1 2 '

metric on C.

Solution. We shall prove only the triangle inequality d( u, v) :s d( u, w) + d( w, v), since other properties can be easily verified. We start with the identity

(u - v)(l + ww) = (u - w)(l + vw) + (w - v)(l + uw)

for u,v,w E C.

This implies

lu - vl(l + Iw1 2 ) :s lu - will + vwl + Iw - vIII + uwl. (1.6)

For 11 + vwl we have the following equality

which is equivalent with the inequality Iv - wl 2 ~ O. Hence

Analogously we obtain

(1 + vw)(l + vw) :s (1 + IvI 2 )(1 + Iw1 2 )

(1 + IvI2)(1 + IwI 2 ).

(1.7)

Using the previous two inequalities in (1.6) we obtain the desired inequality

lu - vi lu - wl Iw - vi -/(1 + luI2)(1 + Iv12 ) :s -/(1 + luI2)(1 + Iw1 2 ) + J(1 + IwI2 )(1 + Iv1 2) •

It is easy to prove that d( u, v) E [0, 1] for every u, v E C.

Chapter 2

Sequences and se ries

2.1 Sequences

2.1.1 Preliminaries

Let {Zn} be a sequence of complex numbers. A sequence {zn} is bounded if there exists M > 0 such that IZnl < M for all n E N. A complex number V is an accumulation point of {zn} if for every c > 0 there exist infinitely many integers n such that

Definition 2.1 We say that {zn} converges to w E C, in the notation

lim Zn = W n-+oo

if for every c > 0 there exists an integer no such that for all n ?: no

We say that {Zn} is a Cauchy sequence if for every c > 0 there exists an integer no such that for all n, m ~ no

Theorem 2.2 A sequence {zn} of complex numbers is convergent if and only if it is a Cauchy sequence.

37

38 CHAPTER 2. SEQUENCES AND SER1ES


Example 2.1 Find which of the following sequences are bounded:

b) {C ~zf};

c) {G::f}; d) {2n:1 +zn:1};

Solution. a) The sequence is bounded since Iznl = Izln = l.

b) The sequence is bounded, since

I( 1 )nl_ ( 1 )n _ 1 < 1 r -- --- --- -- wr every n E N. 1 + z - 11 + zl - (v'2)n - ..;2'

c) The sequence is bounded, since I (i :t !fl = 1, for every n E N.

d) The sequence is bounded since

I n n- 1 1 V n n-l 5 --+ z-- = (--)2 + ( __ )2 < - for every n E N. 2n + 1 n 2n + 1 n - 4

e) The sequence is unbounded since n i= 4k, k = 1,2, ... ,

Example 2.2 1f the sequence {an} is bounded, prove that the following sequences are also bounded

{

n } LPiai

b) i=~ , Pi > 0;

LPi i=1

Solution. We shall use that there exists M > 0 such that lail :s M for every i E N. a) We have

11 nil n 1 - Lai :s - L lail:S - nM = M. n ;=1 n ;=1 n

2.1. SEQUENCES

b) We have

c) We have

Example 2.3 Find the accumulation points of sequences fram Example 2.1.

Solution. The accumulation points are

a) l,-l,z,-lj b) Oj c) 1,-1,z,-zj d) ~+Zj e) O.

Example 2.4 Which sequences fram Example 2.1 converge?

Solution. a) Does not converge.

39

h) The sequence is hounded and it has one accumulation point O. Therefore it converges to O.

c) Does not converge. It is hounded hut with few accumulation points.

d) The sequence is bounded and it has one accumulation point ~ + z. Therefore it converges to ~ + z.

e) Does not converge. It has one accumulation point hut it is unhounded ..

Example 2.5 Prave that ifthe sequence {wn } converges to w, then {Iwnl} converges Iwl. The opposite is not true.

Solution. First we shall prove that the inequality

lu - vi ~ Ilul-lvll for u,v E C. . (2.1)

Since u = (u - v) + v we have

lul ~ lu - vi + lvi·

Hence lul-Ivl ~ lu - vi.

40 CHAPTER 2. SEQUENCES AND SERIES

Starting from v = (v - u) + u, analogously we obtain

Ivl-Iul ~ lu - vi,

-lu - vi ~ lul-Ivl ~ lu - vi·

Since {wn } converges to w, we have that for every e > 0 there exists no E N such that IWn - wl < e for every n ~ no. Putting v = Wn in (2.1) we obtain

for every n ~ no, i.e., Iwnl -7 Iwi as n -7 00. The following example shows that the opposite is not true.

Exercise 2.6 Prove that a sequence {wn} converges to w, if and only ifthe sequence {Re wn} converges Re wand the sequence {Im wn) converges to Im w.

Exercise 2.7 Find a sequence with one accumulation point which does not converge.

Answer. {(( _l)n + l)n}.


imply

lim Iwnl = r (r > 0) and lim argwn = cp n-+oo n-+oo

lim W n = re"", n-+oo

where arg W is the main value.

Solution. For W n = U n + ZVn , we have

Therefore

Hence

un = Re W n = Iwnl cos{arg wn)

Vn = Im W n = Iwnl sin(arg wn).

lim Un = lim Iwnllim cos(arg wn) = rcoscp, n ....... oo n-+oo n-+oo

lim Vn = lim IWn I lim sin{ arg wn ) = r sin cp. n--+oo n--+oo n-+oo

lim W n = r(coscp + zsincp). n-+oo

We have used that {wn } converges, if the sequences {un } and {vn } converge.

2.1. SEQUENCES 41

Example 2.9 Let lim W n = wo. Is it true then that also n .... oo

lim arg W n = arg wo? n .... oo

Solution. It is not true ! Counter-example: Wo = 0, Wn = -1 + (_l)n;. We have

1 1 Ji..~ Wn = -1, and arg W2k = 7r - arctan 2k' arg W2k+l = 7r + arctan 2k + 1 .

Hence {arg wn } diverges.

Remark. If {wn } converges to Wo =I 0, then for every value 'P = IArgwo there exists a sequence 'Pn = Arg Wn which converges to Arg wo. If Wo =I 0 is not a negative number, then we have also limn .... oo arg W n = argwo.

Example 2.10 Prove that if lim W n = wand lim w~ = w' then for an arbitrary n-+oo n-oo

but fixed k E N :

a) lim (wn)k = w\ n-+oo

k k

c) lim "w1' = "w1' • n-+oo L...J n L...J 1'=0 1'=0

Solution. a) and b) The proof goes by induction. We known that for k = 2

lim (wn )2 = w2 and lim (Wn ) 2 _ (:::.-) 2 n-+oo n-+oo W~ - W' .

Suppose that the desired equalities hold for k - 1,

k 1 k 1 (Wn)k-l (W )k-l lim (Wn ) - = W - , lim - = - . n-+oo n-+oo w~ w'

Then we have for k

( Analogousl y also for b) ).

c) By induction we can prove that the surn of finite number of convergent sequences converges to the sum of their limits. Therefore by a) we have

k k k

lim "w1' = " lim w1' = "w1' • n-+oo L..J n L...J n-+oo n L.J p=o 1'=0 1'=0


Example 2.11 If lim Wn = w, and {U n } does not converge, can the following n--+oo

sequences can be convergent?

) Wn C -.

Un

Solution. a) Never. If {un } is unbounded then the inequality

implies that IWn + U n I is also unbounded, and the sequence {wn + un } is not convergent. Examine the case when {un } is bounded but it has few accumulation points.

b) Take U n = n and W n = A. Then {wn un } converges to O. n

c) Take U n = n and {wn } arbitrary convergent sequence. Then {wn un } converges to O.

Example 2.12 Find

a) one value of

J -1 + (1- z)J-1 + (1 - 2z)V-1 + (1 - 3z)vf-1 + ... ;

b) one real value of

zJ -1 + 2zJ -1 + 3zV-1 + 4zvf-1 + ....

Solution.

a) and b) are complex generalizations of the equality

a) For n E N we have

n(n - 2z) nV-1 + (n - z)(n - 3z)

nJ -1 + (n - z)J-1 + (n - 2z)(n - 4z)

nJ-1 + (n - z)J-1 + (n - 2z)J-1 + (n - 3z)(n - 5z) .

2.1. SEQUENCES 43

Letting this proceded to infinity we obtain

n(n - 2z) = nJ-l + (n - z)J-l + (n - 2z)V-l + (n - 3z)V-1 + ....

Putting n = 1, we obtain

J -1 + (1 - z)V -1 + (1 - 2z)V-l + ... = 1 - 2z.

b) For z E C we have

z(z - 2z) = zV-l + (z - z)(z - 3z)

= zJ -1 + (z - z)V-l + (z - 2z)(z - 4z)

= zV -1 + (z - z) J -1 + (z - 2z )V -1 + (z - 3z)( z - 5z) .

Letting this proceded to infinity we obtain

z(z - 2z) = ZV-l + (z - z)J-l + (z - 2z)V-l + (z - 3z)V-l + ....

Putting z = z we obtain

zV -1 + 2zJ -1 + 3zV -1 + 4zV-I + ... = -3.

Remark. For real case we have that

Jl + xVI + (x + 1)~ = f(x) (x E IR)

satisfies the functional equation

1 + x f(x + 1) = f2(x),

whose solution is f(x) = x + 1.

Example 2.13 Let the members of the sequence ao, al, ... , an, ... can take only the values -1,0,1. Prove that


Solution. The form

is meaningful, since

By induction we prove the equality

• ( 11" n aOal··· ak ) Zn = 2 sm "4 E 2k •

k=O

We have

( (11" n aOal··· ak ))

sign Zn = sign 2 sin 4" E 2k = ao·

For ao ":f 0 we have

Z~ - 2 = al J 2 + a2,/2 + ... + an V2 , and on other side by (2.2)

Z~ - 2

Exercise 2.14 Prove that for x, -2 :$ x :$ 2 we have

x = ao ,/1 + an/2+~ , where ao, ab ... ,an , ••• take only one of the value either -1 or 1.

2.2 Series

2.2.1 Preliminaries

(2.2)

An infinite series L:~=l Zn is called (ordinary) convergent if the sequence {Sn} of its partial sums given by

2.2. SERIES 45

converges. If E~=l IZn I converges we say that E~=l Zn absolutely converges.

A series E~=l Zn is convergent if and only if for every c > 0 there exists no E N such that for all n, m ~ no

Theorem 2.3 (Cauchy criterion) IflimsuPn--+oo yllznl < 1, then the series

E~=l Zn converges absolutely. If limsuPn--+oo y'lznl > 1, then the series E~=l Zn

diverges absolutely.

Theorem 2.4 (D'Alembert criterion) IflimsuPn--+oo IZ:!11 < 1, then the series

L~=lZn converges absolutely. Ifliminfn--+oolz:!ll > 1, then the series E~=lZn diverges absolutely.

We have for any sequence {zn}

liminf IZn+ll :$ liminf y'lznl :$limsup yllznl :$limsup I. Zn+ll· n-+oo Zn n-+oo n-...oo n--+oo Zn

Theorem 2.5 (Raabe criterion) IflimsuPn--+oo n (IZ~n-ll-1) < -1, then the series E::1 Zn converges absolutely.

Für two series E~=o U n and E~=o Vn we define their multiplication as aseries


Example 2.15 Examine the absolute and ordinary convergence of the following series with respect to the real parameter a.

00 1 + zn a) E~();

n=2 n n

00 an + zn c) E-,-;

n=O n. 00 ,

d) E n. n=O (n + z)n'

00 eIn

g) E-; n=l n

00 zn h) E n·

n=lln nTI


Solution. a) an b). Diverges absolutely and ordinary for all values of a. c) and d). Converges absolutely for aB a. e) Diverges absolutely, but converges ordinary, since

as k ----+ 00.

f) and g). Diverges absolutely, but converges ordinary. h) Diverges.

Example 2.16 Let

( w) = w( w - 1) ... (w - n + 1) n n!

for w E !C, n E N. Find for which complex w the series

converges.

Solution. Let w = x + zy. We obtain by D' Alembert criterion

lim n--+oo

w(w - 1) .. · (w - n - l)(w - n) (n + I)!

w(w-1)· .. (w-n-l) n!

I. IW - nl 1m --n--+oo n + 1

. Jx2 - 2xn + n2 + y2 = IIm --------

n--+oo n + 1 1.

Therefore we can not decide by D'Alembert criterion about the convergence of the given series except we aBow that w depends of n. Namely, taking w(n) = n + zy for an arbitrary but fixed real y the series

converges absolutely. Taking the Raabe criterion we obtain

n(I:~~I-l) _n _ (J'-x-2 ---2-n-x-+-n-2 -+-y-2 - (n + 1)) n+l

2.2. SERIES

n = n+l

1

Iwl 2 + n2 - 2nx - (n2 + 2n + 1)

Vlwl2 + n 2 - 2nx + n + 1

Iwl 2 1 n - 2(x + 1) - n

47

as n -t 00. By Raabe criterion the series absolutely converges for -(x + 1) < -1, i.e., Rew > O. The series converges ordinary for -1 < Rew ~ O.

Remark. Analogously the hypergeometrie series

absolutely converges for Re (w + U - v) < O.

Example 2.17 Prove that the series

converges if the se ries Ek:OUk absolutely converges and the sequence {ak} monotonically decreasingly converges.

Solution. We have

= I 'f uk(ak - ak+l + ak+1 )1 k=n+l

 0 there exists no E N such that E~!~+1lukl < e for all n ~ no(E) and pE N. Therefore

(2.3)

since lan+11 ~ lan I for k = n + 1, ... , n + p. The series Ek~O( ak - ak+d absolutely converges by

n+p n+p L lak - ak+ll = L (ak - ak+d = an+l - an+p+1 -t 0 as n -t 00,

k=n+l k=n+l

since the sequence {an} converges and therefore it is a Cauchy sequence. Hence by (2.3) the series

is convergent.

Exercise 2.18 Prove that if the series

converges and I arg ukl :::; <P < ~, then this series absolutely converges.

Exercise 2.19 Let 00 00

LUn and LU! n=l n=l

be convergent series. Prove the for Re U n ;::: 0 the series

00

converges.

Exercise 2.20 Examine the convergence of the following series

a) 00 III

L~j n=l n

b) c)

Answers. a) Diverges. b) Diverges. c) Converges absolutely.

Example 2.21 Prove that the set of all complex series S with respect to addition of series, i.e.,

00 00 00

LUk+ LVk = L(Uk+Vk), k=O k=O k=O

is an abelian group.

00 00

Solution. By the definition the sum of series E Uk and E Vk is again a complex 00

series E (Uk + Vk). k=O

k=O k=O

2.2. SERIES 49

Associativity:

E Uk + (E Vk + E Wk) 00 00

L Uk + :~:JVk + Wk) k=O k=O 00

k=O 00

k=O 00 00

L(Uk+Vk)+ LWk k=O k=O

(E Uk + E Vk) + E Wb

where we have used the associativity of the addition of complex numbers. The neutral element is the zero series, L:~=o 0 = o. The inverse element of the series L:k=O Uk is the series L:~o( -Uk).

Example 2.22 11 one series converges and the other diverges can their sum converges?

Solution. No. Apply Example 2.11.

Example 2.23 11 both series diverges can their sum converges?

Solution. Yes, e.g., L:k=o t and L:~o( -t)· Example 2.24 Find two convergent (divergent) series whose sum is the series

Solution. We have

00 (l-z/ 00 cos 71rk 00 sin 71rk

L k =L~+zL~' k=O 2 k=O 2 2 k=O 22

where on the right side are two convergent series.

We have

where on the right side are two divergent series.

Exercise 2.25 Prove D'Alembert and Cauchy criterions for the absolute convergence of aseries.

Example 2.26 Prove that the condition limsuPn->oo IW~±l1 < 1 in D'Alembert cri-n

terion of the convergence of series is not necessary.

Solution. Take the real series

a + b2 + a3 + b4 + ... for 0 < a < b< 1.

Then lim SUPn->oo I W~;;l I > 1 hut the series converges, since

00 00

a + b2 + a3 + b4 + ... = E a2k+1 + E b2k ,

k=O k=l

and hoth series on the right side converge.

Exercise 2.27 Apply Cauchy and D'Alembert criterions on the series

Z (Z)2 (Z)3 (Z)4 2+ 3" + 2 + 3" + ....


a) The series E ~ not converges absolutely, but it converges ordinary. k=O k + 1

b) Using

for p ~ k, prove that k 1 E >1.

p=oV(k-p+l)(p+l) -

c) Multiplying the series E ~ by itself we obtaining a divergent series. k=O k+l

Solution. a) Series Lk>O k not converges ahsolutely, since I kl - k+l k+l

and the series Lk::l t diverges.

The ordinary convergence follows by I Lk=O zk I ~ .;2, and k '\. o.

1 > 7C'

2.2. SERIES 51

b) We obtain

t 1 ~ t ~ = 2( k + 1) ~ 1 for k > 2. p=oy'(k-p+1)(p+1) p=02"-1 k-2

c) Since

we obtain by b) for the member of series

Hence it does not converge to zero what is a necessary condition for the convergence of aseries.

Example 2.29 Find the sums 0/ the /ollowing series:

Solution. a) Since

~ (~+ l)k = ~ (~) cl I~I ~ 2k ~ 2 ,an 2 < 1, k=O k=O

the series converges and has the sum 1 ~ ~ = 1 + ~. b) Multiplying the series

with itself we obtain


Therefore by a)

c) Using a) and b) we obtain

00 ( + 1)" 00 (. + 1)"-1 00 k-l (. + 1)"-1 1 00 ( + 1)"-1 E l k • E k· • = E E p. • = - E k(k - 1 )-,--l----o--''----"=1 2 k=1 2"-1 k=2 p=1 2k - 1 2 "=2 2k - 1 '

where we have used the equality

n(n + 1) 1+2+···+n= .

2

Hence 00 (l + 1)"-2 E k( k - 1) k-2 = -4 + 4l.

k=2 2

d) Since

z + 1 (f k(k + 1) (z + 1)"-1 _ f k (z +"~lk-1) = f k2 • (z +,,1)k, 2 k=1 k - 1 k=1 2 k=1 2

we obtain by b) and c)

fk 2 .(z+1)" =-3-z. "=1 2k

Remark. The preceding sums can be obtained also using the power series

see Chapter 6.

Exercise 2.30 For a real find:

zk E k for z = 1 + z, k=O 2

Ek2 a 00 ( )k k=1 a + z

Answer. a(a + z)( -2a2 - 2az + 1).

Exercise 2.31 Prove that there exists a sequence {zn} of complex numbers such that the senes

n=O

converge and absolutely diverge for all k.

Answer . Zn = e27r1a / ln( n + 1) for airrational.

Chapter 3

Complex functions

3.1 General Properties

3.1.1 Preliminaries

Let z(t) = x(t) + zy(t), a ~ t ~ b. The curve z(t) is a path if z'(t) = x'(t) + zy'(t) exists and it is continuous on each subinterval of a finite partition of [a, b], and z'(t) "1= 0 except at a finite number of points.

Let J : A ...... C and Zo is an accumulation point of A. We say that w is a limit of J at Zo through A, in the notation

lim J(z) = w, z.....,.zo,zEA

if for every e > 0 there exists 6 > 0 such that if z E A and Iz - zol < 6, then IJ(z) - wl < e. We usually omit the symbol z E A under the limit sign. Let Zo E A. We say that J is continuous at Zo if lim J(z) = J(zo).

z-+zo

Definition 3.1 A complex Junction J defined in a neighborhood oJ a point z E C is differentiable at z iJ

r J(z+h)-J(z) h~ h

exists and then is denoted by f'(z).

If J and gare both differentiable at z then also the functions J + g, Jg, ~ (for g(z) "1= 0) are differentiable at z and

(J + g)'(z) = j'(z) + g'(z),

53

54 CHAPTER 3. COMPLEX FUNCTIONS

(fg)'(z) = J'(z)g(z) + J(z)g'(z),

(L)' (z) = J'(z)g(z) - J(z)g'(z). 9 g2(z)

Theorem 3.2 IJ J = u+w is differentiable at z = (x, y), then there exist the partial derivatives

äu äu äv äv äx ' äy , äx ' äy

and they satisJy the Cauchy-Riemann equations at z

or, equivalently

If the partial derivatives

äu äv äx äy'

äu äv äy äx

,

äu äu äv äv äx ' äy , äx ' äy

are continuous at z and satisfy the Cauchy-Riemann equations at z, then J is differentiable at z.

Definition 3.3 A Junction is analytic at z iJ J is differentiable in a neighborhood oJ z. J is analytic on a set Ace iJ J is differentiable at all points oJ an open set containing A. An everywhere differentiable Junction is called an entire Junction.


Example 3.1 Find the real and imaginary parts oJ the Jollowing Junctions: a) Z2 + Zj b) v'Z+Tj c) zn j d) yZj e) 2z2 - 3z + 5j J) (z - l)/(z + 1).

Solution. a) We write the function J(z) = Z2 + z in the form

J(z) = u(x, y) + w(x, y),

where u(x, y) = x 2 - y2 and v(x, y) = 1 + 2xy for z = x + zy.

b) We write the function J(z) = v'Z+T in the form J = u + w, where u and v are given by the system of equations

u2 - v2 = X + 1 and 2uv = y

3.1. GENERAL PROPERTIES

for z = x + zy.

c) We have f( z) = zn = u(p, tJ) + zv(p, tJ) for z = p( cos tJ + z sin tJ). We have

u(p, tJ) = pR cos ntJ and v(p, tJ) = pR sin ntJ.

d) Taking f(z) = vrz = u(p, tJ) + zv(p, tJ),

for z = p( cos tJ + z sin tJ), we obtain

tJ + 2k1r . tJ + 2k1r u(p, tJ) = ~ cos and v(p, tJ) = ~ sm ---

n n

for k = 0,1,2, ... , n - 1.

e) Taking f(z) = 2z2 - 3z + 5 = u(x,y) + w(x,y)

we obtain u(x,y) = 2x2 - 2y2 - 3x + 5 and v(x,y) = 4xy - 3y,

where z = x + zy.

f) Taking

we obtain

for z = x + zy.

z-l f(z) = z+ 1 = u(x,y) + zv(x,y),

z-l z+l =

(z-l){z+l) Iz + 11 2

Example 3.2 Find the inverse mappings for the following functions a) l/z, z 1= 0; b) (z -l)/(z + 1), z 1= -1; c) 2z2 + zz - z + 1; d) Z7 + 1 + z.

55

Solution. a) The inverse function for the function 1/ z, z 1= 0, is the same function.

b) The inverse function for the function (z - l)/(z + 1), z 1= -1, is the function (-z -l)/(z - 1), z ~ 1.

c) The inverse mapping for the function 2z2 + zz - z + 1 is the multi-valued function


where the square root is taken in the complex plane.

d) The inverse mapping for the function Z7 + 1 + z is the multi-valued function

w=V'-1-z+z

for k = 0,1, ... ,6.

Exercise 3.3 Let I(z)~. a) Find the images by the lunction 1 01 the lollowing lines: 1. x = C (constant), 2. y = C (constant), 3. Izl = R, 4. argz = a, 5. Iz -11 = 1. b) Find the lines which are mapped by Ion: 1. u = C (constant), 2. v = C (constant).

Hint. Find the functions u and v in the representation 1 = u + zu. Answers. a)

1. Circle u2 + v 2 - V = 0 for C # 0, and for C = 0 the axis u = o.

2. Circle u2 + v 2f; = 0 for C # 0, and for C = 0 the axis v = o.

3. Circle Iwl = l 4. argw = a.

5. The straight line u = ~.

b)

1. x2 + y2 - V = 0 for C # 0, and x = 0 for C = o.

2. x2 + y2 + -t = 0 for C # 0, and y = 0 for C = o.

Exercise 3.4 Explain geometrically the lollowing curves:

a) z = ut + vt\ 0 ~ t ~ 1;

b z= l+zt ) tEIl~:,

1 + t + t 2 '

{ V;t+1+,V;, 0~t~1

c) z(t) =

v'2 v'2 T t + 1 - 'T' -1 ~ t ~ O.

3.1. GENERAL PROPERTIES

Answers.

a) The straight line segment [0, u + v].

b) A closed curve.

c)

57

Example 3.5 Find the images of the circle Izl = R by the functions fl(z) = z + } and fl(z) = z - }.

Solution. The function !t maps the circle Izl = R :f:. 1 on an ellipse

u 2 v 2

(R+ *)2 + (R- *)2 = 1,

and the circle Izl = 1 on the straight line segment v = 0, -2 :::; u :::; 2. The function f2 maps the circle Izi = R :f:. 1 on an ellipse

u2 v 2

( 1)2 + ( 1)2 = 1, R-R R+R

and the circle Izl = 1 on the straight li ne segment u = 0, -2 :::; v :::; 2.

Example 3.6 Suppose that there exists limz ..... zo fez) = w, then

a) lim fez) = Wj b) lim Ref(z) = Rew and lim Imf(z) = Imwj %-+%0 %-+%0 %-+%0

c) limz ..... zo If(z)1 = Iwl·

Solution. The condition limz ..... zo fez) = w implies that for every c: > 0 there exists S> 0 such that Iz - zol < Simplies If(z) - wl < c:.

a) We have

c: > If(z) - wl = If(z) - wl = If(z) - wl for Iz - zol < S(c:).

b) We have

c: > If(z) - wl ~ IRe (f(z) - w)1 = IRef(z) - Rewl,

and

c: > If(z) - wl ~ Im (f(z) - w)1 = IImJ(z) - Im wl, for Iz - zol < d(c:).

c) We have

c: > If(z) - wl ~ IJ(z)I-lwll for Iz - zol < S(c:).

Example 3.7 Suppose that there exists limz ..... zo Ij(z)1 = Iwl. For which values there exists also limz ..... zo j(z)?

Solution. We have that limz ..... zo j(z) exists for w = 0, since e > Ilf(z)11 = If(z)1 for Iz - zol < b(e).

We shall show that this is the only case in general. Suppose that there exists limz ..... zo j(z) = A for w ~ O. Then by Example 3.6 we have lAI = w. We shall examine when we have equality in the following inequality

Ilf(z)l- All::; If(z) - AI

(since then t: > Ilj(z) -lAll = If(z) - AI for Iz - zol < b(t:)). The equality would imply

(lf(z)l- IAI)2 = (f(z) - A)(f(z) - A) = If(zW + IAI2 - 2Re (f(z) . A),

Re (f(z)A) = Ij(z)AI = Ij(z)AI,

and therefore Re (f(z)A) ~ 0 and Im (f(z)A) = O.

Then f(z)A have to be real and non-negative, which is impossible for a general complex function f. This is possible specially when j(z) = A or f real function with a constant sign.

Example 3.8 Find m and n such that there exists

Solution. Taking z = l/w we obtain

For the limit to exist we have to have p - m ::; p - n, m ~ n.

Example 3.9 Find which of the following limits exist .

l-z a) lim-l -; z ..... 1 - z

b) lim-l z _; z-+1 + z

Z2 _ Z2 c) lim--.

z-+o z

3.1. GENERAL PROPERTIES 59

Solution. a) The limit does not exist, since for z = 1 - x we have

1· 1 - z 1· 1 - 1 + x 1 Im--= Im = , z--+ 1 1 - z z--+o 1 - I + x

and for z = 1 + zy we have

lim 1 - z = lim I - I + zy = -1. z--+ 1 1 - z y--+O 1 - I - z y

b) The limit exists and it is 1 (prove that using the polar form of the complex number).

c) The limit exists and it is 0 (prove that).

Exercise 3.10 Can we modify the following functions at z = 0 such that the new functions will be continuous at z = 0 ?

a) Rez; z

Answers. a) No. b) No. c) Yes, f(O) = O.

) zRez c -I-zl-.

Exercise 3.11 Examine the continuity of the following functions in the unitdisc

a) _1_. b) _l_ 1- z' 1 + Z2·

Answers. a) and b). They are continuous.

Example 3.12 Prove that x 2 - y2 + zxy is a polynomial in z = x + zy (analytic polynomial) and x 2 + y2 - 2zxy is not.

Solution. We can write

x 2 _ y2 + 2zxy = (x + zy)2.

On the other side, if we suppose that

x 2 + y2 - 2zxy = an(x + zyt + ... + al(x + zy) + aa,

then for y = 0 we would obtain

(3.1 )

Hence aa = 0, al = 0, a3 = 0, ... , an = 0 and a2 = 1. Therefore we would have by (3.1)

wh at is a contradiction. Second method. Apply the Cauchy-Riemann equations on the previous polynomials.


Example 3.13 Let f : Q -4 C be an analytic function, where Q is a region in C. Prove that the functions u(x,y) = Ref(x + zy) and v(x,y) = Imf(x + zy) for x + zy E Q satisfy

a) the Cauchy-Riemann equations

8u 8v 8u 8v - and - . 8x - 8y 8y - - 8x'

b) the Laplace equation 82u 82u 8x2 + 8y2 = 0,

ij, additionally, u and v are from the class C2(Q).

Solution. a) Since the function fis analytic, there exists

J'(z) = lim fez + h) - fez) h-O h

for z = x + zy E Q. We evaluate the preceding limit in two ways. First let h -+ 0 for real h. We have for h 1- 0

fez + h) - fez) h

Taking h -+ 0 we obtain

= fex + h + zy) - fex + zy)

h u(x + h,y) - u(x,y) v(x + h,y) - v(x,y)

= h +l h .

f '( ) _ 8u(x,y) 8v(x,y) z - 8x + z 8x .

Now let zh -+ 0 for real h. We have for h 1- 0

fez + zh) - fez) fex + zeh + y)) - fex + zy) = zh zh

u(x,y+h)-u(x,y) v(x,y+h)-v(x,y) = -t h + h .

Letting h -+ 0 we obtain

f '( ) 8u(x,y) 8v(x,y) z = -z 8y + 8y .

(3.2)

(3.3)

Since both the real and imaginary parts of (3.2) and (3.3) must be equal, we obtain the Cauchy-Riemann equations.


b) Differentiating the first Cauchy-Riemann equation with respect to x and the second one with respect to y we obtain

(PU (Pv - and

ox2 - oxoy

Adding the obtained equalities we obtain that the function u satisfies the Lap1ace equation. Differentiating now the first Cauchy-Riemann equation with respect to y and the second one with respect to x and repeating the preceding procedure we find that the function v satisfies the Lap1ace equation, too.

Example 3.14 Let a complex function f be dilferentiable at z. Find

1. f(z) - f(w) 1m '--'---'---'--'--'-z-+w Z - W

when z belongs to the straight line Re z = Re wand then when z belongs to the straight line Im z = Im w. Prove that J = u + zv satisfies the Cauchy-Riemann equations.

Solution. If z = x + zy and w = a + zb, then the limit on Re z = Re w is given by

1· J(z)-J(w) l' u(a,y)+zv(a,y)-u(a,b)-zv(a,b) 1m = 1m --'--:"'::-'-_-O...~'"--_":"""";--:-'-_---''-'---'-%-+w z - W y_b a + zy - a - zb

1· u(a,y)-u(a,b) l' y(a,y)-v(a,b) 1m + 1m ::""':"-:"::-'---::--'-'--'-y_b z(y - b) y-+b Y - b

= _z(OU) + (OX) . oy x=a,y=b oy x=a,y=b

Analogous1y we obtain on Im z = Im w :

1. J(z) - J(w) 1m '--'---'---'--'--'-z-+w Z - W

1. u(x, b) + zv(x, b) - u(a, b) - zv(a, b) Im~~-'---o"'~~-~~----'~~

x-+a X + zb - a - zb

= (OU) + z(OX) . ox x=a,y=b ox ",=a,y=b

We obtain the Cauchy-Riemann equations taking equal the real and imaginary parts, respective1y.

Example 3.15 Prove that the Junction VlZ2 - z21 satisfies the Cauchy-Riemann equations at z = 0, but it is not dilferentiable at this point.

Solution. The function VIZ2 - z2 1 is identically zero on the real and imaginary axes and therefore trivially satisfies the Cauchy-Riemann equations at z = O. Taking z = r( cos 6 + z sin 6) and r -+ 0 we obtain

. VIZ2_Z21 . VI4cos6sin61 hm = hm .' r_O Z r-O (cos6 + zsm6)

Since the last limit depends on (), take () = 0 and ()7r / 4, we conclude that the function

VIZ2 - z2 1 is not differentiable at z = o.

Example 3.16 Find by the definition the derivatives (in the regions where they exist) of the following functions

z a) Z2 + 1 j

b) l+z. 3 ' z

n

c) EUiZij d) Imzj e) Zj J) 14


for all w =I- z, -z.

i=O

z w

lim Z2 + 1 w 2 + 1 %-+w Z - w lim -wz{z-w)+z-w z-+w (z - w)(z2 + 1)(w2 + 1)

= lim (z - w){l - wz) %-+w (z - W)(Z2 + 1)(w2 + 1) 1- w 2

= (w2 + 1)2

Exercise 3.17 Find which of the following functions can be real or imaginary part of a complex function f which is differentiable in the region Izl < 1.

Solution. a) We have for w(x, y) = x2 - axy + y2 that

Therefore ()2w 82w 8x2 + 8y2 = 2 + 2 =I- 0,

and the given function w can not be the real or imaginary part of a differentiable function.

b) No. c) No. d) Yes.

Exercise 3.18 Find the constants a, b, and c such that the following fundions would be analytic

a) fez) = x + ay + l(bx + cY)j b) fez) = cosx(coshy + asinhy) + lsinx(coshy + bsinhy).

Answers.

a)c=l,b=-a, then f(z) = (l-al)z; b)a=b=-l, then f(z) = eU •

Exercise 3.19 Find the region of analyticity of the function

Answer . The function is analytic for

0< argz < 7r/4 and 7r < argz < 57r/4,

then fez) = Z2 j and for

7r/2 < argz < 37r/4 and 37r/2 < argz < 77r/4

and then fez) = _z2.

Exercise 3.20 If we represent a complex analytic function f as

( z + i z - i) _ w(z) = f(x,y) = f -2-'~ = F(z,z),

and considering z and z as independent variables prove that

aw 1 (aw aw) -=- --l-az 2 ax ay

and aw _ ~ (aw aw) ai - 2 ax + l ay .

Example 3.21 (Hadamard) Show that the function u = u(x,y) given by

e ny _ e-ny

u(x,y) = 2n2 sinnx

for n E N is a solution on D = {(x, y) I x2 + y2 < I} of the Cauchy problem for the Laplace equation

u(x,O) = 0, au(x,O) _ sin nx

ay - n

Prove that this problem is not well-posed, i.e., a problem with a partial differential equation is well-posed in a dass of functions C, if the following three conditions are satisfied:


(i) there exists a solution in Cj

(ii) the solution is uniquej

(iii) the solution is eontinuously dependent on the given eonditions, e.g., initial~

values, boundary eonditions, eoeffieients, ete.

Solution. It easy to check that the function U n for an arbitrary but fixed n E N

given by e ny _ e~ny

un(x, y) = 2 sin nx 2n

is a solution of the given Cauchy problem. Letting n ---t 00 we obtain for x, y E D, x f- 0, y f- 0

IUn(x, y)1 ---t 00.

On the other side, the given Cauchy problem for n ---t 00 reduces to the problem

82u 82u 8x2 + 8y2 = 0,

() 8u(x,0) =0 U x,O = 0, 8y ,

which has only a trivial solution U = O. Therefore the considered solution of the given Cauchy problem does not depend continuously of the initial condition. Hence it is not well-posed (it is ill-posed).

Remark . In contrast with this simple problem of a partial differential equation which is not well-posed, let us remind some weIl known results from the theory of ordinary differential equations where for general classes of problems with ordinary differential equations the well-posedness can be ensured. For example, the wellposedness of the Cauchy problem

y' = J(x,y), y(xo) = Yo

is ensured supposing that the function J is continuous and satisfies the Lipschitz condition in some region which contains the point (xo,Yo).

3.2 Special Functions

3.2.1 Preliminaries

The function eZ •

The function eZ is the analytic solution of the functional equation

3.2. SPECIAL FUNCTIONS

which for every z = x real reduces on the function eX • This gives

eZ = eX cos y + zeX sin y

for z = x + zy. The function has the following properties

(i) lezi = eX ;

(ii) eZ =I- 0;

(iii) e'Y = cosy + zsiny;

(iv) the equation eZ = a has infinitely many solutions for any a =I- O.

(v) (ez )' = eZ •

The functions sin z and cos z. We define

1 cos z = -( elZ + e-'Z).

2

We have (sinz)' = cosz and (sinz)' = -cosz.


65

Example 3.22 Prove that f(z) = eX cos y + zex sin y is the only analytic solution of the functional equation f(zdf(Z2) = f(ZI + Z2) which satisfies the condition f(x) = eX fOT all real x.

Solution. It is easy to check that the given function satisfies the functional equation f(ZI)f(Z2) = f(ZI + Z2) and f(x) = eX for all real x. Suppose now that f is an analytic solution of the functional equation which satisfies the condition f(x) = eX for all real x. Then

f(z) = f(x + zy) = f(x)f(zy) = eXf(zy).

Taking f( zy) = u(y) + w(y) we obtain

f(z) = eXu(y) + zeXv(y).

Since f is analytic it satisfies the Cauchy-Riemann equations and therefore u(y) = v'(y) and u'(y) = -v(y). Hence u" = -u. This ordinary differential equation has the general solution

u(y) = a cos y + b sin y


for a and b real constants. Since f(x) = eX and

v(y) = -u'(y) = asiny - bcosy

we obtain for y = 0 u(O) = a = 1 and v(O) = -b = O.

Therefore v(y) = sin y and finally

f ( z) = eX cos y + zex sin y.

Exercise 3.23 Prove that eZ is an entire function.

Exercise 3.24 Show that lezi = eX and eZ =I- O.

Hint. The equality follows by Example 3.22. For the second property use the preceding equality and that eX =I- 0 for x real.

Example 3.25 Find the real and imaginary parts and modulus of

Solution. Since

We have arg e(2+"·/4)2 = 7r. Since e'" = cos 7r + z sin 7r = -1 we obtain

Exercise 3.26 Solve the equations: a) eZ = 1; b) eZ = 2; c) e% = 1 + 1.

Answers. a) (7r/2 + 2h)1, k = 0, ±1, ±2, ... ; b) In2 + 2hz, k = 0, ±1, ±2, ... ; c) ~(ln 2 + 7r /2) + 2h1, k = 0, ±1, ±2, ....

Exercise 3.27 Prove the identities

a) sin2 + cos2 z = 1;

b) sin2z = 2sinzcosz;

c) (cos z)' = - sin z.

3.2. SPECIAL FUNCTIONS 67

Example 3.28 Find the real and imaginary parts, and modulus of the following functions: sin z, cos z, tan z, cot z.

Solution. For sin z and cos z see chapter 3. It is easy to prove that

Therefore

( ) tan u + tan v tan u + v = ------

1 - tan u . tan v for u, v E (C.

tanx + tan iy tan z = tan( x + zy) = ,

1 - tan x . tan zy

and since tan zy = z tanh y we obtain

Re( ) tanx(1+tanh2y) tanz = 2 ,

1 + tan2 xtanh y

I ( ) tanhy(1-tan2x) m tanz = 2 ,

1 + tan2 xtanh y

I tan zl = tan2 x + tanh2 y.

Exercise 3.29 Prove that

sin(zz) = zsinhz, cos(zz) = coshz,

tan(zz) = danhz, cot(zz) = -zcothz.

Exercise 3.30 Solve the equation sin z = 3.

Hint. Solve first the equation v = e'Z for v.

Exercise 3.31 Prove that for every u, v and z

a) sin( u + v) = sinu cos u + sin v cos Uj

b) cos(u + v) = cosucosv - sinusinv.

Hint. a), b) . Putting

elZ _ e- IZ elZ + e-'Z

smz = and cosz = ----2 2z

we directly verify the desired equalities.

3.3 Multi-valued functions

3.3.1 Preliminaries

A point Zo is a branching point of a multi-valued function w = j(z) if taking a closed path around Zo the image of this path in the w-plane is not a closed path. If after going n times around a branching point Zo the path in w-plane is closed, then Zo is an algebraic branching point of order n. If after going an unlimited number of times around a branching point Zo the path in w-plane is never closed, then Zo

is a transcendental branching point of order n. Cutting is the straight line segment connecting two branching points ( we can go through z = 00).

The junction Log z The multi-valued function Logz as an inverse mapping of eZ is given by

Logz = logz+ 2kn, k = 0,±1,±2, ... ,

where log z is the logarithm principal value (branch) of Logz given by

log z = log Izl + zarg z, -7r < arg z ~ 7r.

for -7r < arg z < 7r when the cutting is the negative real axis or 0 < arg z < 27r

when the cutting is the positive real axis.

For each k we have a branch of the multi-valued function Log z, which is a function.

We define for w f- 0 and z the multi-valued function


Example 3.32 Find Log z and log z, jor

a) 1+z; b) 1-z; c) -1+z; d) -1-z.

Solution. In the complex plane without the positive real axis we obtain:

1 7r a) log(l + z) = "2 log 2 + "4z,

Log (1 + z) = ~ log 2 + (~+ 2h) z (k = 0, ±1, ±2, ... ).

1 77r b) log(l - z) = "2 log 2 + TZ,

3.3. MULTI- VAL UED FUNCTIONS 69

Log (1 - z) = ~ log 2 + C: + 2k1r) z (k = 0, ±1, ±2, ... ).

1 311" c) log(-l + z) = 2 log2 + 4z,

Log ( -1 + z) = ~ log 2 + (3: + 2k1r ) z (k = 0, ±1, ±2, ... ).

1 511" d) log(-l-z) = 2 1og2 + 4z,

1 (511" ) Log(-1-z)=2Iog2+ 4+2k1r z (k=0,±1,±2, ... ).

Example 3.33 Find the mistake in the J.I.Bernoulli paradox chain of reasoning

i) Log (_Z)2 = Log(Z2).

and

ii) Log (-z) + Log (-z) = Logz + Logz.

iii) 2Log (-z) = 2Log z.

iv) Log (-z) = Log z for z =I- 0.

Solution. We have

Logz = log Izl + zarg z + 2k1l"z,

Log( -z) log I - z I + zarg z + zarg ( -1) + 2k1rz log Izl + zarg z + (2k + 1 )1I"Z, k = 0, ±1, ±2, ...

We see that no value of Logz coincides with any value of Log( -z).

The mistake was made in going from ii) to iii) since Log(-z) + Log(-z) is not equal 2Log( -z). Namely, Log( -z) + Log( -z) is sum of any two numbers from the set of values Log(-z), and 2Log(-z) is a sum of the same value of Log(-z) with itself.

We can illustrate this property with the following simple example. Take A = {O, I}, then the set A + A consists of three elements: ° + ° = 0, 0+1 = 1 and 1 + 1 = 2 and the set 2A consists of two elements: 2· ° = ° and 2· 1 = 2.

Exercise 3.34 Is it true the equality log( u . v) = log u + log v? Find log( -1 - Z)2. Is it true the equality log( -1 - Z)2 = 2log( -1 - z)?

Example 3.35 Prove that wz, w =I- 0, for z = p/q rational has most q values.


Solution. Follows from

Example 3.36 Find F(O), F(l), F( -1) where F(z) is the bmnch of the function ~z - z which for z = 1 + z takes the value 1.

Solution. First we shall find the desired branch. We have

for k = 0,1,2,3. The desired branch for which Fk (l + z) = 1 is given by

Therefore

Fo(z) = e(ln!z-.!+arg (z-.).)/4.

Fo(O) = e37r'/8 ;

Fo(l) = e(1n2)/8 e71n/16;

Fo( -1) = e(1n2)/8 e57r./16 •


1 Arccos z = -Log z(z ± Vz2=1).

z

Find Arcsin z.

Exercise 3.38 Prove that all bmnches of Arccos z are given by

Arccos z = arccos z + 2br for k = 0, ±1, ±2, ....

Hint. Use Example 3.37.

Example 3.39 Find the real and imaginary parts, and modulus of the following functions: sin z, cos z, tan z, cot z.

Solution. For sin z and cos z see chapter 3. It is easy to prove that

t ( ) tan u + tan v an u + v = .,.------

1 - tan u . tan v for u,v E C.

3.3. MULTI- VALUED FUNCTIONS

Therefore tanx + taniy

tan z = tan( x + zy) = 1 ' - tanx· tan zy

and since tan zy = z tanh y we obtain

Re( ) tanx(I +tanh2y) tanz = 2 ,

1 + tan2 x tanh y

I ( ) tanhy(I-tan2x) m tanz = 2 ,

1 + tan2 x tanh y

J tan zJ = tan2 x + tanh2 y.


sin(zz) = zsinhz, cos(zz) = coshz,

tan(zz) = ztanhz, cot(zz) = -zcothz.

71

Exercise 3.41 Find the branching points 0/ the /ollowing /unctions a) VZj b)

Vz2=1j c) Arcsinzj d) Logz(1 + zz)J(I- zZ)j e) J(z - 5Hz - zHz - 2z + 3).

Answers. a) z = 0 and z = 00 are algebraic branching points of the order n.

b) z = 1 and z = -1 are algebraic branching points of the order 2.

c) z = 1 and z = -1 are algebraic branching points of the order 2, and z = 00

is a transcendental branching point.

d) z = z and z = -z are transcendental branching points.

a) z = 5, z = z, z = 2z - 3 and z = 00 are algebraic branching points of the order 2.

Exercise 3.42 Find the sets where are the branches 0/ the multi-valued /unction

W = J(1 - z2)(1 - p z2), 0< k< 1.

Answer. The branching points are ±I and ±IJk. The point z = 00 is not a

branching point. We can take the cuttings [-l,-I] and [1, l] from the complex

plane. The second solution: we can take the cuttings [00, -7}], [-1,1] and [l,oo] from the complex plane.


z z + z Arctgz = -Log--.

2 z - z


Example 3.44 The brunches of the function

z-l Log-

z+l

are separuted by the cuttings [=, -1] and [1, =]. Let w( z) be the brunch of the preceding function which is real on the upper part of the cut [1, =]. Find w( -2z).

Solution. To find w( -2z) we have to calculate arg (z-l) and arg (z+ 1) at z = -2z when z moves through a path from a point from (-1, 1) up to the point - 2z not crossing the cuttings. So we obtain arg(z - 1) = 57r/4 and arg(z + 1) = -7r/4. Therefore

Hence w( -2z) = In 2 + Z~7r.

z -1 3 arg-- = -7r.

Z + 1 2

Exercise 3.45 The brunches of the function ZZ are separuted by the cut [-=,0]. Let w(z) be the brunch of the preceding function for wh ich w(l) = 1. Find w( -e+zO) and w( -e - zO), where -e + zO and -e - zO denotes the value at the brunch upper and down of the cut, respectively.

Answer. w( -e + zO) = a-e(l+l) and w( -e - zO) = e-e(l-I).

Chapter 4

Conformal mappings

4.1 Basics

4.1.1 Preliminaries

Let z(t) = x(t) + ty(t), a $ t $ b, be a path. We suppose throughout this chapter that z'(t) #- 0 for all t. Let 1 be defined in a neighborhood of a point Zo E ((:. 1 is conformal at Zo if 1 preserves angles at Zo, for any two paths PI and P2 intersecting at Zo the angle from PI to P2 at Zo , the angle oriented counterclockwise from the tangent line of PI at Zo to the tangent line of P2 at Zo, is equal to angle between I(Pd and I(P2 ). The function 1 is conformal in a region 0 if it is conformal at all points from O.

Theorem 4.1 111 is analytic at Zo and f'(zo) #- 0, then 1 is conlormal at zoo

Definition 4.2 A 1 - 1 analytic mapping is called a conlormal mapping. Two regions 0 1 and O2 are conlormally equivalent il there exists a conlormal mapping lrom D 1 onto D2 •

Theorem 4.3 (Riemann Mapping Theorem) For any simply connected region o #- IC and Zo E 0, there exists a unique conlormal mapping 1 010 onto the unit disc D(O, 1) such that I(zo) = 0 and f'(zo) > O.


Example 4.1 Examine the conlormality ollunctions a) I(z) = z2 j b) eZ •

73


74 CHAPTER 4. CONFORMAL MAPPINGS

Solution. a) The function j(z) = Z2 is not conformal at z = 0, it maps the angle between positive real axis and positive imaginary axis on the angle between positive real axis and negative real axis. Since 1'(z) = 2z -:j; 0 für z -:j; we conclude by Theorem 4.1 that f(z) = Z2 is conformal for all z -:j; o. Second method. Taking f = u + zu we have that the preimages of the curves u = rl

and v = r2 for rl and r2 different from zero have to be orthogonal. Namely, since we have for z = x + zy

u(z) = x 2 - y2 and v(z) = 2xy,

the desired preimiges are the orthogonal families of hyperbolas

b) Since 1'(z) = eZ and it is always different from zero we conclude by Theorem 4.1 that j(z) = eZ is everywhere conformal.

Exercise 4.2 Prove that if f is analytic at Zo and !'(zo) -:j; 0, then j is conformal at zoo

Example 4.3 Prove that a 1 - -1 analytic function a region 0 is conformal as weil its inverse function f- I which is also analytic.

Solution. Since f is 1 - -1 we have l' -:j; O. Hence f is conformal and f- I IS

analytical. Since (1-1)' = 1/1' we obtain (I-I)' -:j; O. Hence f- 1 is conformal.

Exercise 4.4 Prove that the relation conformal equivalence is an equivalence relation.

4.2 Special mappings

4.2.1 Preliminaries

1) Linear transformation W = az + b

can be viewed as a composition of three mappings: WI = lalz (magnification for laD; W2 = e·9z, () = arga (rotation through the angle theta); and W3 = Z + b (translation for b); as W = W3 0 W2 0 WI.

4.2. SPECIAL MAPPINGS 75

Power transformation w = za, areal.

It maps the wedge {z I ()l < arg z < ()d onto the wedge {w I a()l < arg w < a()2}. For ()2 - ()l ::; 2: it is a conformal mapping.

3) Exponential transformation

It maps the strip {x + ty I Yl < Y < Yd onto the wedge {wl Yl < arg w < Y2}. For Y2 - Yl ::; 21T it is 1 - -1.

4) The bilinear (Möbius) transformation

az + b w - -- ad - bc -:j:. O.

- cz + d'

It is conformal and 1 - -1.

Theorem 4.4 Bilinear transformation maps circles and lines onto circles and lines.


Example 4.5 Consider the linear fractional transformation

t-z w=--.

t+z

Find the regions in w-plane which are images of the following regions in the z-plane a) Imz ~ 0; b) Imz ~ O,Rez ~ 0; c) Izl < 1.

Solution. The linear fractional transformation w = (t -z)j(t+z) or represented by the matrix

can be written in the form

or in the matrix form

[~1 :] 2t

w=-l+-z+t

[ ~ 1 :] = [~t ~ 1 ] [~ ~] [~ :].


a) First the region {z I Im z :::: O} (Figure 4.1)

o Figure 4.1 Imz :::: 0

is translated for l (Figure 4.2) by the transformation

/ I

./

'\ \

I /

Figure 4.2 Im Wl :::: 1

u,


in the wrplane.

Then the transformation

maps the Ene Im Wl = 1 on the circle ( Figure 4.3)

IW2 + z/21 = 1/2, since the line has common point z = z with the unit circle (this point goes in the point z = -z) and the point z = 00 go es in the point O.

Figure 4.3 IW2 + z/21 = 1/2

The image in the w2-plane is

(Figure 4.4). Finally, the transformation

maps the region

on the unit disc Iwl ~ 1 ( first rotating for ~ the starting region, then multiplying it by 2 and translating it by -1, Figure 4.4).

w

~"-,.'-.Q----------- -- --.

Figure 4.4 IW2 + z/21 ~ 1/2 and Iwl ~ 1

b) In the analogous way as in the preceding example we obtain

Figure 4.5 Imz ~ O,Rez ~ 0

that the linear fractional transformation

z-z w--- z+z·

maps the region Imz ~ 0, Re z ~ 0 ( Figure 4.5) in the z-plane on the half-disc Iwl < 1,Im ~ 0 (Figure 4.6).

U

4.2. SPECIAL MAPPINGS

0, i

Figure 4.6 Iwl < 1, Imz ~ 0

c) The unit disc Izl ::; 1 (Figure 4.7)

"'y !

Figure 4.7 Izl ::; 1

first is translated by z with the transformation

79

w

u

z

x


(Figure 4.8) in the wl-plane.

u,

Figure 4.8

The transformation

[~z ~1] maps the circle (Figure 4.8) IWI - zl = 1 on the line Im W2 = -1/2 since the circle cross the point 0 (and this point goes to z = 00) and met the unit circle in the points ../3/2 + z/2 and -../3/2 + z/2 which are transformed in the points ../3/2 - z/2 and -../3/2 - z/2, respectively.

Figure 4.9 Imw2 ~ -1/2

In this way the disc IWI - zl ~ 1 is mapped on the region Im W2 ~ -1/2 (Figure 4.9). Finally, the transformation

maps the region Im W2 ~ -~ on the half-plane Re w ~ 0 ( rotated by ~, then multiplying it by 2 and then translated for -1, Figure 4.10).

Figure 4.10 Rew ~ 0

Exercise 4.6 Find the regions in w-plane which are images of the following regions in the z-plane by the following maps

a) Half-disc Izl < 1, Imz > 0 by transformation w = ~z+-;;.

b) Half-disc Izl < R, Imz > 0 by transformation w = ~:; ~. c) The region 0 < 0 < 'Ir / 4 by transformation w = z ~ 1 .

Figure 4.11 Izl < 1, Imz > 0

d) The ring 1 < Izl < 2 by transformation w = z/(z -1).

Answers. a) The half-disc Izl < 1, Imz > 0 (Figure 4.11) is transformed on the region which contains the point w = 0 and it is bounded by circles Iwl = 1 and Iw + 5t/41 = 3/4 (Figure 4.12).

w

1 u

Figure 4.12

b) The half-disc Iz/ < R, Imz > 0 (Figure 4.13) is transformed

Figure 4.13 /z/ < R, Imz > 0

on the region Rew < 0, Imw > 0 (Figure 4.14).

w

u

Figure 4.14 Re w < 0, Im w > 0

c) The region 0 < 0 < 7r/4 (Figure 4.15) is transformed

o Figure 4.15 0 < 0 < 7r/4

on the half-plane Im w < 0 without the disc Iw -1 + 2-1 ::; {} (Figure 4.16).

Figure 4.16

d) The ring 1 < Izl < 2 (Figure 4.17) is transformed

Figure 4.17 1 < Izl < 2


on the region bounded by the line Rew = ~ and the circle Iw - 4/31 = 2/3 (Figure 4.18).

v

f o

Figure 4.18

Exercise 4.7 Using the progmm Mathematica or Maple V (or some similar computer program) plot the image of the lines z = x + zy, x = k!:l.x, y = k!:l.y for fixed !:l.x and !:l.y in the w-plane by the function a) w =~; b) w = 2z/{1 + Z2).


Answer. a) w = 1/ z

Figure 4.19


Example 4.8 Prove the uniqueness of the bilinear (Möbius) transformation which maps the points Zb Z2, Z3 on the points Wb W2, W3 (i =/:- j => Zi =/:- Zj 1\ Wi =/:- Wj), respectively.

Solution. We introduce the so called anharmonic relation between four points Zi, i = 1,2,3,4:

Z3 - Zl Z4 - Zl (Zl>Z2,Z3,Z4) = ---: ---.

Z3 - Z2 Z4 - Z2 Abilinear transformation saves the anharmonic relation.

Namely, let

aZi + b. az + b Wi = ---d for z = 1,2,3,4, where W =--

CZi + CZ + d

is the given bilinear transformation.

Then we have

W3- W1 =

Therefore

(ad - bC)(Z3 - Zl) (CZ3 + d)(CZ1 + d)

and (ad - bC)(Z3 - Z2)

W3 - W2 = (CZ3 + d)(CZ2 + d)'

W3 - W1 _ (Z3 - Zl)(CZ2 + d) W3 - W2 - (Z3 - Z2)(CZ1 + d)'

We obtain analogously

W4 - W1 (Z4 - Zl)(CZ2 + d) W4 - W2 = (Z4 - Z2)(CZ1 + d)'

The last two equalities imply (Zl,Z2,Z3,Z4) = (W1,W2,W3,W4)' Putting Z4 = Z (z is an arbitrary point) and exchange the positions of Z3 and Z4 we obtain

Z - Z2 Z3 - Z2 W - W2 W3 - W2

Th" I' h h b'I' 1" az + b . . lS 1mp les t at tel mear translOrmatIon W = --d 1S umque. CZ+

Example 4.9 a) Find the image of the unit circle Izi = 1 by the following transformation

Z-v w=u·----1' u,vEC and Izl=/:-l. vz -

b) Which bilinear transformation maps the unit circle Izi = 1 on the unit circle Iwi = I? What is the image of the unit disc Izi :5 1 by this transformation?

88

Solution. a) Since

CHAPTER 4. CONFORMAL MAPPINGS

zz - vz - vz + vv w·w=u·iJ.·------

vvzz - vz - vz + 1 '

for z . z = 1 we have that w . W = u· iJ.. Therefore the unit circle Izl = 1 is mapped on the circle Iwl = lul.

b) U sing a) we obtain that the bilinear transformation

z-v w=u·-_-- v=/:-l,

vz -1

maps the unit circle Izl = 1 on the unit circle Iwl = 1, if and only if lul = 1, for u = eO. (0 :::: B :::: 211"). The desired bilinear transformation is

0. z - v w = e . -- lvi =/:- 1.

vz -1'

The unit disc Izl :::: 1, for lvi< 1, is mapped on the unit disc Iwl :::: 1, and for lvi> 1 on the region Iwl 2 1.

Exercise 4.10 Find abilinear transformation whieh maps points -l,z, 1 +z on the following points: a) 0, 2z, 1 - Zj b) z, 00, 1, respectively.

Hint. See Example 4.8.

Answers.

a) w = -2z(z + 1) j

-4z - 1 - 5z b) w=(1+2z)z+6-3z.

5(z - z)

E · A b '1'·1 az + b 1 a b 1-1 d f xerClse 4.11 z mear transjormation w = --d' d -r 0 is perio ie i ez + c

there exists a natural number n such that

[; ~r=[; ~r=[~ ~]. Find whieh of the following bilinear transformations are periodie

a) u

b) z+l

c) 2zz - 3

- , ---j z z z-z

d) w = az + b, cz+d I; ~ 1 =/:- 0, (a + d? = 3(ad - bc).


Answers.

a) Yes. b) Yes. c) No. d) Yes.

Example 4.12 Find all bilinear transformations with fixed points z = 1 and z = -1.

Solution. Using the anharmonic relation from Example 4.8 we find that the desired transformations are

w-l z-l --=k·--w+l z+l

(k # 0 and k E C).

Example 4.13 Prove that two bilinear transformations are commutative (with respect to function composition) if they have a common fixed point.

Solution. Let

and

with I:: ~ 1·1:: ~ I # o. A fixed point for both transformations will be determined from

CIZ2 + (d - al)z - b1 = 0, C2z2 + (d2 - a2)z - b2 = o. The zeroes of these quadratic equations will be equal if and only if

Cl dl - al bl

C2 = d2 - a2 = b2 •

This condition implies

what we had to prove.

Exercise 4.14 Prove that J(z) = l/z maps circles and lines on cireles and lines, respectively.


Example 4.15 Prove that the bilinear transformation maps cireles and lines onto circles and lines.

Solution. We have for c =1= 0

f(z) = az + b = ~ (a _ (ad - bC)) , cz+d c cz+d

i.e., f = ho f2 ° f1, where

f1(Z) = CZ + dj h(z) = -j h = - - z. 1 a (ad - bC) z c c

Then use Exercise 4.14.

Exercise 4.16 Let us consider the following sets of bilinear transformations

G={W= u~+v I U,V,EIC,luI2+lvI2=1}, -vz+u

{ () . () cos 2 . z - z . sm 2 I

G2 = W = . () () -zsm 2 + cos 2

a) Prove that (G,o) is a group, and (G1 ,o) and (G2,o) are its subgroups, where ° is the operation of the composition of functions.

b) Prove that for every T E G there exist Tl, T3 E Gi and T2 E G2 such that

c) Apply b) on

[ l+z 1-Z] -l+z 1-z .

Answer.

b) We have

[ ~ ~] = [ e,,,,,j2 -v u 0

o ]. [ cos ~ Z· sin ~ j. [ e''''2/2 0 ] e,,,,,j2 . () . 7) 0 e''''2/2 ,

-zsm2 Z· sm2

where

and

Exercise 4.17 Find the images oj the jo/lowing regions by the given junetions:

a) The angle 0 < argz < 0, Figure 4.20, by thejunetion w = z'1f/8.

b) The strip 0 < Imz < 71', Figure 4.22, by the junction w = eZ •

e) The region between two eireles between points Zl and Z2, Figure 4.24, by the z - Zl

junction w = --. Z - Z2

d) The region between two eireles, jrom whieh one is inside oj other and they

have one eommon point zo, Figure 4.26, by the junction w = _1_. Z - Zo

Answers.

z

o

Figure 4.20 0 < arg Z < 0

a) The image is the region Im w > 0, Figure 4.21.

Figure 4.21 Im w > 0

b) The image is

y z

Figure 4.22 0 < Im w < 1f'


the region Im w > 0, Figure 4.23.

Figure 4.23 Im w > 0

c) The image is

y

o x

Figure 4.24


the angle on Figure 4.25.

v w

u

Figure 4.25

cl) The image is

z

x

Figure 4.26

the strip between two parallel stright lines on Figure 4.27.

Figure 4.27

Example 4.18 Construct a Junction w = J(z) which maps the region Oz : /z/ < 1, Imz > 0 on the region Ow : /w/ < 1.

Solution. Up to now our task was to find the image Ow of a region Oz under the given mapping f.

Now we have two given regions Oz and Ow, and we have to construct a mapping from Oz onto Ow.

The Riemann Mapping Theorem ensures the existence of the desired conformal mapping and with some additional suppositions (that a point from 0% and a line through it are mapped on a given point and a line in Ow), we have uniqueness.

There is no general algorithm for finding the desired conformal mapping. We can use for the construction of the desired function the functions from Exercise 4.17.

z

-1 o 1 x Figure 4.28 /z/ < 1, Imz > 0

For our case this can go in the following way. First we map with the function

z-l w=--

z+l

(see Exercise 4.17 c) ) the region Izl < 1, Imz > 0 (Figure 4.28) on the inner part of angle 7r /2 with the corner at the point (0,0)

V,

w,

o u, Figure 4.29 RewI < 0, Im WI > 0

in the WI- plane and because of WI(O) = -1, on the region ReWI < 0, ImwI > 0, Figure 4.29.

Then we map with the function W2 = w; the region Re WI < 0, Im WI > 0 on the half-plane Im W2 < 0, Figure 4.30.

Now using the function

+Vz I

Figure 4.30 Im W2 < 0

W2 + t W=--, W2 - t

we map the half-plane Im W2 < 0 on Iwl < 1. So finally, we obtain the desired transformation

( z - 1)2 z:tT + Z Z2 - 2z + z w- -

- (Z - 1)2 _ Z - Z2 - 2zz + 1 . z:tT

-1

Figure 4.31 Iwl < 1

Example 4.19 Construct a Junction w = J(z) which maps conJormally the region 0% : Izl < 1, Imz > 0 Rez > 0 on the region Ow : Iwl < 1.

Solution. We map with the function WI = Z2

z

x

Figure 4.32 Izl < 1, Imz > 0" Rez > 0

the quarter of the unit disc Izl < 1, Im z > 0" Re z > 0, Figure 4.32, on the half-disc IWll < 1, Im Wl > 0, Figure 4.33.

U1

Figure 4.33 IWll < 1, Im Wl > °

Using now the function from Example 4.18 we obtain the function

Z(Z2)2 - 2z2 + Z ZZ4 - 2z2 + Z

W = J(z) = (Z2)2 _ 2zz2 + 1 = Z4 - 2zz2 + 1·

Example 4.20 Construct afunction W = J(z) which maps the angle 'Ir/6 < argz < 'Ir /3 on the unit disco

Solution. Firstly we rotate the given angle (Figure 4.34)

y

z

o x~ Figure 4.34 'Ir /6 < arg z < 'Ir /3


for -'Ir /6 by the function Wl = e- l1r/ 6 . z, Figure 4.35.

o u, Figure 4.35

Next we map with the function W2 = w~

o Uz

Figure 4.36 Im W2 > 0

the angle obtained on the half-plane Im W2 > 0, Figure 4.36.

Finally, we map by the function w = W2 - l (see Example 4.5) the region Im W2 > W2 + l

o onto Iwl < 1, Figure 4.37.

w

~~~~*M~-----~ -1 1 u

Figure 4.37 Iwl < 1

So we obtain

w = f(z) =

Exercise 4.21 Construct a function w = f(z) which conformally maps

a) the region Izl < 1, Iz - 1/21 > 1/2 onto the half-plane Im w > 0;

b) the region Izl < 1,0< argz < rr/n (n E N) on the half-plane Imw > 0;

c) the region Izl < 1,0< argz < rr/3 on the unit disc Iwl < 1.

Answers.

a) w = e-'7l"(Z+l)/(Z-l);

Example 4.22 Prove that the function w = cos z maps the lines Re z = const on hyperbolas in the plane w, and the lines Imz = const on ellipses in the w-plane.

Solution. We have by definition

e'(X+'Y) + e-'(x+,y) w=u+w= ,

2

for z = x + zy. This implies

eV + e-Y u = cosx

2 and

e-Y - eV • v = 2 smx.

Then the lines x = const are mapped in w-plane onto the hyperbolas

u 2 v 2 ----.--=1, cos2 x sin2 x

and the straight lines y = const onto ellipses

u2 v2

(eV ile-vr + (e-Y :2 eyr = 1.

Exercise 4.23 Find the sur/ace area 0/ the region onto which the /unction w = eZ

maps the rectangle

K = {z I a - b ~ Re z ~ a + b, - b ~ Im z ~ b},

where a is real and 0< b< 'Ir.

Find the limit 0/ the /raction 0/ the sur/ace area 0/ the obtained region and sur/ace area 0/ rectangle as b _ O.

Answer: The desired region is bounded by straight lines arg w = band arg w = -b and by circles w = eaH , w = ea- b• The limit fraction of surface areas is e2a as b_ O.

Exercise 4.24 Find cireles which are invariant under the mapping z I-t w given by

1 1 ---= --+a, W-Zo Z-Zo

where Zo and a are arbitrary but fixed points on C.

Answer. The desired circles crosses the point Zo where they have a tangent given by 0.

Example 4.25 Find the curves on which the /unction w = (z+ 1/z)/2 maps circles Izl = r, r < 1.

Solution. Since we have for z = re,a that

l+r l-r q = Tcosa-zTsina,

we obtain that the circles Izl = r, r > 1, are mapped on the ellipses with axes (l/r + r)/2 and (l/r - r)/2 and same focuses 1 and -1.

Exercise 4.26 Prove that the function w = (e-·az)21r/(b-a) for 0 :::; a < b < 271', maps the angle a < arg z < b onto the whole w-plane without the non-negative part of the x-axis.

Hint. We have

2 argw = -b -(argz - a), 0< argw < 271'.

-a

Exercise 4.27 Find the function which maps the annulus 0< rl < Izl < r2 on the region between two ellipses Iz - 21 + Iw + 21 = 4b1 and Iw - 21 + Iw + 21 = 4b2 for 1<b2 <b1 •

Hint. Use Example 4.25. The desired function is w = cz + ~, for cz

Chapter 5

The Integral

5.1 Basics

5.1.1 Preliminaries

For a continuous complex-valued function J(t) = u(t) + w(t), a S; t S; b, we take

l b J(t) dt = t u(t) dt + z t v(t) dt.

Definition 5.1 The (line) integral oJ a continuous function J at all points of a path P given by z(t), is given by

h, fez) dz = t f(z(t))z'(t) dt.

Two curves PI : z(t), a S; t S; b, and P2 : w(t), c S; t S; d, are smoothly equivalent if there exists an 1-1 and Cl mapping s: [c,d] ~ [a,b] such that s(c) = a,s(d) = b, S'(t) ~ 0 for t E [c, d], and w = z 0 s. We have

( fez) dz = ( fez) dz }p, }P2

for smoothly equvalent paths PI and P2 . We have for a path P : z(t), a S; b, continuous function J on P

{pJ(z)dz = - h,J(z)dz,

where -P is defined by z(b+ a - t),a S; t S; b. We have for a path P: z(t),a S; b, continuous functions J and 9 on P

k(f(z) + g(z)) dz = k J(z) dz + kg(z) dz.

103


104 CHAPTER 5. THE INTEGRAL

We have for a path P: z(t),a S; b, continuous functions f on P and A E IC

We have for a path P : z(t), a S; b, continuous functions f on P

Theorem 5.2 If {fn} is a sequence of continuous functions which uniformly converges to f on a path P, then

Ern r fn(z)dz= r f(z)dz. n-+oo}p }p

Theorem 5.3 If f is the derivative of an analytic function F (analytic on a path P) then

i f(z) dz = F(z(b)) - F(z(a)).

Theorem 5.4 (Cauchy's Theorem) Let f be analytic in a simply connected region 0 and P is a closed path contained in O. Then

i f(z) dz = O.

Theorem 5.5 (Generalized Cauchy Theorem) Let f be analytic inside of a region 0 bounded by a closed path P. If f is continuous on C and in 0, then

if(z)dz=O.


Example 5.1 Find the following integrals:

a) 11 e'!· cosatdt, a E lW.;

j 1 dt b) _1t2 +Z'


11 e'! . cos at dt = 11 cos t . cos at dt + z 11 sin t . cos at dt.

5.1. BASICS

Since

we obtain

cost· cosat = (cos(t + at) + cos(t - at))/2,

sin t . sin at = (sin( t + at) + sin( t - at)) /2,

11 e't . cos at dt

= ~ (sin(a + l)t + sin(l - a)t) 11 _ z~ (cos(a + l)t + cos(l - a)t) 11

2 a+1 1-a 0 2 a+1 1-a 0

105

= ~ (sin(a + l)t + sin(l - a)) _ ~ (cos(a + 1) + cos(l - a) __ 2_), a E IR. 2 a + 1 1 - a 2 a + 1 1 - a 1 - a2

b) We take 1 A B

t2 + l t - 4 + z4 + t + 4 -z4' Then we obtain

../2 + z../2 -../2 - z../2 A = and B 4 = 4 .

Therefore

11 dt

-1 t2 + Z

= ../2 + z../2 log (t _ ../2 + z ../2) 11 _../2 + z../2 log (t + ../2 _ V2) 11 .

4 2 2 -1 4 2 2 -1

Example 5.2 For which reals a and b do there exist the following integrals

a) 11 cos it dt. ta ' -1

11 ta c) --dt?

o z + t b

Solution. The integral f~l c~sa it dt is improper because of the point O. It con-

verges absolutely far a < 1, what follows by

-- t< -- t<2 -11 1 COS it I d 11 cosh t d 11 dt -1 t rl - -1 Itrll -1 t rl '

where we have used I cos ztl ::::; cosh t < 2 far -1 ::::; t ::::; 1. The last integral converges for a < 1.

We remark that in a quite analogous way we can consider also the general case when a is a complex number using that W I = tRe a. The integral converges in this case for Re a < 1.

100 ta b) The integral --b dt converges absolutely for a - b < -1, what follows

1 z + t from

[00 I~I dt < [00 ta-bdt, or a < -1, 1t z + t J1

smce

I~I ,...., ta as t --t 00. Z + t

For complex numbers a and b using the fact that Ita-bl = tRe(a-b) we obtain the convergence of the integral for Re (a - b) < -1 or Re a < -1.

11 ta c) The integral --b dt converges absolutely for a > -1, since

o z + t

[11~1 dt < [1 t a dt or a - b > -1. Jo z + t Jo

Investigate the convergence of this integral for a, b E !C.

Example 5.3 Find for which complex numbers z the integral 11 e dt exists, and

give abound for the modulus of this integral.

Solution. Since we have

the considered integral converges for Re z > -1. We have for the modulus of the integral

1 [1 t z dtl:::; [1 WI dt = [1 tRez dt = tRez+I 11 = 1 , Jo Jo Jo 1 + Re z 0 1 + Re z

for Rez > -1.


[b [b_ Ja f(t) dt = Ja f(t) dt,

for any integrable function f : [a, b] --t !C, what will imply that a complex function of real variable 7 is integrable on the interval [a, b] if the function f is integrable on [a,b].

5.1. BASICS

Solution. Taking J(t) = u(t) + zv(t), we obtain

t J(t) dt t u(t) dt + z t v(t) dt

= l b u(t) dt - z l b v(t) dt

= l bCuCt)-zvCt))dt

= l b JCt) dt.

Example 5.5 Find the values 0/ the /ollowing integrals

107

(i) fL Re zdz; (ii) fL z2dz; (iii) fL dZ, Jrom the point Zl = 1 to the point Z2 = i z

in the positive direction on the /ollowing curves L :

a} the boundary 0/ the square: 0 :::; x :::; 1, 0 :::; y :::; 1 (Figure 5.1);

-t

-o~--~~--------- .~-.

Figure 5.1

b} the part 0/ the circle: z = e't, 0 :::; t :::; i (Figure 5.2);

~----~-------+ o X

Figure 5.2


c) the straight line segment z = (1 - t) + zt, 0:::;: t :::;: 1, Figure 5.3.

+y

o x Figure 5.3

Solution. (i) a) We have

1 Rezdz = ll.zdt+ 1°Re (z+t)dt=z-ltdt=-1/2+z;

(i) b) We obtain

(i), c) We have

I Re z dz = Z 1,,/2 Re (e,t)e,t dt

r/2 = ZJo cost(cost+zsint)dt

r/2 ["/2 Z Jo cos2 t dt - Jo cos t sin t dt

Z7I" 1

lRezdz = lRe[(l-t)+zt](z-l)dt

= (-1 +z) 11(1-t)dt

(-1+z)/2;

(ii), a) We obtain

(ii), b) We have

[ ["/2 r/2 JL z2dz = Z Jo e2t'et' dt = Z Jo (cos 3t + Z sin 3t) dt = -(1 + z)/3.

5.1. BASICS

(ii), c) We obtain

kZ2dz = k((1-t)+zt)2(-1+Z)

(-1 + z) (l dt - 211 t dt + 2z j1 Ot dt - 2z l t2 dt)

= -(1+z)/3.

109

We remark that in (i) we have obtained different values of the integral on different curves in the contrast to the case (ii) where we have always obtained the same value. Why? The case (iii) can be considered in an analogous way as the previous two ones.

Exercise 5.6 Prove that k zndz = 0, for every n ~ 0 and any closed path L.

Example 5.7 Find the integral k Izlz dz on the path L which consists of the half

circle z = Ret', 0 ~ t ~ 7r, and the straight line segment: -R ~ Rez ~ R, Imz = 0, Figure 5.4.

-R R

Figure 5.4

Solution. We have

k Izlz dz

Example 5.8 Find the integral k z dz on the closed path L wh ich consists 01 the

semi-circle L1 : z = e,t, -7r /2 ~ t ~ 7r /2 and the straight Une segment L2 : z =

zt, -1 ~ t ~ 1, Figure 5.5.

L,

x

-i

Figure 5.5

Solution. We have

Ir 11'/2 1-1 Z dz = e-Itze,t dt + (-zt)z dt = 7rZ.

L -,,/2 1

Example 5.9 Let a lunction 1 be continuous in a neighborhood 01 z = O. Prove that 121'

a) lim I(re't) dt = 27r 1(0); T-+O 0

b) lim f I(z) dz = 27rZ/(0), T-+olL z

where L is the circle Izl = r.

Solution. Since the function 1 is continuous in a neighborhood of z = 0 we have that for any c: > 0 there exists a 8 > 0 such that

I/(r. ed ) - 1(0)1 < 2:' for r < 8(c:), 0 ~ t ~ 27r.

a) Therefore

5.1. BASICS

for r < 6(e), 0::; t::; 211".

b) Analogously as in a), putting z = ret', 0 ::; t ::; 211", we obtain

11 /~z) dz - 211"z/(0)1 = 112,. /~.·e~~') r· ehz dt -12

" /(0) dtl

= I, 102,. (J(r . et') - /(O))dtl

< 102,. I/(r . et') - /(0)1 dt

< - dt e 102,.

211" 0 = e,

for r < c5(e), 0 ~ t ~ 211".

111

Remark. We have a more general result if we assume continuity of the function / in a neighborhood of a point z = w. In this case we have

lim r /(z) dz = 211"/(w), for L: Iz - wl = r. r ..... O}L Z - W

Example 5.10 Let / be a continuous /unction in the region D,

D = {z 10 < Iz - wl < R, 0 ~ arg (z - w) ~ 0 (0 < 0 ~ 211")},

and there exists lim(z - w)/(z) = k. Prove that then liml /(z)dz = zOk, where L z-+w r-+O L

is an arc 0/ the circle Iz - wl = r, which lies in the region D, positively oriented, Figure 5.6.

x Figure 5.6

Solution. The condition lim(z - w)J(z) = k implies that for every c > 0 there z .... w

exists b > 0 such that c (j> I(z - w)J(z) - kl = IrehJ(w + ret.) - kl

for r < b(c), 0::::; t ::::; 271". Therefore we have

liJ(Z)dZ-dJkl = 11° J(w+ret·).et·zdt- z1° kdtl

Iz 1° (J(w + reti)ret• - k) dtl

for r < b(c), 0::::; t ::::; ().

< 1° IJ(w + r· eti)re tl - kl dt

< ~ fO dt () Jo c,

Example 5.11 Let J be a continuous Junction in the region

IJthere exists lim zJ(z) = k Jor z in the region D, then z .... OO

lim f J(z)dz = i()k, R'-+OO JL

where L is the part oJ the circle Izl = R' which lies in the region D.

Solution. The condition lim zJ(z) = k means that for every c > 0 there exists '-00 zED

Ro > 0 such that

~ > IzJ(z) - kl = IR'rt'J(R'rto ) - kl, for R' > Ro, 0::::; t ::::; (). Therefore

1 i J(z) dz - Z()kl 119 J(R' etl)R' et·z dt - Z()kl

< 1° IJ(R'et')R'et• - kl dt

c f9 < (j Jo dt

c,

for R' > Ra, 0 ::::; t ::::; ().

5.1. BASICS 113

Example 5.12 Let J be a continuous function in the region Oz containing a path L connecting the points z(a) and z(b). If a function 9 is a bijection 0/ a path C /rom a region Ow (for z = g( w) on a path Land 9 and g' are continuous on C) prove that then 1 J(z) dz = fc J(g(w))g'(w) dw,

where w(a) = g-l(z(a)) and w(b) = g-l(z(b)); g-l is the inverse /unction 0/ the function g.

Solution. The proof follows by the definition of the integral of a complex function.

Let (Zk), k = 0,1, ... , n, be a partition of the path L. The function w = g-l(Z) maps it on partition (Wk), k = 0,1, ... n, of the path C. Then we have

n-1 n-1

L f(Zk)(Zk+1 - Zk) = L J(g(Wk))(g(Wk+1) - g(Wk)) k=O k=O

We shall show that the last sum tends to zero as n -+ 00 for finer partition. Then the preceding equality will imply the desired conclusion.

Since g' is continuous we have that in every closed region which lies in Ow by uniform continuity for every c > ° there exists 8 > ° such that

Ig(Wk+d - g(Wk) '()I < - 9 Wk C, Wk+1 - Wk

whenever IWk+1 - wkl < 8(c). This is possible always to achieve for any small 8 > 0, since max IWk+1 - wkl-+ ° as n -+ 00, together with IZk+1 - zkl,since

!Zk+1 - Zk! = 19(Wk+1) - g(Wk)I!Wk+1 - Wk!, and exists g'(w). wk+1 - Wk

Let M = max If(z)l. Then zEL

IE J(9(Wk))(g(w~:~~ = ~~Wk) - 9'(Wk)) (Wk+1 - Wk)1

n-1

< € L !f(9(Wk))!!Wk+1 - Wk! :::; €. M . C', k=O

where C' is the length of the path C.

Example 5.13 Let L be a path and L the path which is the image oj L by the junction z f-4 z. Let j be a continuous junction on L. Prove:

a) The junction z f-4 j(z) is continuous on the path Land

[j(z) dz = k j(z) dz.

b) Ij L is a circle z = elt , 0 :::; t :::; 271", positively oriented, then

i j -dz j(z) dz = - j(z) 2"

L L Z

Solution. The function z f-4 j(z) is continuous on L, since for z E L we have z E L, and by the supposition j(z) is continuous on L we have that j(z) is also continuous as composition of continuous functions. We have used that the map z f-4 Z is a bijection and continuous function. This follows by the fact that for every c > 0 there exists 8 > 0 such that

Iz - zol = Iz - zol = Iz - zol < c for Iz - zol < 8 = c.

Now we shall prove the desired equality. We remark that by the preceding result follows the existence of the integral from the right part of the desired equality. We can put w = z in the integral in the right part of the desired equality

k j(z)dz = [j(w)dw = [j(w)dw.

Putting w(t) = z(t) = x(t) + ly(t) and j(z(t)) = u(t) + zv(t), we obtain

[j(z)dz = l b j(z(t))z'(t) dt

= l (u(t) + iv(t))(x'(t) + iy'(t)) dt

= l (u(t) - zv(t))(x'(t) - ly'(t)) dt

= l b(u(t)x'(t) - v(t)y'(t)) dt + l l(v(t)x'(t) + u(t)y'(t)) dt

= l(u(t) +zv(t))(x'(t) +ly'(t))dt

= [j(z)dz.

5.1. BASICS

b) By the proof of a) we have

Example 5.14 Let

1 J(z) dz = 121r J(et')ze" dt

f21r-= - Jo J(et')ze-hdt

121r_-zeh = - J(eh )- dt

o e2t•

/, -dz = - L J(Z)-;2.

and let J be a continuous Junction on Izl ::; r. Prove that

f J(z)dz = lim f J(z)dz. JL n-+oo JL n

Solution. We have

= 1 J(z)dz,

where we have used the continuity of the integration.

Exercise 5.15 Find the integral

on the Jollowing paths:

a) semi-circle Izl = 1, Y 2': 0, VI = 1;

b) semi-circle Izl = 1, Y ::; 0, VI = -1;

c) semi-circle Izi = 1, Y ::; 0, VI = 1.

115


Answers.

a) -2(1 - t); 2) 2(1 - t); c) -2(1 + t).

Exercise 5.16 Prove that Jor lai =j; R we have

Hint. Put in the integral z = Re'8 (0 ~ () ~ 27r) and use the inequalities for the modulus.

Exercise 5.17 Prove:

IJ J is a continuous function in the region {zllzl ~ Ra, Imz ~ a} (a is a fixed real number) and in this region it holds J(z) --+ 0 as z --+ 00, then Jor every positive integer m

lim 1 e,mz J(z)dz = 0, R_oo eR

where CR is the part oJ the circle Izl = R which belongs to the considered region (Figure 5.7).

y

x

Im 2 - Cl

Figure 5.7

5.1. BASICS 117

Hint. To find the bound for modulus of the integral on the semi-circle Izl = R, Im z > 0 use the inequality

• () 2() sm >

- 'Ir

'Ir for 0< () < -, - - 2

and the paths of the circles L l and L 2 which lies in lower the half-plane, see Figure 5.7 . In the case a < 0 use that the lengths of each of them tends to lai as R ~ 00).

Example 5.18 Using the connection between integral on a closed path and the double integral in ]i2 prave

Let J be an analytical Junction in the simple connected region 0 C C with a continuous derivative in this region. Then Jor every closed path L C 0 we have

L J(z)dz = o.

Solution. The proof follows by Green's formula which relates the line integral on a closed path Land the double integral on the region D which is bounded by the path L. Putting J(z) = u(x,y) + w(x,y), we obtain

L J(z) dz = L (u(x, y) dx - v(x, y) dy) + z L (v (x, y) dx + u(x, y) dy),

where dz = dx + zdy.

Since f' is continuous, we have that also the partial derivatives of the functions u(x,y) and v(x,y) are continuous in D. Applying Green's formula we obtain

f J(z) dz = _]. f (8v + 8U) dx dy + z]· f (8u _ 8v) dxdy. JL JD 8x 8y JD 8x 8y

Therefore, by the Cauchy-Riemann equations

we obtain

8u 8v 8x = 8y

and 8u 8v -=--, 8y 8x

L J(z)dz = O.

Remark. The proved statement is known the name of the Cauchy theorem. The condition of the continuity of the first derivative f' is not necessary (Goursattheorem). Pollard proved that even the differentiability on the path L is not necessary


(it is enough to assume only continuity).

Figure 5.8


[00 sin x dx, Jo x

taking for the path for the integration that from the Figure 5.8.

Hint. The given path can be replaced with the following path:

- R ~ z ~ -rj Z = reD., 0 ~ () ~ 71'j r ~ Z ~ Rj 8i and z = Re , 0 ~ () ~ 71',

Figure 5.9.

Figure 5.9

5.1. BASICS 119

. 100 sinx Then, we obtam --dx = 'Ir /2. a x

Find the given integral just on the given path, directly.

Example 5.20 Let J be an analytical Junction on the strip 0 ~ Im z ~ k, and lim J(x + zy) = 0 on this strip. Prove that iJthe integral [00 J(x)dx exists, then

x-+±oo 1-00 also the integral i: J(x + zk)dx exists, and these two integrals coincides.

Solution. By the analycity of the function J we have fL J(z) dz = 0, where L is the path from Figure 5.10.

ki ~----.....(I------..... ---

I

t I t t

--~------~----~---~ -R 0 R X

Figure 5.10

We have

[ J(z)dz = jR J(x)dx+z [k J(R+zy)dy+ r R J(x+zk)dx+z [a J(-R+zy)dy. JL -R Ja JR Jk

Letting R - 00, the second and fourth integrals are zero (by the condition lim J(x + zy) = 0).

x--+±oo

Therefore

Since we have

1 00 J(x)dx= 100 J(x+zk)dx+z1k J(zy)dy,

and the function J is regular on the given strip the last integral is finite. Therefore by the existence of the integral i: J(x) dx, the integral 100 J(x) dx,


it follows the existence of the integral

10''''' J(x+zk)dx.

It can be proved analogously the existence of the integral 1000 J(x + zk) dx.


100 2 Vi b2 e-x cos 2bx dx = -e- , ° 2

iJ it is known that [00 2

i-oo e-t dt = Vi·

Hint. Take the integral of the function J(z) = ez2 on the rectangle:

IRezl ::; R, 0::; Imz ::; b (Figure 5.11).

I , -R ---

Solution. We have

bi

0/ Figure 5.11

-. X R

5.1. BASICS

Since fb 2 fb 2 2) '10 e-(R+.y) dy = z 10 e(R -y e-2R•y dy

we obtain

Iz !ob e-(R+.y)2 dyl < e-R2 !ob ey2 dy < 15M,

as R -+ +00, for arbitrary small 15 > 0, since

/l ey2 dY/ ::::; M, for finite b.

We can prove analogously that

Iz 1° e-(-R+.y)2 dyl-+ 0 as R -+ +00.

Therefore we obtain for R -+ 00

VP 1: e-x2 dx - VP 1: e-(xH)2 dx = 0,

VP 1: e-(x+.bj2 dx = ,fii,

where we have used 100 2

e- t dt = ,fii. -00

Further we have

2 100 2 eb V P -00 e-x (cos 2bx - z sin 2bx) dx = ,fii,

which implies

100 2 2 V P -00 e-x cos 2bx dx = ,fiie-b •

Since

we obtain 1: e-x2 cos 2bx dx = Vii e-b2 .

Since the function e-x2 cos 2bx is even we have

Second method. Using the representation

121


and interchanging the order of sum and integration (why is it possible?) we obtain

Since

t'" 2 10 xe- X X 2n- 1 dx

1 _x2 2n_1 100 2n - 1100 _x2 2n-2 d -e x +-- e x x

2 0 2 0

2n - 1 100 _x2 2n-2 d -- e x x 20'

2n -1 Jn = -2- Jn - 1 for n = 1,2, ... ,

and so J = (2n -l)!!J = (2n -I)!! . V7r

n 2n 0 2n 2 .

Finally, we obtain

[00 2 10 e- X cos 2bx dx

Third method. The desired real integral we can find also using only the real analysis, differentiating the following integral with respect to the parameter z.

100 2

J(z) = -00 e-X coszx dz for z E IR.

5.1. BASICS 123

It is easy to check that the conditions for such differentiation are satisfied. Therefore we have

j'(z) = i:(-xe-X2 )sinzxdx

1 2. 100 z 100 2 = 2"( e-X SIn zx) -00 - 2" -00 eX cos zx dx

z = -2"f(z).

The general solution of the obtained differential equation j'(z) = -~f(z) is

where C is an arbitrary real constant.

Putting z = 0 we have C = f(O) = 1: e-x2 dx = ..;:;r, and so we obtain

Putting z obtain

2b and using that the function under integral is even we finally

e-X cos 2bx dx = _eb • 100 2 y'7i 2

o 2

Example 5.22 Taking the function ez2 , and choosing a convenient path and knowing that

p1'Ove that

(Fresnel integrals).

100 2

-00 e-t dt = ..;:;r,

[00 cos x 2 dx = [00 sin x 2 dx = ~ 10 10 2v2

Solution. We take the following path

124

z = x, 0 ~ x ~ R; (see Figure 5.12).

CHAPTER 5. THE INTEGRAL

y

~~~------~---~ o -- R X

Figure 5.12

By the analicity of the function e- z2 in the region bounded by the path L we have

Therefore we obtain

{R 2 { f 2 2<, (o 2 Jo e-X dx + Jo e-R e zR eh dt + JR e-·t e'7r/4 dt = O.

We have for the second integral

["/4 22·

~ R Jo le- R e t'ldt

["/4 ~ R Jo e- R2 cos2t dt

["/4 ~ R Jo e k(1-4t/7r) dt,

5.1. BASICS

since cos2x ~ 1 - 4x/7r for 0 ~ x ~ 7r/4 (see Figure 5.13).

Therefore

Figure 5.13

I 11r/4 e- R2e2" zRet. dt I < Re-R2 11r/

4 e4R2t/1r dt

< Re-R2 ~(eR2 - 1) 4R2

< ~(1 _ e- R2 ) 4R

- 0,

as R - 00. Letting R _ 00 we get

e-1r/4 100 e-x2 dx - 100 e-·t2 dt = o.

Separating on real and imaginary parts and taking into account that

we finally get

100 -x2d .Ji e x=-o 2 '

100 cos e dt = 100 sin t 2 dt = v;". o 0 2y2

Example 5.23 Knowing that

1, dz = 2n L Z

125

x

on the circle L : z( t) = r . et., 0 ~ t ~ 27r, find a path which will enable us to prove that 1211" ___ d_t __ -,,- _ 27r

o a2 cos2 t + b2 sin2 t - ab·


Solution. We shall choose for the path the following ellipse

C:x=acost, y=bsint, 0~t~2'11".

Using theorem on the equality of integrals on these paths we have

f dz = f dz = 2'11"z, Je z iL z

f21r -a·sint+zbcost f21r sint cost(b2 -a2 ) f21r abdt Jo a cos t + z b sin t dt = Jo a2 cos2 t + b2 sin2 t dt + z Jo a2 cos2 t + b2 sin2 t

2'1TZ.

Taking equal separately the real and imaginary parts, respectively, we find that the imaginary part of the integral on the left side is equal to 2'11". Hence

Example 5.24 Prove the basic theorem oj algebra using the Cauchy theorem.

Solution. Take an arbitrary but fixed polynomial P(z) = Lk=O ak zk. Suppose the contrary, that P(z) f: 0 for every z E c. We introduce the following polynomial Q(z) = Lk=Oak zk. Then also Q(z) f: 0 for every z E C (in the opposite it would be P(z) = 0). We apply Cauchy's theorem on the following regular function

1 j(z) = P(z)Q(z) '

on the path L given on the Figure 5.14.

y

Figure 5.14

5.1. BASICS 127

Therefore r dz JL P(z)Q(z) = 0,

r dz jR dx r Reet dO JL P(z)Q(z) = -R [P(x)]2 + Z Jo P(ReO·)Q(ReOt )·

The second integral tends to zero as R -+ 00 (prove !). Hence

j oo dx -00 (P(x))2 = 0,

which is impossible, since the function under integral is positive for all 00 < x < 00.

Exercise 5.25 Let log z be the principal value oJ the complex logarithm and L ett,O :::; t :::; 27r. Find h log zdz Jor starting points: a) z = 1, b) z = Z, c) z = -1.

Chapter 6

The Analytic functions

6.1 The Power Series Representation

6.1.1 Preliminaries

Theorem 6.1 Let J be an analytic Junction in the disc D(zo, r). Then Jor every a E D{zQ,r)

(i) there exist analytic Junctions in D Fand G such that

F'(z) = J(z) and G'(z) = J{z) - J{a) ; z-a

(ii) Jor every closed path P contained in D(zQ, r)

f J(z) dz = f J(z) - J(a) dz = O. }p }p z - a

Theorem 6.2 Let J be an analytic Junction in D(zo, r), 0 < r < R, and la- zol < r. Then

J(a) = _1 1 J(z) dz, 211"~ er z - a

where er is the circle ZQ + re'B, 0 :::; () :::; 211".

Theorem 6.3 (Cauchy's Integral Formula) Let J be analytic in the disc D(zQ, r), 0 < p< r, and la - zol < p. Then

J(a) = _1 f J(z) dz. 211"z }av(zo,p) z - a

129

130 CHAPTER 6. THE ANALYTIC FUNCTIONS

Theorem 6.4 (Taylor Expansion) If f is an analytic function in D(zo, r), then there exist constants Ck, k E EN u {O}, such that it represented by power series

00

f(z) = L Ck(Z - zo)k k=O

for alt z E D(zo,r). Moreover, f is infinitely differentiable at Zo and

f(k)(zo) 1 1 f(z) Ck = k' = -2 (_ )k+l dz,k E ENu {O}.

• ll'Z er Z Zo

Ifmaxlzl=r If(z)1 = M(r), then the coefficients Ck satisfy Cauchy's inequality

M(r) lekl ~ -n-' r < R.

r

Theorem 6.5 (Uniqueness Theorem) If two functions fand g, analytic in a region D are equal on a subset of D with an accumulation point in D, then fand g are equal on the whole D.

Theorem 6.6 (Liouville's Theorem) A bounded entire function is constant.

Theorem 6.7 If f is an entire lunction and I( z) -+ 00 as z -+ 00, then I is a polynomial.

Theorem 6.8 (Mean Value) Let I be an analytic lunction in a region D and Zo E D. Then

lor D(zo,r) C D.

Theorem 6.9 (Maximum Modulus) III is a non-constant analytic lunction in a region D, then for every z E D and every c > 0 there exists z' E D(z, c) U D such that

I/(z')I> I/(z)l·

Theorem 6.10 (Open mapping theorem) The image of an open set under a nonconstant analytic mapping is an open set.

Theorem 6.11 (Schwarz Lemma) Prove that if I is an analytic function on Izl < 1, I/(z)1 ~ 1 and 1(0) = 0, then If(z) I < Izl for Izl < 1 and 11'(0)1 ~ 1, excluding the case f(z) = e,8z (0 real).

Theorem 6.12 (Morera's Theorem) Let f be continuous on an open set O. il

kl(z)dz=O

whenever R is the boundary of a closed rectangle in 0, then f is analytic on O.

6.1. THE POWER SERIES REPRESENTATION 131


Example 6.1 If exists lim I

W n+ll, n--+cx:> W n

then this limit is the reciprocical value of the radius of convergence of the power series a L:~=o wn(z - W t.

Solution. The power series L:~own(z - wt. absolutely converges by D'Alembert criterion when

( )n+l lim IWn+l z-w I< 1, n~oo wn(z-w)n

Jz - wJ lim I wn+1 1 < l. n----+(X) W n

Therefore we have for the radius of convergence

1 Jz-wJ< r JWn +1 J =R.

Iffin -+oo W n

If we suppose that for sorne z such that Jz - wJ > R the power series converges we obtain a contradiction with the previously proved part.

Exercise 6.2 Find the radius of convergence for the following power series

00 (z wt 00 (z w)n a) L -a ,aE~; b) L -2 ,aE~;

n=l n n=2 In n

00 00 n' c) L(3 + (-ltt(z - wt; d) L -;;-(z - wt·

n=O n=l n

Answers. a) The radius of convergence is 1 for all a E ~.

b) The radius of convergence is 1 for all a E ~.

c) The radius of convergence is 1/4.

d) The radius of convergence is e.

Example 6.3 Find the radius of convergence for the following power series

L w n 00 ( )

n=l n z

with respect to the complex parameter w.

Solution. By D'Alembert's criterion we have R = 1. On the boundary circle Izl = 1 of convergence the power series converges absolutely for Re w > 0, and ordinarily for -1 < Re w ::; O. The result about ordinary convergence follows from the following facts. For z = emS , -7[' < () < 7[', we can write

It Ukl = letS I:( _l)ke.ksl = 11 - (_l)n-le,(n-l lSI < _1_ k=1 k=O eSt/22 cos ~ - I cos ~I

the sum I EI:=1 ukl is bounded for -7[' < () < 7['. Since

converges by Raabe's criterion we obtain the ordinary convergence for -1 < Re w ::;

o of the series E~1 (:)zn.

Example 6.4 Compare the radiuses 01 convergence lor the power series

00 00

Lun(z-wt and LnB.un(z-wt, n=1 n=1

where a is real.

Solution. Let R be the radius of convergence of the power series E~=1 un(z - w)n. Then En~l un(z - w)n

independently of a. We have used that

limsup( vnt = 1 B = 1. n-+oo

Exercise 6.5 Find the radius es 01 convergence 01 the lollowing power series

00 1 1 oon-1 A = L(-; + _)zn and B = L __ zn.

n=l 2 n n=1 n

Then find the radii 01 convergence lor A + Band A - B.

6.1. THE POWER SERIES REPRESENTATION

Answer. We have R(A) = R(B) = R(A + B) = R(A - B) = 1.

. 1 Example 6.6 Takzng r;:,o zn = -1 - prove that

-z

~n(n -1)· ··(n - k+ l)zn-k = (l_k~)k+l' for Izl < 1.

133

Solution. The proof goes by induction. For k = 0 we obtain r;~=ozn = 1/(1- z). Supposing that the desired equality is true for k and differentiating it we obtain

'"' n(n - 1)··. (n - k + 1)(n _ (+1) + 1) . zn-(k+l) = k!(k + 1) L...J (1 _ z)(k+l)+l ' n;?:k+l

the desired equality for k + 1.

Example 6.7 Prove that the radiuses of the following power se ries 00 00

A = ~ un(z - w)"j B = ~ vn(z - w)"j n=O n=O

00 00

c = ~(Un)P(Z - w)"j D = ~ unvn(z - W)", n=O n=O

satisfy the equalities

a) R(G) = RP(A)j b) R(D) = R(A) . R(B).


RtC) = lie.s~p v'lu~1 = li~-!~p ( v'lunlY = (lie.s!P v'lunlY = RP~A)' since lim SUPn_oo v'lun I = RlA)"

b) We have

1 R(D)

= limsup v'lunvnl n_oo

< limsup v'lunl ·lirrisup v'lvnl n-+oo n-+O

1 1 = R(A) . R(B)'

Example 6.8 Let Sn = LUk.

k=O

Prove that if the radius of convergence R(A) of the power se ries A = ~k=O ukzk, is one then the radius of convergence R(C) of the power series C = ~k=O Skzk is also one.

Solution. Since

00 00 001;; 00

L zk . L Uk zk = L L Ui zk = L Sk zk = C, k=O k=O k=O i=O k=O

we have R(C) 2:: 1, since both series which are in the product have radius one. It can not be R(C) > 1. Namely, supposing ISnl = ~k=O IUkl (prove without this supposition!), we have

n+p n+p t> L ISkllzl k 2:: L (n + p - k + 1)lukllzlk for R > Izl > 1

k=n+l k=n+l

and for all p E 11:1 and n 2:: no(c). Hence

n+p n+p L (n + p - k)lukllzl k ::; L (n + p - k + 1)lukllzlk < c (6.1)

k=n+l k=n+l

for R> Izl > 1. Therefore for R > Izl > 1

n~ n~ n~

t> L (n + p - k + 1)lukll z lk = L (n + p - k)lukllzlk + L IUkllzl k k+n+l k=n+l k=n+l

for p E 11:1 and n 2:: no(c). Then by (6.1) we obtain

n+p L IUkllzl k < c

k=n+l

for R > Izl > 1 and all p E 11:1 and n 2:: no, what is impossible, since R(A) = 1. Contradiction. Therefore R( C) = 1.

Exercise 6.9 Let the radius of convergence of the power series ~~=o unzn be R, 0 < R < 00. Find the radiuses of convergence of the following series

00

a) L(2n - l)un zn j n=O


Answers. a) R/2; b) 00; c) O.

Example 6.10 Find two power series A and B such that the radius 0/ convergence 0/ their product AB is strictly greater than their radiuses R(A) and R(B).

Solution. We can take the power series of the following functions

1-z 2-z ..,-----:-;----,- and ..,---...,.-:----,-(5-z)(2-z) (1-z)(3-z)

Exercise 6.11 Let U n = an /or n 2: 1 and V n = bn /or n 2: 1, a f: b, and if: a = (1 - vo)(l - b) and b = (uo - l)(a - b). Prove that then the series A . B, /or

00 00

A = L: unzn and B = L: vnzn , converges /or alt z. n=O n=O


z 00 (tz)2k J(z) = "2 {; 2k k!(k + I)!

is the solution 0/ the ordinary differential equation

Example 6.13 Find apower se ries expansion in a neighborhood 0/ zero /or the /ollowing /unctions and find their radius es 0/ convergences.

a) 1

u f: 0; b) 1

c) 1 --,

z2 - 5z + 6' uz +v (z+l)2'

d) sin2 Z; e) 1 1 + z og--' 1- z'

f) v'ZTI, (0=1)j

g) iZ 2 h) iZ sinz d i)

1

o eZ dz; -- Zj er=z· , o z

j) (log(l - Z))2.

Solution. All functions are analytic at the point z = 0 and can be expanded in apower series at the point z = o.

1 _ 1 1 1 00 n (uz)n 00 n un n

a) uz + v -;;. 1 + uz = ;; I)-l) --;.;- = L:(-1) vn+! z , v n=O n=O

Slnce 1 00

- = L zn, Izl< 1. 1 - z n=O

The radius of convergence is I~I.

b) Put 1 1 A B

-::----:---=- = = -- + --. Z2 - 5z + 6 (z - 3)(z - 2) z - 3 z - 2

Then we obtain A = 1 and B = -1. Therefore we have

1 Z2 - 5z + 6

The radius of convergence is 2.

c)

1

00

= L( -lt+1nzn - 1 n=O 00

= L(-lt(n + l)zn. n=O

The radius of convergence is 1. Why is it possible to exchange the order of differentiation and SUffi ?

d) S' h' 2 1 - cos 2z d h . E h f' . mce we ave sm z = 2 ,an t e senes or t e unctlOn cos 2z IS

00 (2z)2n 1 00 22n 2n cos2z = L(-lt -(-)-" we obtain sin2 z = -2 L(-lt +1-( z)'

n=O 2n . n=O 2n .

The radius of the convergence is 00.

e) Since

00 zn 00 zn log(l + z) = L( -lt+1-, and log(l - z) = - L -,

n=O n n=O n


we obtain

logl+z = log(l+z)-log(l-z) 1-z

The radius of convergence is 1.

137

f) The function j(z) = (z + l)a, a E ~, has power series expansion at the point z=O

the binomial series, with the radius of convergence 1. For a E ~ \ Z, the point z = -1 is the branching point of the function (z + l)G. In our case we have a = 1/2. Since

we want to expand the branch for which y'(O+l) = 1, we obtain

Vz+T = f (!)zn, n=O n

with the radius of the convergence 1.

g) Since

we obtain

Z2n

-" n.

1z 2 1Z 00 z2n 00 1 1z 00 z2n+l eZ dz = L -dz = L - z 2ndz = L --:-;-----,-

o 0 n=O n! n=O n! 0 n=O n!(2n + 1)· The radius of converges is 00. Explain why it was possible to exchange the order of integration and SUffi

h) Since sin z 00 z2n -z- = E(-l t (2n + I)!'

we have

= - z 2n dz 00 ( l)n 1z E(2n+1)! 0

00 z2n+l

= E(-l t (2n + 1)!(2n + 1)


The radius of the convergence is 00.

i) The desired expansion is

1 00 ( n 1 (n -1)) e1-z = 1 + L L k' k _ 1 zn.

n=l k=l .

j) The desired expansion is

00 n 1 (Iog(l - Z))2 = !; (; k(k _ n) zn

= 2 (z2 + (1 + ~) z3 + (1 + ~ + ~) z4 + ... + (1 + ~ + ~ + ... + !..) zn + ... ) 2 23 234 23 n z

Example 6.14 Find the power series expansion at the point z = v for the next functions with their radius of convergence

z a) z + z j c) cos Zj e) log(l+z).

Solution. a) To find the power series of the function f(z) = _z_ at thepoint z+z

z = v (v =I- -z) we put z - v = w. Then we have

f(z) = z

z+z l __ z_

z+z = 1----

w+v +z 00 wn

1- z E(-lt (v + i)n+1

00 n (z-v)n = 1-z E(-l) (v+z)n+l'

The radius of the convergence is Iv + zl. b) First we shall find the power series expansion of the function for _z - at the

z+l point z = v. We have z - v = w (v =I- -1)

z z+l

1 1--

z+l 1

1---w+v+1


Then

00 (n (_l)k (_l)n-k) =,?; E (v + l)k+l (v + l)n-k+l wn

Finally, we obtain

2 1 = 1 - -- + ..,.------,c-

z+l (z+l)2

00 (_l)n n + 1 - 1 - " (2 - -) (z - v)n - ~ (v + 1 )n+1 V + 1 .

The radius of the convergence is Iv + 11. c) Putting z - v = w we obtain

smwsmv 00 w2n 00 W2n+1

cosv E(-l t -( )' - sinv E(-lt ( )' n=O 2n . n=O 2n + 1 .

00 n(z-v)2n .00 n{z-v)2n+1 cos v E( -1) (2n)! - sm v E( -1) (2n + I)! .

The radius of the converges is 00.

e) We have for v -:f -1 :

log(l + z) log(l + v + w)

139


= {W 1 dw 10 1 + v + w

l w 00 wn

= I) _l)n ( )n+l dw o n=l V + 1 00 wn+1

~(_l)n (v + 1)n+1(n + 1) 00 wn ~( l)n-l----;-_:-~ n(v + l)n 00 (z _v)n

= ~ (-1 ) n-l ---'-:-----',:--~ n(v+1)n·

The radius of the converges is Iv + 11.


or more general

Solution. Using the power series representation of the function eZ we obtain

and

Example 6.16 Prove that the Junction J(z) = _z_ can be represented by eZ -1

z ~ Bn n --- L.J -z eZ -1 - n" n=O .

where Bn are the Bernoulli numbers Jor which it holds

( n + 1) (n + 1) (n + 1) Bo = 1, 0 Bo + 1 BI + ... + n Bn = O.

Solution. We shall prove that

1 = ( f: B; zn) . (~) n=O n. z

141

with the prescribed properties of Bn • Then we can easily obtain the desired power series as inverse of (eZ - 1)Jz. The inverse exists since Uo = 1 ~ 0 :

00 n B =EE k n n=Ok=O k!(n + 1 - k)!z

=1.

Comparing the coefficients by the same power z we obtain

B - 1 Bo BI Bn = O. 0- , O!(n+1-0)!+1!(n+1-1)!+···+n!(n+1-n)!

Multiplying the last equality by (n + 1)! we have

( n + 1) (n + 1) (n + 1) o Bo + 1 Bb + ... + n Bn = O.

00

Exercise 6.17 Let J be given by J(z) = E unzn, which converges Jor Izl < p. n=O

a) Prove the inequality

00

Izol = ro < Pi

b) prove that the series

has a radius oJ convergence ~ p - ro;

c) prove that

Example 6.18 Let 9 be a continuous Junction on the path Land let the region 0 be in the region whose boundary is the path L such that inf Iz - wl = p =I 0 Jor z E Land w E O. Prove:

a) J(z) = r g(u) du is an analytic Junction in O. Expand _1_ zn apower ftu-z u-z

series with respect to z - w, w E O.

b) J{n}(w) = n! r g(u) du Jor n ~ 1, w E O. JL (u - w)n+1

Solution. a) The analyticity of the function J follows from

J(z) r g(u) du JL U - Z

J, 1 1 g(u)--· -_-du

L u-w ~ u-w

r g(u). 1 du lL U - w 1 _ z-w

u-w

= r Jt0!l f (~)n du J L U - W n=O U - W

00 r g(u)du n

'foJL (u _ v)n+1 (z - w) .

Explain the exchange of L: and f. b) By a) and the power expansion of the function J at 0 we have

J{n}(w) J, g(u) I = ( ) +1 du for n ~ 1, w E O. n. L u - W n

Example 6.19 Let J be an analytic Junction in Izl < R (R > 1). Starting Jrom the integral


where L is the circle z( t) = eh, 0 S t S 211", prove that

2 r21f t :; Ja f(et')cos2 (2) dt = 2f(0) + 1'(0),

~ 121f f(e ti ) sin2 (~) dt = 2j(0) - 1'(0).

Solution. We have

r (2±(z+~))f(z)dz= r (2±z)f(z)dz+ r j(z) dz. JL Z Z JL z-o JL(z-O)2

Therefore

i (2 ± (z + ~)) f~z) dz = (2f(0) ± j'(0))27rZ. (6.2)

On the other side, putting z = et., 0 S t S 211", we obtain

z+ 1 dz cos t = __ z, dt = -

2 zz

Therefore i (2 + (z + ~)) f~z) dz = z 121f (2 + 2cost)j(etl ) dt.

Therefore i (2 + (z + ~)) f~) dz = 2z 1021f 2 cos2 (~)f(et.) dt. (6.3)

Comparing (6.2) and (6.3) we have

We can analogously prove the second desired equality taking in the starting integral minus sign.

Exercise 6.20 How many different values can have the integral

r dz Je jn(z)'

where j(z) = (z - Zl)(Z - Z2)··· (z - zn), Zi ::I Zj for i ::I j, and the path C does not cross any point zi,i = 1, ... ,n

Answer. Using the Cauchy integral formula we obtain that the desired number is 2n - 1.


. . 1 1 eZdz . Exercise 6.21 Fznd the zntegral -2 ( )3' iJ:

1I'Z e z 1 - z

a) the point z = 0 is inside, and z = 1 is outside 0/ the region whose boundary is Cj

b) the point z = 1 is inside and z = 0 is outside the region whose boundary is Cj

c) the points z = 0 and z = 1 are inside oJ the region whose boundary is C.

Answers. Using the Cauchy integral formula we obtain

a) 1j e

c) 1 --2'

where it is used the representation of ( 1 )3 by simple fractions. z1-z

Exercise 6.22 Let / be an analytic /unction in a region containing the point (0,0) and bounded by the closed path C. Prove that Jor any choice 0/ the branch oJ Ln z we have

-21 . [ j'(z)Ln zdz = J(zo) - J(O), 'n Je

where Zo is the starting point 0/ the path C.

Hint. Use partial integration.

Exercise 6.23 Prove (Cauchy integral /ormulas Jor an unbounded region):

Let C be a closed path which is the boundary 0/ a bounded region D. Let the /unction J be analytic outside 0/ the region D and lim J(z) = A. Then

z""'OO

_1_ [ J(u) du = { - J(z) + a, z outside D 211'z Je u - z A tor z E D

The path has positive orientation with respect to the region D.

Hint. First consider the case A = O.

Exercise 6.24 Find the power senes expansion 0/ the /unction

21- ( 3"1 = -1 +2 zJ3) J(z) = -?z VI

at the point z = 1 and find the radius oJ the convergence oJ the obtained series.


Answer.

-1 ~ zV3 E (!) (z -Ir, with the radius of the convergence 1.

Example 6.25 Prove that the coefficients of the expansion

1 00

---- = E anzn 1 - z - Z2 n=O

satisfy the equality an = an-l + an-2 (n 2:: 2). Find an and the radius of the convergence (an are the Fibonacci numbers).

Solution. To solve the difference equation

we put an = qn. This gives q2 - q - 1 = 0, with the solutions ql,2 = 1 ±2.J5. Then

the general solution is given by

_ C (1 + .J5)n C (1 - .J5)n an - 1 2 + 2 2 '

for arbitrary real constants Cl and C2 •

On the other side, we have

11 00

-z2- z +I = (z-z~) (l-ZLy'K) = ~anzn. Then

= _1 ((1 + .J5)n+l _ (1 _ .J5)n+l) > an .J5 2 2' n - 0,

(6.4)

which is of the form (6.4). The corresponding radius of convergence R is given by

R- l = .J5 + 1. 2

Example 6.26 For the functional series

00 zn F(z)=E--

n=l 1- zn 00

find its power series 2: anzk and find its radius of convergence. k=l

zn Solution. The functions i(z) = -1--' n = 1,2, ... , are analytic functions in - zn

the disc Izl < 1.

The series 00 zn

EI-zn

is uniformly convergent on every closed subset F from Izl < 1. This follows for z E F (Izl = p < 1) by the inequality

since the series

is convergent.

I zn I pn pn -- <--<--1 - zn - 1 - pn - 1 - p'

We find for the function in its apower series in the form

00

in(z) = zn + z2n + z3n + ... = L zkn (n = 1,2, ... ). k=l

We see that the coefficient by zr is 0 if n is not a divisor of rand equal 1, if n is a divisor of r.

We shall now find the coefficients ar for the function F 00 00 00

F(z) = L L zkn = L arzr. n=l k=l r=l

By the preceding reasoning ar is equal to the sum of such a number of 1 which is equal to the number of all divisor of the number r. We denote this number by 7(r). Specially,

7(1) = 1, 7(2) = 2, 7(3) = 2, 7(4) = 3, ...

So we have 00

F(z) = L 7(k)· Zk.

k=l

Since F is an analytic function in the unit disc we obtain that the preceding series converges in the unit disc {z Ilzl < I}.

For z = 1 the obtained series diverges, since it reduces on the series whose numbers are natural numbers

00

L 7(k). k=l

Hence the radius of the convergence of this series is one.

00

Example 6.27 Find Jor the Junction J(z) = e1/(1-z) its power se ries L anzn and n=O

find the radius oJ the convergence oJ this series.

Solution. We shall use the equality e1/(1-z) = e·ez/(1-z). We have the expansions

and so

z 00

w = <p(z) = -- = L zr for Izl < 1, 1 - Z r=1

00 wk J(w) = ew +1 = e· L -, for Izl < 00.

k=O k.

We have for Izl < 1

( 2 3 )k Z+Z +Z + ...

Since

we obtain

k(k+1)···(k+n-1) n!

(k + n - 1)(k + n - 2) .. · (n + 1) (k - 1)!

( n + k -1) k -1 '

(z + Z2 + z3 + .. . )k = f: (n + k - 1)zn+k. n=O k - 1

Replacing the expansion of w = <p( z) in J (w) we obtain

( 1 00 1 00

J(z) = e 1 + lf E zn+1 + 2! E(n + 1)zn+1

1 ~ (n + 2)(n + 3) n+3 1 ~ (n + k - 1) n+k ) +-~ z + ... + - ~ z + ... 3! n=O 2! k! n=O k - 1

e 1 + z + - + - z + - + 2- + - z + ... ( ( 1 1) 2 (1 1 1) 3

1! 2! 1! 2! 3!

( 1 (n - 1) 1 (n - 1) 1 + 1! + 1 2! + 2 3! + ...

( n -1) 1 1) n ) + +-z + ... n-2 (n-1)! n! .

Since the function j is analytic in the unit disc {z I Izl < I} the obtained series converges in this disco The point z = 1 is a singular point of the function j and therefore R = 1.

Example 6.28 Let 00

Lan(z- at j (z) = ~n:,,-::=O,----__ _

L bn(z - at n=O

jor Iz - al < R, be such that in the given disc both series converge an the series in the denominator has no zeros and bo -=J O. Express the coefficients Cn in the expansion

00

j(z) = L Cn(z - at n=O

by the coefficients ak and bk in the disc Iz - al < R.

Solution. By the given conditions the fraction

n=O

is an analytic function in the disc {zl Iz - al < R} and therefore there exists its Taylor series

00

The equality 00

Lan(z-at 00

n:,o = L cn(z - at, L bn(z - at n=O n=O

implies

00 00

n=O n=O 00 n

= E ECkbn-k(Z - at· n=O k=O

By the uniqueness of the expansion in power series in the disc {zllz - al < R} the coeflicients by the same powers of z - aare equal

Cobo = ao, Cob l + clbo = ab Cob2 + Cl bl + C2bo + a2,

This is an infinite system of linear equations with unknown Co, Cl, C2, ... , cn • ••

This system is of special form since for every n (n = 0,1,2, ... ) the first n + 1 equations contain only the first n + 1 unknown Ck. Therefore we obtain (the solution by determinants)

bo 0 0 ao bl bo 0 al

1 b2 bl bo Cn = bgH a2

bn bn - l bn - 2 ... an

where bo 0 0 0 bl bo 0 0

~ bl bo 0 = b~H f:. o.

bn bn - 1 bn - 2 bo

Remark. The result obtained can be useful for finding power series of the fraction oftwo analytic functions, e.g., we obtain for the function J(z) = _z_, Co = 1,

eZ -1

1 1 0 2r

1 1 1 3r 21

Cn = (_1)n 1 1 1 4T 31 2r

1 1 1 (n + 1)! nr (n -I)!

where Bn are the Bernoulli numbers.

0

0

0

1 2r

-B n -" n.

Finally, we have

Z Z ~ B2k 2k eZ -1 = 1 - "2 + ~ (2k)!z .

Example 6.29 Let

00

<p(z) = L an(z - zot n=O

be apower series with the radius of convergence R. Let w be a point fram the boundarg and let Zl be an arbitrary point on the straight line segment [zo, w] different from Zo and w. Prave that the point w is singular for <p if

and it is an ordinary point if

1 ß < ------,===;'=;=== ( ) ,

1. n l<p n (Zl)1 1m sup ,

n~oo n.

Hint. Use Figure 6.1 and the expansions of the function <p at the points Zl and w and the fact that for the point Zl

00

<p(z) = L bn(z - zdn n=O

6.1. THE POWER SERlES REPRESENTATION 151

for n = 0,1,2, ....

Figure 6.1

Example 6.30 (Pringshajm theorem) Prove that if for the power series cp(z) = 00

L: anzn we have R = 1 and an ~ n for n = 0,1,2, ... , then the point z = 1 as a n=O singular point for the function cp.

Solution. Let x be an arbitrary point from the open interval (0,1). If, we suppose that z = 1 is not a singular point of the function cp, then by Exercise 6.29 we have

1 A = 1 -Ix - zol = 1 - x< ,

cp(nl(x) lim sup n __ _

n .... oo n!

Further, let w be an arbitrary point on the unit circle Izl = 1, and Zl a point on the straight line segment [0, w) and which is also on the circle Izl = x. Then

R-lzI-Zol = I-x.

On the other side we have

Icp(nl(zl)1 = lan + n ~ 1 an+1 z1 + (n + l~n + 2) an+2Z~ + ... 1

n+l (n+l)(n+2) 2 :5 an + -1-an+1X + 2 an+2X + ... = cp(nl(x).


Therefore 1 1

---r==;;=:==>-----;==::;:=:::::::= • n Icp(n)(zdl . n Icp(n) (x) I

hm sup , hm sup , n--+oo n. n-+oo n.

By the last inequality and the consequence of the supposition that the desired statement is not true we obtain

Since w was an arbitrary point on the circle we obtain by the last inequality and Exercise 6.29 that every point of the circle Izl = 1 is an ordinary point for cp, which is a contradiction with R = 1. Hence the desired statement is true.

Exercise 6.31 Prove that in Example 6.90:

a) It is enough to suppose an ~ 0 for n ~ no.

b) The point w on the unit circle Izl = 1 is singular if anwn ~ 0 for n ~ no.

Hints. a) Represent the function cp in the form

no-l 00

~ akzk + ~ akzk . k=O k=no

b) Follow the proof for z = 1.

Example 6.32 Let 00 2k

cp(z) = ~ ;k2' k=O

prove that

a) the function cp is continuous and differentiable infinitely many tim es on the disc Izl 5 1;

b) the function cp has no ordinary points on the circle Izl = 1.

Solution. a) We have for Izl 5 1

00 1 00 z22

Since the series L "k2 is convergent we have that the series L 2k2 converges ab-k=O 2 k=O

solutely and uniformlyon the closed unit disc Izl s: 1.

Differentiating the function <p (the series under the surn)

This series converges uniforrnly in the disc Izl s: 1, since

b) Since the coefficients in the series of the function 0 for k = 0,1,2, ... , we obtain by Pringshajrn's theorern-Exarnple 6.30

that the point z = 1 is singular for <p. By Exercise 6.31 we obtain that also singular are the points {j = 20, n is an arbitrary natural nurnber, since

for k 2': n.

Therefore the set of all singular points is dense on the circle Izl = 1. Hence on the unit circle there is no ordinary points of <p.

Remark. It is interesting that the singular points frorn the boundary have no influence on the rnain properties of the considered function, it is continuous and differentiable at the boundary points.

Example 6.33 Expand the function e,mArcsinx in apower series with respect to x.

Solution. The function y = e,mArcsinx satisfies the differential equation

y'/ y = ~m/~, which after squaring and integrating reduces on the equation

The given function forces the following initial conditions y(O) = 1 and y'(O) = ~m. This initial problem has a unique analytic solution in the neighborhood of zero

00

and which has the power series form L anxn. So we obtain the following recursive n=O

formula

(n + 2)(n + l)an +1 - n(n - l)an - nan + m 2 an = 0 (n E NU {O}).

Hence n 2 _ m 2

an+2 = (n + l)(n + 2) an (n E NU {O}). (6.5)

Therefore by the initial conditions we obtain

and k ~m(m2 -1)(m2 - 9)··· (m2 - (2k - 1)2)

a2k+l = (-1) (2k + I)! .

The equality (6.5) implies that the radius of the convergence is 1.

Exercise 6.34 Let f be an analytic function in Izl < R ' . Prove that

1 1 R2 - aii. I a) f(a) = -2 ( )(R2 _/(z) dz, for lai< R < R, and C is a circle 7r~ c z - a - za

Izl=R. b) (Poisson's formula)

for r < R.

Hints.a) Apply the Cauchy integral formula to

lai< R< R'.

b) Put a = ree., 0< r < R, in the equality in a).

Exercise 6.35 Let D be an open disc with the center Zo and radius ro. Let Uk(Z)} be a bounded sequence of of analytic functions on D, which satisfy the condition:

For every fixed n ~ 0 the sequence u1n )(zo)} converges. 00

a) Let fk(Z) = L an.k(z-ZO)'" What are the discs of convergence ofthese series'? n=O

Find the upper bound of an.k which is independent of k.

b) Prave that an.k - an.h ~ 0 for a fixed n, as k and h tend to infinite independent of each other.

c) Using b) prave that

lim !fk(Z) - fh(Z)! = 0 for every Z E D. k.h ..... oo

Example 6.36 We know that E~=o zn = 1/(1 - z) for Izi < 1. Examine the situation re'D ~ e'D as r ~ 1 for () =1= 0, 27r.

Solution. We have for Z =1= 1

lim-1- = _1_. r ..... 1 1 - Z 1 - e,D

On the other side the series E:'o eonD diverges.

Example 6.37 (Tauber theorem) The radius of convergence of apower series E~=o unzn is one and limn ..... oo nUn = O. Prove that

b) if there exists lim., ..... 1 f(x) = A, where f(z) = E~=o unzn for Izl < 1, then the series E:'o U n converges (to the sum A).

Solution. We have for every f > 0 nlunl < c/2 for n > N(c). We fixe one n = no which satisfies the preceding condition. Then we have the estimation for

n=1

m

for m> no:

= :..:..n=-=1'---_ + n=no+1 n=1

m m m

no

Lnlunl n=l (m - no)

< + 2 c: m m

For enough big m > M we have

and so

for m > max(n(c:), M).

b) We have

n=l

m

n=l

m

c: < 2'

< c:,

n = < (1 - x) L IUkl(l + x + ... + X k- 1 ) + L IUkl xk

k=O k=n+l

Let mluml < c:/2 and so by a) m

Lklukl k=O c: l' M( ) m < 2 lor m > c.

Then we obtain by the preceding inequality for n > M(c:)

c c: = < n(l - x)- + - L xk

2 2n k=n+1 C 1 xn +1 - (n(1 - x) + - . -) 2 n 1- x c 1

< 2(n(1- x) + n(l _ x))'

Then for 1 - x = 11n and n > M(c:)

It Uk - f(l - ~)I < c, k=O n

which implies

lim t Uk = lim f (1 - ~) = A. n-+oo n-+oo n

k=O

Remark. Hardy and Litlewood proved more general Tauber theorem, where instead of the condition limn -+oo nUn = 0 they required only boundedness of the sequence {nun}.

Example 6.38 (Schwarz lemma) Prove that if f is an analytic function on Izl < 1, If(z)1 ~ 1 and f(O) = 0, then If(z)1 < Izl for Izl < 1 and 1f'(0)1 ~ 1, excluding the case f(z) = e,9z (0 real).

Solution. By f(O) = 0 it follows the existence of an analytic function on Izl < 1 such that f(z) = zg(z). We have for 0< r < 1

M(r) = max Ig(z)1 = max Ifl(zl)1 ~ ~. Izl=r Izl=r z r

The function M = M (r) is monotone increasing by the maximum principle and so

M(r) ~ limM(t) ~ lim ~ = 1. t-+l t-+l t

Therefore Ig(z)1 = 1. By the principle of the maximum there is no interior point where Ig(z)1 = 1, except g(z) = k and Ikl = 1.


a) For every complex number U

(Un )2 1 1 une"Z - = - --dz for n = 0,1,2,···, n! 2n c n!zn+1

where C is a closed path and the region (C) contains the origin.

b)

Hints.a) Use the Cauchy integral formula for the n-th derivative.

b) Expand the function eU / z in apower series with respect to u and then apply

a) on the function !eu(z+1/z ). z

Exercise 6.40 Let f be analytic on Izl :$ r. Prove that

1 (27f f(O) = 271' Jo f(re,t) dt,

i.e., the value of an analytic function in the center of a disc is equal to the arithmetical mean on the boundary of the circle

1. f(r) + f(rwn ) + ... + f(rw~-l) 1m ,

n-+oo n

Hint. Use in the function under the integral the representation

and the equality

00

f(re lt ) = I: anrne,nt n=O

- zkZ""dt = 1 127f { 0 271' 0 1

for k =I m for k = m.

Exercise 6.41 Let f be analytic on D(O, r) = {z Ilzl :$ r} and f(z) =I 0 for z E D(O, r ). Prove that

If(O) 1 = exp (-d; i 27f In f(re lt ) dt)

Ji.~ .jf(r)f(rwn )··· f(rw~-l),

Hint. Apply on the function Inf(z) Exercises 6.40.

Exercise 6.42 Let Un} be a sequence of analytic functions on the disc D(O, r) which is uniformly bounded, i.e., there exists M > 0 such that If(z)1 < M for n E N, Izl < r. Prove that i f the sequence {fn} converges on a set with accumulation point 0, then

b) the power series Ek:o akzk defines in the disc D(O, r) an analytic function fez);

c) limn _ oo fn(z) = fez) uniformlyon Izl ~ rI·

Hints. a) Apply Schwarz' lemma to the function fn(z) - fn(O). Hence

for Izl ~ rI· Choosing a point ZI f:. 0 in which the sequence {in} converges, such that the right side in the last inequality is enough small we can easily prove that laQ' - aöl = Ifm(O) - fn(O)1 tends to zero as m, n -t 00, limn _ oo aö = ao. Applying the preceding procedure on the sequence {(Jn(z)-aö)/ z} we obtain limn _ oo a]'= al. Continuing this procedure we obtain the desired statement.

b) Use the Cauchy inequality and the Weierstrass criterion for uniform convergence.

c) Since rn-I 00

fn(z) - fez) = L (al: - ak)zk + L (al: - ak)zk, k=O k=rn

we have to estimate each sum on the right side.

Exercise 6.43 (Vitali theorem) Let {in} be a sequence of analytic functions on the region 0, which is uniformly bounded on O. Prave that if {In} converges on the set with an accumulation point in 0, then it converges uniformlyon every compact subset of 0 to an analytic function.

Hint. AppIy the Heine-Borel theorem related to compact sets and reduce to Exereises 6.42.

Exercise 6.44 (Hadamard Three Circle Theorem) Let f be analytic on the annulus rl ~ Izl ~ r3, centered at the origin. let M(r) = sUPlzl=r If(z)l. Ifrl < r2 < r3, then

In M(r2) ~ In r2 -In rl In M(r3) + In r3 - In r2 In M(rI), In r3 - In rl In r3 - In rl

i.e, InM(r) is a convexfunction oflnr.

Hint. Consider the function za J(z) on the annulus A = {z I rl ::; Izl ::; r3}' The maximum of modulus of zaJ(z) in Ais either r~M(rd or r~M(r3)' Choose a such that r~M(rl) = rgM(r3), and put this value in the inequality

Exercise 6.45 The function In M(r) form Example 6.44 is strictly convex function of In r, if the function is not of the form bza, where a and b are real constants.

Hint. Use the preceding Example 6.44 and the fact that za J(z) attends its maximum of modulus on the circle Izl = r2 only if za f(z) is a constant.

Example 6.46 (Montel theorem) Let Un} be a sequence of analytic functions on a region 0, which is uniformly bounded on O. Then there exists of it a subsequence which uniformly converges on every compact subset of O.

Solution. Let {zn} be a sequence from 0 which has an accumulation point in O. Since Ifn(Zl)1 < M,n E N, for some M > 0, there exists a subsequence Unl(i)} of {fn} such that there exists

Now in the same way we extract a subsequence {fn2(i)} of Unl(i)} such that there exists

Continuing this procedure we obtain a sequence of subsequences {fnk(i)} such that Unk(i)} is a subsequence of Unk(i)} and there exists

Then the desired sequence is the diagonal sequence {fnk(k)}'

Exercise 6.47 (Riemann's Mapping Theorem) Let 0 be a simply connected region whose boundary is with minimum two points in the extended complex plane. Let Zo E O. We denote by F the set of all functions f which are analytic and univalent, i.e., J(zt) = f(Z2) implies Zl = Z2, on 0 and If(z)1 < 1 Jor z E 0, f(zo) = 0 and !'(zo) > O. Prove that

a) the set F is non-emptYj

b) sUPfE:!" f'(zo) = M < 00 and there exists in F a function with a maximal derivative at Zoj

c) (Riemann's Mapping Theorem) there exists an analytic Junction J wh ich conJormaly and bijectively maps the region 0 on the unit disc D(O, 1) such that J(zo) = 0 and !'(zo) > 0;

d) the Junction Jrom c) is unique.

Hint.

a) Choose the branch of the function Log (z - a), where a f. 00 is the boundary point, which is uniquely determined on O. Let Wo = log(zo - a). There exists s > 0 such that

Ilog(z - a) - Wo - 27rZI ~ s.

Then the analytic function

J(z) = slg'(zo)l(g(z) - g(zo)) , 1 + sg(zo

where g(z) = (log(z - a) - Wo - 2n)-1, belongs to F.

b) Consider the supremum of numbers !,(zo), J E F, and use Example 6.46. Prove that the limit of the obtained subsequence is the desired function, using Rouche's theorem 8.6.

c) Use the function J from b). To prove that J(O) = {w Ilwl < I} suppose that this is not true, i.e., that there exists e E oj(O) such that lei< 1. Then the function

(p(z) - y'-C)e,argFc h(z) = ,

1 - y'-Cp(z)

where p(z) is one branch of the function

belongs Fand

what is impossible.

~ y~=

J(z) - e

l-cJ(z)'

h'(zo) > sup !,(zo), JEF

d) Use that the only automorphism of the unit disc with J(O) = 0 are given by J(z) = e'oz. This fact follows by Schwarz' lemma.

Exercise 6.48 Let A be a simply connected region and 0 f/. A. Taking Zo E A fix a value Log Zo and put

J(z) = 1.z dw + logzo. zo w

Prove that J is an analytic brunch oJ Log z in A.

6.2 Composite Examples

Example 6.49 Prove:

1. 11 1 is an analytic lunction in the region 0 C C and il this lunction and all its derivatives are zero at the point a E 0, then I(z) = 0, z E O.

2. Let M(R) = sup I/(z)l. 11 1 is an analytic lunction in the entire complex Izl<R

plane and lor fixed n we have

then

M(R) -- -+ 0 as R -+ 00, Rn

a) l(k)(O) = 0 lor k ~ nj

b) the lunction n-l

F(z) = I(z) - 1(0) - z 1'(0) _ ... _ z I(n-l)(o) (n - 1)!

and all its derivatives are zero lor z = O.

c) 1 can be only a polynomial 01 order not greater than n - l.

3. Let 1 be an analytic lunction in 0 and the circle C with mdius Rand center a E 0 lies in O.

a) Prove that lor all z which are inside 01 C and z =F a

I(z) = I(a) - (z _ a)f'(a) _ ... _ (z - a)n-l I(n-l)(a) (n - 1)!

(z - a)n { I(u)du - 2n Jc (u - a)n(u - z)"

b) 11 1 has a zero 01 order n in the point a and

1 (z - a)n { I(u)du (z) = 27rz Jc (u - a)n(u - z)'

then ( Iz - al)n R

I/(z)l:::; ~ R-Iz-al M,

where M = sup I/(z)l. zEC

4. 111 has zero 01 order n prove that P(z) has zero 01 order n· k and that

I/(z)1 :::; M(lz - al)n .k! R . R V R-Iz-al

6.2. COMPOSITE EXAMPLES 163

Solution.

1. We shall expand the function j in a neighborhood of the point a in apower series (what is possible since j is an analytic function)

00 j(n)() j(z) = L _,_a (z - at, (Iz - al < r).

n=O n.

Since by the suppositions

j(z) = j'(a) = '" = j(n)(a) = 0

we have j(z) = 0 for Iz - al < r, where r > 0 and the circle C(a, r) is in the region O. Therefore the functions j and O(z) = 0 are equal on the whole region 0 since both are analytic on 0 and equal on C(a, r).

2. a) Let r < R. Since j is analytic in the entire complex plane we have

Therefore

Ij(n)(O)1 < n! r2" Ij(r eBi)1 dB< M(R) n! = M(R) n!, (6.6) - 271" Jo rn - rn (R - 8)n

where r < Rand R = r+8 (8 > 0). Since

M(R) (R - 8)n

M(R)

Rn - (7) Rn-18 + ... + (-8)n

MAl!) 1 - (~) R + ... + 7r {j ( 8)n,

M(R) M(R) .. we have for R ........ 00 by ~ ........ 0 that also (R _ 8)n ........ O. Then by the mequahty

(6.6) we have j(n)(o) = O.

Since for k ;::: n we have

M(R) = M(R) _1_ ........ 0 Rk Rn Rk-n

as R ........ 00, we obtain in an analogous way as before for n that j(k)(O) = 0 for k ;::: n.


2. b) and c) By a) we have

n-l fez) = f(O) + z 1'(0) + ... + z f(n-l)(o).

(n -1)!

Hence

F(O) = F'(O) = ... = 0

and the function f is a polynomial of the order not greater than n - 1.

3. a) We have for z =I- a :

{ f(u)du Je (u - a)n(u - z)

~ (z-a)n { ~ du

2n Je u - z (z - at fez)

(z - a)n = fez).

We have on the other side for z =I- a :

{ f(u)du Je (u-a)n(u-z)

~ = (z-a)n { ~ du

27r~ Je u - z

= (z - a)n (f(U) ) (n-l)

(n - 1)! u - z u==a

( )n-l f(a) - (z - a) f'(a) _ ... _ z - a f(n-l)(a).

(n - 1)!

So we have proved the both equalities.

3. b) Since the function f has a zero of order n at the point a we have in a neighborhood of a that

fez) = (z - at h(z),

where h is an analytic function and h(a) =I- O. Then

hz _ _ 1_ { f(u) du ( ) - 27r~ Je (u - a)n(u - zr

Further we have

l1(z)1 < Iz - al n f If(u)lldul - 27l' Jc lu - aln lu - zl

Mlz - al n f21r IzR eO'dBI :S 27l' Ja la+Reo'-alnla+Reo'-zl

_ (Iz-al)n R M - R R-Iz - al '

where M = sup If(z)l. zEC

4. If f has a zero in a of order n, then f(z) = (z - a)n h(z). Therefore

where hk is an analytic function and hk(a) i=- 0, what follows by the same properties of h. The last equality implies that fk has zero at a of the order nk.

Applying the inequality from 3. b) on the function r we obtain

Since Ir(z)1 = 11(z W, taking the k- root ofthe last inequality we obtain the desired inequality.

Example 6.50 I 1. Prove that the integral

fc Idzl

is the length 01 the path C.

I 2. Let f be an analytic function in the region 0 which contains the disc Izl :S R. Suppose that 1 maps the circle Izl = R onto bijectively the path L. Prove that the length L 01 the path L is given by

II 1. Suppose that the sequence 01 complex numbers {un } converges to u. Prove that the sequence {u n } converges to U.

00

II 2. Let the function 9 has apower senes representation g( z) = 2: Un zn, n=O

which converges for Izl < r. 112. a) Prove that for r < R

II 2. b) Find for which r the function is J(r) is continuous. Explain.

II 3. Let M(r) = max Ig(z)l, where 9 is a function given in II 2. Izl=r

II 3. a) Prove that IUnl S; M(r) for r < R, where M(r) = max Ig(z)l. r n Izl=r

II 3. b) If for sorne n = k is IUkl = M~r), then g(z) = Uk zk (for that proof use r

II 2. a)).

Solution. I 1. Let C: z(t) = x(t) + ly(t), t 1 S; t S; t 2 • Then

fc Idzl = l21(x'(t) + l Y'(t))ldt = l2 V(x'(t))2 + (y'(t))2 dt = C

where C is the length of the path C.

I 2. By the given conditions, w = f(R eO.), 0 S; B < 27l", gives the path L. By the analyticity of f in the region 0 we have

By I 1, we have - f f2~ L = JL Idwl = R Jo 1f'(R eBi)ldB.

(f maps bijectively the circle Izl = R on the path L). Cauchy's formula implies

F(O) = _1 f F(z) dz. 27l"z Jlzl=R Z

Taking F(z) = J'(z) we obtain

1'(0) = _1_ f2tr f'(R eO.) lR eO. dB = ~ f2tr f'(R eO')dB. 2n Jo R eO. 27l" Jo

Since

finally we obtain

11 1. Let c > O. Then by the convergence of the sequence {un } there exists no E N such that JUn - uJ < c for n ~ no. Then it follows c > JUn - uJ = JUn - uJ = JUn - uJ for n ~ no(c).

00

11 2. a) Since g(z) = 2: unzn for JzJ < R, we have

Then

n=O 00

g(r e9.) = 2: unrnenO, for r < R. n=O

Jg(r eO·W = g(r eO') g(r eO,) 00 00

= 2: unrnenO, 2: unrne-nO, n=O n=O

00 00

= 2: 2: Un_krn-ke(n-k)O·ukrke-kO. n=O k=O

Applying the integral with respect to () we obtain

J(r) = ~ [2" Jg(r eO'Wd() 27l" Jo 1 2,,0000

= - [ (2: 2: Un_kukrne(n-2k)O,) d(). 27l" Jo n=O k=O

We can exchange the order of integration and summing since the series under integral converges in disc JzJ ~ r < R.

Then

sm ce

00

= 2: JU n J2 r 2n ,

n=O

for n = 2k for n =I- 2k.

11 2. b) The function J(r) is continuous for r < R, since for such real numbers r, the function J(r) has a convergent power series representation.

00

11 3. a) We have for the coefficients Un of g(z) = L unzn n=O

U n = ~ f g~:~ dz for r < R. 21Tt Jlzl=r Z

Hence 1 f21r M(r) D, M(r)

IUnl $ 21T Jo rn+Ile(n+I}D'1 r I z e I d() = --;:;;-.

11 3. b) The equality

implies

J.- f21r M(r?d() ~ f: lunl2r2n . 21T Jo n=O

Since for some n = k we have IUkl = M~r) , we obtain r

k-l 00

(M(r) f ~ L lunl2r2n + IUkl2r2k + L lunl2r2n n=O

~l 00

= L lunl2r2n + (M(r)? + L lunl2r2n . n=O n=k+l

Since r > 0, the preceding inequality reduces on equality for IUnl = 0, n i:- k. This implies g(z} = Uk zk.

Example 6.51 I. Let f an analytic on the disc Izl $ Rand on the circle Izl = R we have

M If(z)1 $ TYf for z = x + zy.

I 1. Prove that on the disc Izl $ R

I 2 Using I 1. prove that on the disc Izl < R/2 we have

If(z)1 $ ~ ~

6.2. COMPOSITE EXAMPLES

I 3. Let f be an entire function and

M(r) = sup Iy f(z)l· Izl=T

Using I 2. prove that if f is not a polynomial then

M(r) -- -+ +00 as r -+ +00,

rP

for an arbitrary real number p ~ 1.

Hint. Suppose the contrary and then use the Cauchy inequality.

II Suppose that the function 9 is analytic for Izl < R.

II 1. Prove that

1 1 R2 - aa g(a) = -2 ( )(R2 _) g(z) dz

1I"Z C Z - a - za

lai< R < R' and C is the circle Izl = R.

II 2. Prove that for r < R

1 1211" (R2 - r 2)g(R eICI') g(r eS') = - ----'------'-'('-'---)-'--- dcp.

211" 0 R2 - 2Rr cos () - cp + r 2

Solution.

I 1. We have for z = R e's = x + zy

Therefore by the given conditions on f on Izl = R

Then by the maximum principle

I(Z2 - R 2)r(z)1 :::; 2 M R on the disc Izl:::; R.

I 2. We have for Izl :::; R/2 that

Then by I 1. we have

Iz2 _ R2 1 ~ 3:2•

2MR 8 M If(z)l:::; ~ = 3· ii"

3 4

169

I 3. Suppose that M(r) does not converges to +00 as r ---. +00. Then there r P

exists a sequence {rn} such that r n ---. +00 as n ---. +00 and

for some C > o.

M(rn ) < C (n E w), r~ -

Hence by I 2. we have for R = r n

Ij(z)1 ~ 8~ r~-l on Izl ~ T n /2.

00

Take j(z) = L: amzm (J is an analytic function on the disc Izl ~ R). The last m=O

inequality and the inequality on Izl ~ ~ r n imply

8C la I< - r P- m - 1 2m (m nE W) m - 3 n ,.

Letting n ---. +00 we obtain am = 0 for m ;::: p. This shows that j is a polynomial (of the order less or equal then p - 1).

11 1. Use the definition of the integral and in 11 2. put a = res •.

Chapter 7

Isolated Singularities

7.1 Singularities

7.1.1 Preliminaries

A function 1 has an isolated singularity at Zo if 1 is analytic in {z I 0 0 hut is not analytic at zoo The isolated singularities of 1 are classified in the following way.

(i) If there exists an analytic function at Zo, denoted by g, such that I(z) = g(z) for all z E {z 10 < Iz - zol < r} for some r > 0, we call Zo a removable singularity of f.

(ii) If 1 can be written in the form

I() =g(z) z h(z)

for z =I zo, where 9 and h are analytic functions at Zo, g(zo) =I 0 and h(zo) =I 0, we call Zo a pole of f. If h has zero of order n at Zo, we call Zo a pole of order n of f.

(iii) If 1 has neither a removable singularity nor a pole at Zo, we call Zo an essential singularity of f.

Theorem 7.1 I1 a lunction 1 has an isolated singularity at Zo and

lim(z - zo)l(z) = 0, z-+zo

then the singularity Zo is removable.

171

172 CHAPTER 7. ISOLATED SINGULARITIES

Theorem 7.2 11 1 is analytic in {z I 0 < Iz - Zo I < r} lor some r > 0, and there exists n E N such that

lim (z - zot I(z) f. 0 and lim (z - zo)n+1 I(z) = 0, z~zo z--+zo

then the singularity Zo 01 1 is a pole 01 order n.

Theorem 7.3 11 1 has an essential singularity at Zo, then the range

{f(z) I z E {z I 0 < Iz - zol < r}

for r > 0, is dense in the complex plane.

We eall 1 meromorphic in a domain A if 1 is analytic there exeept at isolated poles.


Example 7.1 Find the singular points in the extended complex plane (with z = 00)

of the following lunctions

1 a) Z-Z3;

Z5

c) (1 _ z)Z' eZ

d) 1 + zZ'

Solution. a) z = 0, z = 1 and z = -1 are poles of first order, z 00 IS

removable singular point.

b) z = l~Z and z = -~ Z are poles offirst order, z = 00 is removable singular

point.

e) z = 1 is pole of first order, z = 00 is pole of third order.

d) z = Z, Z = -1 are poles of first order, z = 00 is essential singular point.

Example 7.2 Find the singular points in the extended complex plane (with Z = 00) of the following functions

1 1 b) ----.

eZ - 1 z'

Solution. a) z = 00 is an essential singular point.

1

e·-1

e) eZ _ l'

b) z = 2k1ri, k = ±1, ±2, ... , are poles of first order, z = 00 is a limit point of poles.

7.1. SINGULARITIES 173

c) Z = (2k + 1)7l"i, k = 0, ±1, ±2, ... are poles of first order, Z = 00 is a limit point of poles.

d) Z = 0 is a essential singular point, Z = 00 is a removable singular point.

e) Z = 1 is an essential singular point Z = 2k7l"i, k = 0, ±1, ±2, ... , are poles of first order, Z = 00 is a limit point of poles.

Example 7.3 Find the singular points in the extended complex plane (with Z = 00) of the following funciions

a) 1 . , b)

smz cos Z

Z2 ' c) tan Z; d)

1 sm--'

1- z'

Solution. a) z = k7l", k = 0, ±1, ±2, ... , are poles of first order, z = 00 IS a limit point of poles.

b) z = 0 is a pole of second order, z = 00 is an essential singular point.

c) z = (2k + 1)7l" /2, k = 0, ±1, ±2, ... , are poles of first order, z = 00 is a limit point of poles.

d) z = 1 is an essential singular point, z = 00 is a removable singular point.

2 e) z = (2k + 1 )7l"' k = 0, + 1, +2, ... , are essential singular points, z = 0 is the

limit of essential singular points, Z = 00 is the removable singular point.

Example 7.4 Prove that if apower series has the radius of convergence one and on the unit circle it can have of singular points only poles of first order, then the sequence of coefficients of the power series is bounded.

00

Solution. The power series L an zn which satisfies the described condit.ions n=O

has the following representation

~ n b1 bp ~ n L..J anz = --- + ... + + L..J cnz , n=O 1 - ZIZ 1 -zpz n=O

where

IZil = 1 (i = 1, ... ,p) and limsup ~ < 1. n-+oo

Therefore

an = blZ~ + ... + bpz; + Cn ·

The last equality enables to prove easily the boundedness of the sequence {an}.


Example 7.5 Let J be an analytic Junction in annulus

A = {z 11' ::; Izl ::; R}.

Prove that the surJace area 0/ the image oJ the annulus by the /unction J(z) = 00

L anzn (counting the parts 0/ the regions as many timesas they are covered) is n=-oo

given by 00

7r L nlanl2(R2n - r 2n ). n=-oo

Solution. Let J(z) = u + zv for z = x + zy. Then the desired area is

where z = pe,a. Since

18(u'v)1 = 1J'(z)12 , 8(x, y)

we have

i R f21r 1 . 12 r Jo J'(pe la

) p dp da 27r iR ( f n2IanI2p2n-l)dp

n=-(X)

00

n=-oo

Example 7.6 Prove that i/ the /unction J has zero 0/ order k, k 2: 2 at z = w, then l' has at z = w zero 0/ the order k - 1.

Solution. The function / has in the neighborhood of w the representation J(z) = (z - w)kg(z), k 2: 2, where 9 is an analytic function in the neighborhood of wand g(w) :I O.

Then J'(z) = (z - w)k-l(k g(z) + (z - w) g'(z)),

which implies the desired statement.

Example 7.7 Prove that iJ the Junction / has pole 0/ the order k at z = w, then l' has pole 0/ the order k + 1 at z = w.

7.1. SINGULARITIES 175

Solution. The function 1 has in the neighborhood of w the representation

I(z) = ( g(z) )k' where 9 is an analytic function in the neighborhood of z = w and z-w

g(w) ::I O.

Then

J'(z) = (z _ ~)k+l (-k g(z) - g'(z) (z - w)),

which implies the desired statement.

Example 7.8 Prove that ill is an entire lunction, then all its derivatives and the

integrals on arbitrary path L i I( U )du are entire lunctions.

Solution. Since the function 1 has apower series representation with infinite radius of convergence the corresponding power series of derivatives and integrals have also infinite radius of convergence.

Example 7.9 Prove that ill is a meromorphic lunction on the region 0, then I' is also meromorphic on 0 and has the same poles as f.

Solution. The function 1 has the following representations in the neighborhoods of its poles Ul, U2, • .• , U n

j(z) = ( 9k(Z\ ' k = 1,2, ... ,n, z - Uk Pk

where gk are regular functions and gk( Uk) ::I 0 for k = 1,2, ... , n, and Pk E N is the order of pole.

Then

j'(z) = (z _ :k)Pk+1 ( - Pk gk(Z) - (z - Uk) g'(z»)

in the neighborhood Uk for k = 1,2, ... , n, which implies the desired statement.

Example 7.10 Let 1 be an analytic lunction, which is not dejined at the point z = wand which the is analytic in the neighborhood 01 w. Prove that a sufficient and necessary condition that the lunction 1 has a removable singularity at z = w isthat it is bounded in so me neighborhood 01 the point w.

Solution. If the function 1 has a removable singularity at the point z = w, then it has apower series representation in a disc Iz - wl < R. Hence for some M > 0

I/(z)1 ~ M (Iz - wJ < R).

Suppose now that J is a function such that IJ(z)1 ::; M for Iz - wl < R. Then J has no singular point at z = w, since in the opposite it would be unbounded in the neighborhood of this point.

Example 7.11 Prove that an entire Junction which at the point z = 00 has a pole oJ order k is a polynomial oJ the order k.

Solution. Since the function J is entire it has apower series representation 00

J(z) = ~ unzn which converges for every z E !C. n=O

On the other side the function J has at z = 00, a pole of the order k, which implies that the function

J( ~) = t u: z n=O Z

will have a pole of the order k at z = O. Then by the definition of pole we have U n = 0 for n = k + 1, k + 2, ...

Therefore k

J(z) = ~ U n zn. n=O

Example 7.12 Let J be an entire Junction and let there exist a natural number n and real numbers M > 0 and R > 0 such that

IJ(z)1 :s M Izln Jor Izl;:::: R.

Prove that J is a polynomial 0/ the order not greater than n.

Solution. We have for the entire function J

_ J(k)(z) _ 1 1, J(u) Uk - -k-' - - -2 k+l du,

• 'lrZ L U

where L is the circle z = R et., 0 ::; t :s 2 'Ir •

Then IJ(u)1 :s M luln , for lul ;:::: R implies

1/(k)(Z) I MR"-k

IUkl = k!::; In - kl .

For k > n the right hand side can be managed small enough . Therefore Uk = 0 00

for k = n + 1, n + 2, .... Hence in the representation J(z) = ~ Uk zk we have k=O

Uk i- 0 for k ::; n.

7.2. LAURENT SERIES 177

Example 7.13 Find the order of the zero z = 0 for the following functions

Answer.

a) Fourth. b) 15th.

Example 7.14 Find the zeros and their orders for the following functions

Z2 + 9 b)

4 ' Z c) zsinz.

Answers.

a) z = ±3z are zeros of first order.

b) z = ±3z are zeros of first order, z = 00 is a zero of second order.

c) z = 0 is zero of second order, z = k-rr, k = ±1, ±2, ... are zeros of first order.

Example 7.15 Find the zeroes and their orders for the following functions

Answers.

a) z = ±7r are zeros of third order, z = k7r, k = 0, ±2, ... , are zeros of first order.

b) It has no zeros.

c) z = 0 is zero of third order, z

±1, ±2, ... , are zeros of first order.

{jh and z

7.2 Laurent series

7.2.1 Preliminaries

If both series 00 00

LZn and LZ-n n=O n=l

~ ifh (1 ± zV3), k

00

converge, then we define L Zn as -00

00 00 00

LZn = LZ-n+ LZn. -00 n=1 n=O

Theorem 7.4 We have that L:~oo anzn converges in the domain

A = {z 1 r < Izl, Izi < R},

where 1 1

R = 1. 1 11/ and r = 1· 1 11/. 1m SUPn-+oo an n 1m SUPn-+oo a_n n

Moreover, ij r < R, then A is an annulus and J given by

-00

is an anytic junction in A.

Theorem 7.5 IJ J is analytic in the annulus A = {z 1 r < Izi < R}, then J has a (unique) Laurent expansion

n=oo

J(z) = L anzn -00

in A.

Theorem 7.6 IJ J has an isolated singularity at Zo, then Jor some c > ° we have

00

J(z) = L an(z - zo)" n=-oo

on A = {z 10< Izl < c}, where

1 f J(z) d an = 27l"z Je (z _ zo)n+I z

and C = C(O, R), ° < R < c.

Characterization of singu1arities by Laurent expansion:

(i) If j has a removab1e singu1arity at zo, then all coefficients an, n < 0, of its Laurent expansion about Zo are zero.

(ii) If f has a pole of order k at Zo, then all coeflicients an, n < -k, of its Laurent expansion about Zo are zero and a-k -=I- O.

(iii) If f has an essential singularity at Zo, then infinitely many coefficients an, n < 0, of its Laurent expansion about Zo are nonzero.

Theorem 7.7 (Mittag-Leffier) Let {zn} be a sequence 0/ distinct complex numbers such that IZnl --+ 00. If {Pn } is a sequence of polynomials without constant term, then there exists a meromorphic function f whose only poles are at {zn} with

All such functions f have the following representation

00 1 f(z) = L (Pn(--) - P:(z)) + e(z),

n=O Z - Zn

where P;:' is some polynomial, and e is an entire function. The series converges absolutely and uniformlyon every compact set not containing the poles.


Example 7.16 Expand in the Laurent senes the following functions in the annulus where they satisfies Laurent theorem:

a) 1

b) 2z~

Z2 - 3z + 2' (1-z)(Z+t)2'

c) 1

d) 1

(zz+2)n' sm-j

z

e) 1

sinz sin-, z

first in a neighborhood of zero and then zn a neighborhood of an arbitrary point wEC,

Solution. a) Since

1 1 1 ---- = -- - --, Z2 - 3z + 2 z - 2 z - 1

we obtain for Izl < 1

1 1 1 1

Z2 - 3z + 2

which is the desired expansion in the neighborhood of the point w = o. For 1 < Izl < 2 we have

1

Z2 - 3z + 2 1 = 1 1 1 1 --L:-._--.-z z 1- 1 2 1-!

n=O z 2

the expansion in w = o. For Izl > 2 we have

1 1 1 1 1 Z2 - 3z + 2 -;·--1 +;·--2

1-- 1--z z

1 f= In + ~ f= (~r z n=O Z Z n=O Z

-1 1 = - L: (2n+I -1)zn,

n=-oo

expansion in the neighborhood of the point w = o. Case I: For arbitrary w (w i- 1,2) we have for

Iz - wl < min(lw - 11, Iw - 21),

Figure 7.1,

0 ~ 0/ 2 X

Figure 7.1

1 1 1 = -----Z2 - 3z + 2 1-z 2-z

1 1 1 1 1-w·1-z -2-w· 2-z

l-w 2-w 1 1 1 1

-----,- - -- . ---1-w 1_ z - w 2-w 1_ z - w

l-w 2-w =

1 t(Z-Wr 1 t (Z-Wr 1 - w n=O 1 - w 2 - w n=O 2 - w

=

00 (1 1) = E (1- w)n+1 - (2 _ w)n+1 (z - wt·

for min(lw -11, Iw - 21) < Iz - wl < max(lw -11, Iw - 21).

Case 11: For

min(lw -11, Iw - 21) < Iz - wl < max(lw -11, Iw - 21).

Figure 7.2

Find the expansions for cases Iw - 21 < Iz - wl < Iw -11 and Iw -11 < Iz - wl < Iw - 21, Figure 7.2 .

Case III: For Iz - wl > max(lw - 11, Iw - 21),

Figure 7.3,

Figure 7.3

we have 111

Z2 - 3z + 2 = - z - 1 + z - 2

1 1 1 1 = --_. +-_._.....".--

z-w l_l-w z-w 1_ 2 - w z-w z-w

= - z ~ w f: (~ = :r + z ~ w f: (! = :r n=O n=O

-1 1 1 = n200 ( - (1 - W)n-1 + (2 _ w)n-J . (Z - wt·

Find the expansions for w = 1 and w = 2.

b) The consideration is analogous to the previous example a). For arbitrary w (w =11, -z) the regions are:

Case I: Iz - wl < min(lw - 11, Iw + zl)j

Case 11: min(lw -11, Iw + zl) < Iz - wl < max(lw -11, Iw + zl)j

Case 111: Iz - wl > max(lw - 11, Iw + zl).

Find the expansions for w = 1 and w = -z.

c) We have for n = p :

dP ( 1) ( -1 )PzPp! dzp zz + 2 = (zz + 2)P+1 .

For Izl < 2 we have

Therefore

dP 1 -(-) -dzp zz + 2 -

Comparing the obtained results for n = p, we have

1 ( -l)P 00 (k + p) (k + p - 1) ... (k + 1) . zk

(zz + 2)P+1 = zpp! E 2k+p+1zk+p ,

for Izl < 2.

We have for Izl > 2


Using this result we obtain

-1 k(k-1) ... (k-p+1)zk- P

= - E 2k-1zk k=-oo

Finally for n = p we have

1 _ (-1)1'+1 -EI (k+p)(k+p-1)· .. (k+1)zk (zz + 2)1'+1 - Zpp! k=-oo 2k+p+1 zk+p ,

for Izl > 2.

Find the Laurent expansion at the point w (w ~ 2z).

Example 7.17 Prove that the principal value 0/ log _z - satisfies the Laurent thez-l

orem in the region Izl > 1 (prove that log _z_ is analytic in this region and on any z-l

path which lies in this region we stay always on the same branch) and expand it in Laurent series in w = o.

Solution. The function log _z_ is regular in the region Izl > 1, since it can z-l

be expand in power series at an arbitrary point w (Iwl > 1) :

z log-

z-l log z - log( z - 1)

z-w z-w logw(l + --) -log(w -1)(1 + --)

w w-1

I w ~ (-l t (z-w)n ~ (_1)n(Z-w)n og--+ L.,;-- -- - L.,; -- --w - 1 n=1 n w n=1 n w - 1

w 00 (_1)n 1 1 log-+" -(-- )(z-wt. w - 1 L.,; n w R (w - l)n n=1

The branching points of the function

z log -- = log z -log(z - 1)

z-l

are z = 0 and z = 1. Since they are outside of the considered region Izl > 1, we stay always on the principal value going through any path in this region.

The Laurent expansion is obtained by

for Izl > 1.

z log-

z-l z-l

-log-z

-log (1 - ~) 00 (l)n 1 -L - -

n=1 Z n -1 n

L~ n=-oo n

Example 7.18 Prave that on both circles which are boundaries ofthe annulus where the series

n=-oo

converges is surely a singular point of the function which is represented by this series.

Solution. We have for r < Iz - ul < Rj

00 -1 00

L un(z - ut = L un(z - ut + L un(z - ut n=-oo

1 where we have put w = --.

z-u

n=-oo n=O

1 00

= ~U-n(z_u)n + ~ un(z-ut 00

00 00

= L u_nwn + L un(z - u)n n=1 n=O

The desired statement now follows from the property of the power series that on the boundary circle of the convergence there is always a singular point.

Example 7.19 Find by the definition the coefficients in the Laurent series of the function

the point u.

f(z) = cos z (z - U)2

Solution. Using the formula

- 1 i f(w) d U n - - w, n = 0,±1,±2, ... 271"Z L(W-U)n+1

on the path L : z(t) = U + reh, 0 ~ t ~ 271", we obtain the Laurent series

00 ()2n cos Z _ '" (_l)n+I z - U r I I L...J lor z - U > O. (z-u)2 - n=-I (2n+2)!'

Exercise 7.20 Find the Laurent series of the function ~ eI/z2 in the region Izl > O. z

Answer.

~ eI/z2 = ~ f: (~)n 1 f: 1 for Izl > O. z Z n=O Z2 n! = n=O z2n+I n!'

Example 7.21 Let

00

for a E ~. Find the Laurent series ~ UkZ k of the function f using the following k=-oo

steps.

a) Find the annulus of convergence of this series;

b) Prove that Uk = U-k, k ~ 1;

1 121< c) Prove that Uk = - eacostcos ktdt,

271" 0 k ~ 0;

d) Prove that

(k + 2p)! forn=k+2p, pENU{O}

p!(k+p)!

otherwise

e) Expand the coefficient Uk in apower series with respect to a.

Solution. a) The desired annulus is the whole complex plane without zero and z = 00;

1 00

b) Putting z = - in ~ Ukzk, we obtain w k=-oo

00 1 a 1 ~ Uk k = exp ( - ( w + - ) ) .

k=-oo W 2 w


Comparing these two expansions (using the uniqueness of the Laurent expansion) we obtain

00 00

E 'Uk zk = E 'U-k w k , k=-oo k=-oo

'Uk = 'U_k, for k 2: 1.

c) We have by the definition

'Uk = 2~t k (w ~(:;k+1 dw, k = 0, ±1, ±2, ....

Let 'U = O. We take for the path L the unit circle z( t) = et., 0 ~ t ~ 271". Then

= -.!... r'" eacost cos kt dt - t r'" eacost sin kt dt. 271" Jo Jo

By b) we have 'Uk = 'U_k and hence

d) By the definition 21 { 71" J1z1=1

(z2+1)n zn+k+1 dz is the coefficient 'Uk of the Laurent

series for the function

The expansion

gives Uk i= 0 only as a coefficient by zk for n = k + 2p and i = k + p. Then

e) We have

U = (k+2P) = (k+2p)! > 0 k k + p p!(k + p)!' p - .

00 an e acost = ~ nt

L..J '" cos . n=O n.

Therefore

1 12". 1 12" 00 an - eacost cos kt dt = - L , cosn t cos kt dt. 271" 0 271" 0 n=O n.

We can exchange the order of integration and surn since the series converges uniforrnly in every closed region ( for Izl > 0) by the estirnation

We have

By d) we have

I an I an , cosnt cos kt ::::; , (n E N). n. n.

1 00 an 12" Uk = - L , cosnt coskt dt.

271" n=O n. 0

{ (k+2p)!

(Z2 + l)n dz = p!(k + p)!' zn+k+l

o

for n = k + 2p, p ~ 0

otherwise

= ~ [2" 2n COSn t( COS kt - z sin kt) dt. 271" Jo

Cornparing the obtained results we have

1 12" n (k + 2p)! - cos t cos kt dt = n '(k )' 271" 0 2 p. + p .

for n - k = 2p, p ~ 0, and for other cases the integral is zero. Therefore

00 an n!

Uk= E n! 2n (~)! (~)!' where n runs through natural nurnbers n for which n - k = 2p, pE NU {O}.

Example 7.22 Find the Laurent se ries for the function

z2 - 2z + 5 (z - 2)(Z2 + 1)

in the annulus 1 < Izl < 2. Then find the expansion at z = 2.

Answer. The desired Laurent series on annulus 1 < Izl < 2 is

00 1 00 zn

2 I:(-lt--I:-. n==l z2n n==O 2n+ 1

The expansion at Z = 2, or more precise for 0 < Iz - 21 < v's is

1 ~ ( )n (2 + zt+ 1 - (2 - i)n+1 ( 2)n --+z L.J -1 z- . z - 2 n==O 5n+1

Exercise 7.23 Prove the Mittag-Leffier theorem.

Hints. Suppose that Zn i 0 for all n. Expand Pn ( z ~ zJ in apower series of z / Zn

at the origin.

Exercise 7.24 Prove that for given entire functions fand 9 without common zeroes there exist entire functions hand e such that

hf + eg = 1.

Hint. Use the Mittag-Leffler theorem to obtain a meromorphic function F whose part with negative powers (principal part) occur only at zeroes of g, and the principal part of F at a zero Zn of 9 is the same with the principal part of 1/ f 9 at Zn' Take h = Fg.

Example 7.25 Find the Laurent series for the function

1 f(z) = (Z2 + 1)2

in the neighborhood of the points z = z and z = 00.

Answer. The expansion in 0 < Iz - zl < 2 is

The expansion for Izl > 1 is 00 n I: (-1t-· n==l 2n + 2


Example 7.26 Examine which o/the/ollowing multi-valued/unctions have branches which can be expanded in Laurent se ries (especially in power series) in the neighborhood 0/ the given point

a) yZ, z = 0; b) vz(z - 1), z = 00;

c) f}(z - 1)(z - 2)(z - 3), z = 00; d) VI + yZ, z = 1.

Answers.

a) Not possible;

b) Both branches can expand;

c) All three branches can expandj

d) Two branches can expand, and they are given by the condition

Chapter 8

Residues

8.1 Residue Theorem

8.1.1 Preliminaries

Definition 8.1 The residue oJ J at Zo given by the Laurent expansion 00

J(z) = L an(z - zot n=-oo

is the coefficient a_l> denoted by Res(f(z))z=zo'

If J has a simple pole at Zo (pole of first order),

g(z) J(z) = h(z)'

where 9 and h are analytic at Zo and h has a simple zero at Zo, then

. g(zo) a_l = hm(z-zo)J(z) = -h( )'

Z--+ZQ , Zo

If J has a pole of order k at zo, then

1 dk - 1

a_l = (k -1)! dZ k - 1 ((z - zo)k J(z))z=zo·

Theorem 8.2 (General Residue Theorem) Let J be analytic in a region D except Jor isolated singularities at Zl> Z2, ... , ZS' IJ C is a closed path not intersecting any oJ the singularities, then

10 J(z) dz = ; 10 z :ZZn Res(f(z))Z=Zn'

where C is a closed path arround Zl, Z2, . .. ,Z •.

191


192 CHAPTER 8. RESIDUES

A closed path C is called regular if C is a simple closed path with

Then we call

the inside of C.

1 dz -- = 0 or 1 for a ~ C.

ez-a

aj- ---1 { I 1 dz } 27rt e z - a

Theorem 8.3 (Residue Theorem) Let J be analytic in a region D except Jor isolated singularities at Zl, ZZ, • .. ,ZS' IJ C is a closed regular path not intersecting any oJ the singularities, then

j J(z) dz = 27n t 1 ResU(z))Z:=Zn, e n:=l e

where C is a closed path arround ZI, Zz, ... ,Zs'

Theorem 8.4 Let C is a regular closed path. IJ J is meromorphic inside and on C and contains no zeros or poles on C, then

1 ( J'(z) 27rt Je J(z) dz = ZU) - PU),

where ZU) and PU) are the number oJ zeros (counted with their multiplicity) and poles oJ J inside C, respectively.

Theorem 8.5 (Argument Principle) Let C is a regular closed path. IJ J is analytic inside and on C and contains no zeroes on C, then

_1 1 J'(z) dz = ZU), 27rt e J(z)

where ZU) is the number oJ zeroes of J inside C.

Theorem 8.6 (Rauche) Let J and g be analytic inside and on a regular closed path C and jJ(z)j > jg(z)j for all z E C. Then

ZU + g) = ZU) inside C.

Theorem 8.7 (Generalized Cauchy Integral Formula) Let J be analytic in a simply connected domain A and let C be a regular closed path contained in A. Then for every Zo inside C

f (n)( ) - ~ 1 J(z) d r -Zo - ( ) +1 Z lOr n - 0,1,2, ....

27rt e z - Zo n

8.1. RESIDUE THEOREM 193


Exercise 8.1 Find the residues of the following functions at the isolated singular points in the extended complex plain (with (0) :

1 a) -3--5; z - z

Z2n b) nE N;

Z2 + z -1. c) Z2(Z _ 1) , (1 + z)n'

Answerf' 1 1 a) Res(--) = --, Res (--) = 1, Z3 - Z5 z=±1 2 Z3 - z5 %=0

Res(+s) =0. z - z z=oo

b) Re ( Z2n) (lt+1 (2n)! s (l+z)n z=-1 = - (n-l)!(n+l)!'

R ( z2n) (lt (2n)! es (1 + z)n z=oo = - (n - 1)!(n + I)!"

) Re ( Z2 + z - 1) _ 0 R (Z2 + z - 1 ) _ 1 R (Z2 + z - 1) - -1 es -,es -,es -. Z2(Z - 1) %=0 z2(Z -1) z=1 Z2(Z - 1) z=oo

~ 1 ~ 1 d) Res( 2( 2 9») = -, Res( 2( 2 9») = --5 (sin3 - z cos3), z z + z=o 9 z z + z=3. 4

~ 1 ~ 1 Res( 2( 2 9») = --(sin3 + zcos3), Res( 2( 2 ») = 27(sin3 - 3). z z + z=-3. 54 z z + 9 z=oo

Example 8.2 Find Res (~{:j) z=w if:

a) w is a zero of n-th order of the function f;

b) w is a pole of n-th order of the function f.

Solution. J'(z) n g'(z). . .

a) We have by -f( ) = -- + -( )' where 9 IS a regular functlon m the z z-w 9 z

neighborhood of wand g(w) -=F 0, that Res (~{:n z=w = n.

J'(z) -n g'(z) b) We have by f( ) = -- + -( )' where 9 is a regular function in the z z-w 9 z

neighborhood of wand g( w) -=F 0, that Res (j~ n z=w = -no

1" { 'In tan(t + az) dt = o -:rrz

for a E lR.

for a > 0 for a < 0,

Solution. It is easy to prove that

1" 1 12" tan(t + at) dt = - tan(t + at) dt. o 2 0

Taking z = eh, 0 ::; t ::; 2:rr, we have

. 1 1 1 1 dz sm t = - (z - -), cos t = - (z + -), dt = -,

2t z 2 z zz

and we obtain

1" 1 1 z2 - e2a dz tan(t+at)dt==-- 2 2·-· o 2 JzJ=l Z + e a Z

We have for a > 0 inside of the path /z/ = 1 only one pole at the point z == 0 ;

Z2 _ e2a

Res ( ) == -l. (Z2 + e2a )z z=o

Therefore 1

tan(t + az) dt == -2" 2:rrz( -1) == 1rZ for a > O.

We have for a < 0 inside of the path /zl == 1 three poles of first order at the points

( Z2 _ e2a )

Res (Z2 e2a )z == l. + z=:b eG

Therefore

r tan(t + az) dt == _! 2:rrz( -1 + 1 + 1) == -1rZ for a < O. Jo 2

Exercise 8.4 Find the integrals

a) 121r (1 + 2 cos tt cos nt dt, n E Nj 12" dt b) , 0< a < 1j

o 1 - 2a cos t + a2

c) r" dt for a > b. Jo a + bcost

8.1. RESIDUE THEOREM

Answers.


a)

27r b) 1 _ a2;

27r c) ~.

(n) __ 1 [(1 + z)n dz k - 27rz Je zk+l '

where C is a simple path surrounding the originj

b)

195

Solution. a) (~) is the binomial coefficient of zk in (1 + z)n and by the theorem on residue

b) By a) we have

(n) __ 1 r (1 +z)n dz k - 2n Je zk+! .

(2n) = _1 r (1 + z)2n dz. n 2n Je zn+!

Choosing for C the unit circle we obtain

00 (2n) 1 Example 8.6 Find L n --;. n=O 7

Solution. By Example 8.5 we have

(2n) = _1 r (1 + z)2n dz, n 27rz Je zn+l

where C is an arbitrary contour surrounding the origin. Therefore

00 (2n) 1 1 00 r (1 + z)2n dz ~ n 7n = 2n ~Jo (7z)n -;.

Taking for C the unit circle surrounding origin we have

1(1+Z)21 < ~ C 7z - 7' z E .

Then the convergence of the considered series is uniform and therefore

f (2n) ~ = .!...- f dz = 7Res ( 1 ) n=O n 7n 27rZ Jlzl=l 5z - 1 - Z2 5z - 1 - Z2 z=(5-01)/2·

Find this residium.

Exercise 8.7 Prave that

) 100 cos ax d 1r -aV2/2 (aV2 aV2) Eor a -- x = - e cos -2- + sin -2- l' -00 1 + x4 V2

1000 Xp-l 21r cos(2p + 1)~ b) VP dx = -. for 0 0;

~---.----~--------~ -R ~ 0 R ~ Figure 8.1

8.1. RESIDUE THEOREM

p-l

b) Use the function 2 Z on the path in Figure 8.2 ; Z +z + 1

-R R

Figure 8.2

Figure 8.3

e2z1 - 1 c) Use the function 2 on the path in Figure 8.3.

z

Example 8.8 Taking the integral

f In2 Z d lL (z2 + a2)2 z fo1' a > 0,

197

x

x

and the path L f1'om Figu1'e 8.4 find which real integrals can be calculated fo1' R -t 00

and l' -t O.

y

-R R x

Figure 8.4

Solution. We obtain by Theorem 8.3 on residues

1, In2 z l R In2 x 1211" (In Re tl )2 tI

(2 2)2 dz = (2 2)2 dx + Z (R2 2t 2)2 Re dt L Z +a T x +a 0 e '+a

We estimate the second integral:

<

<

<

as R ~ 00. We obtain in an analogous way for the fourth integral

I [21f (In r + (t + 211")z)2 reh dtl 10 r 2e2h + a2)2

< {21f (llnr+(t+211")zl)2 rdt 10 la2 + re2t.12

< [21f (Iln rl + It + 211"1)2 Jo (a2 - r2 )2

~ 0

as r ~ O. The desired residues are

=lim.!!...- ~ ( )2

z->.a dz z + za

z ( 11"2 ) = -3 2ln a - In 2 a + - + n - 1I"Z In a , 4a 4

R2 = Res ((Z2 ~2a2)2) z=-.a = - 4:3 (2 lna -ln2 a + a:2

+ 3n - 311"z lna) .

Therefore we have for R --+ 00 and r --+ 0 :

l OO Zn x 100 dx -2z V P (2 2)2 dx + 211" V P (2 2)2 = z(R1 + R2 ).

o x +a 0 x +a

Putting the values of residues R1 and R 2 into the last equality and comparing the real and imaginary parts we obtain

100 In x 11" (2 2)2 dx = --3 (1 -lna);

o x + a 4a

since the integrals obtained converge absolutely.

r x sin x d tor a > o. Jo 1 + a2 - 2a eos x x

Hint. Take the integral of the function t(z) = z on the reet angle: a - e-az

-7r :::; Rez :::; 7r, 0:::; Imz:::; h,

and let h -t 00, Figure 8.5.

-11

7r Answer. - In(l + a).

a

hi

o Figure 8.5

f

100 eax

Exercise 8.10 Calculate the value ot the integral -00 (ex + 1)( eX + 2) dx, tor 0 < a < 2.

Hint. Take the integral of the function

eaz

t(z) = (ez + l)(ez + 2)

on the rectangle with the vertiees - R, R, R + 27rz, - R + 27rZ. 7r(1 - 2"-1)

Answer . --'-----'sm a 7r

Example 8.11 Find the integral _1_ r sin ~ dz, where C is the circle Izl = r. 27rz Je z

Solution. The point z = 0 is an essential singularity of the function sin~. The z

corresponding Laurent series is

(_1)n ~ (2n + 1)! z2n+t'

00

Hence U-l = 1. Therefore

_1_ [ sin ~ dz = Res (sin~) = 1. 211"z Je z z z=o

Exercise 8.12 Find the value of the integral -21 [ ~z , where aZ = eZ In a 1I"Z Je aZ sm 1I"Z

for a > 0, and C is the straight line x = d (0 < d < 1), oriented from below to above.

Hint. Take first the integral on the path from Figure 8.6 and let b -+ 00.

.0IIII ~y d+ib ~

~ ,

0 d 1 ---~

1+<1 X

d-ib I

Figure 8.6

1 Answer . ----:-:------:-

11"(1 + a)'

Exercise 8.13 Find the real integral

for -1< p < 2.


Hints. Take the integral of the function

on the path C given on the Figure 8.7 and then take R --+ 00.

Answer. 1rp(l ~ p) 23- p sm 1rp

~y

-_R+---HlIJi(4~~R ---~

Figure 8.7


11 dx n~' n=2,3, ... -\}l-xn

Hints. Take the integral of the function

1 J(z) = vr=zn

on the path C given on the Figure 8.8 which consists of the cuttings on radius through the points 1, W, w2 , ..• , wn , where w = e 2:' and the circle Izl = R > 1.

-R

Answer. ~. nstn

n

Figure 8.8

Example 8.15 Let J be an analytic Junction in the region which contains the disc Izl S 1 Prove that

Jor lai< 1

Jor lai> 1,

where L : z(t) = eh, 0 S t S 211".

Solution. By Example 5.13 b) we have on the path L : Izl = 1

/, /, -dz J(z)dz = - J(Z)2'

L L z

Therefore

1=_1_ ( J(z) dz = _1_ I J(Z)Z2 . dz = __ 1_ ( J(z) . Z2 dz. 2n lL z - a 2n lL z - a Z2 211"zlL z - a

Since z2 = e-2ts = _1_ = ..!.., Ost 5 211", we have e2t• z2

I __ 1 f j(z) dz 211"zJL z2(z - a)

= __ 1 f j(z) dz 211"zJL z(l - äz)

__ 1 f j(z) dz 21l"ZiLäz(k- z ) ,

where Z· z = 1 on the path L.

Since

we obtain

Finally

Res (_ ~(z) ) = _j(~)j az(- - z) -1. a a z_"

Res ( j(z) ) = ( j(z) ) = j(O) äz(k - z) z=o z(l - äz) z=o '

for lai< 1, {

211"zj(0) j1(z) dz =

1 äZ(a - z) 2 .. (1(0) - 1 m) for lai> 1.

for lai< 1,

for lai> 1.

Example 8.16 Find the number oj zeroes oj polynomials in the disc Izl < 1 :

a) Z4 - 5z + 1 = 0; b) Z9 - 2z6 + Z2 - 8z - 2 = Oj

c) z6 - 6z + 10 = 0; d) z7 - 5z4 + z2 - 2 = O.

Solution. We shall use Rouche's theorem. We put in Izl < 1 that j(z) = -5z and g(z) = Z4 + 1. We have on the circle Izl = 1 polynomial Ij(z)1 = 15z1 = 5 and Ig{z)1 = Iz4 + 11< Iz41 + 1 = 2. Since the polynomial (-5z) has one zero in Izl < 1, then also the (Z4 - 5z + 1) has one zero in the disc Izl < 1.

b) One zero.

c) There are no zeros in Izl < 1. Put j(z) = 10, and g(z) = Z6 - 6z.

d) Four zeros.

Example 8.17 Prove that the equation zn = ez- k for k > 1, has exactly n zeros in the disc Izl < 1.

Solution. It follows by Rouche theorem putting fez) = _zn and g(z) = eZ-k, where on the unit circle Izl = 1 we have If(z)1 > Ig(z)1, i.e., 1 > ex - k for k > 1 and -1 ~ x ~ 1. Since f has n zeros in Izl < 1, the considered equation has also n zeros in Izl < 1.

Example 8.18 Prove that the equation z = h(z), where h is an analytic function in the disc Izl ~ 1 such that Ih(z)1 < 1, has in the disc Izl < 1 exactly one zero.

Solution. It follows by Rouche theorem putting fez) = z and g(z) = h(z), since If(z)1 > Ig(z)1 on the circle Izl = 1, and f has one zero in Izl < 1.

Exercise 8.19 How many zeroes has the equation

eZ - 4zn + 1 = 0 (n E N)

in the disc Izl < 1?

Answer. It has n zeroes.

Exercise 8.20 How many zeroes in the disc Izl < R has the equation

eR for lai> Rn?

eZ = a zn, n E N,

Answer. It has n zeros.

Example 8.21 Prove that for enough small p > 0, and enough big n all zeros of the function

are in the disc Izl < p.

111 fn(z) = 1 + - + - + '" + -

Z 2!z2 n!zn

Solution. Since the sequence of functions {In} converges to the function e~ everywhere except of the point z = 0 we can find for every disc K, with the center in z =F 0 and which does not contain z = 0 for every c > 0 a natural number no such that

for all points from K and n > no. The desired conclusion follows by Rouche theorem taking

c < rnin le1/ z I zEC '

where C is a path in K.

206 CHAPTER 8. RESID UES


Example 8.22 Let A be an open subset of C and f a continuous function on A.

Let 1+ be the region Im z > 0, and 1- the region Im z < O. Let f be analytic on A n 1+ and A n 1- .

1) Prove that for every real number a E A there is a disc D( a, r) with the center at a and radius r, wh ich lies in A.

2) Let C+ be the path

z(t) = a + (2t + l)r, -1:::; t:::; 0; z(t) = a + r et ... , 0:::; t:::; 1.

Let C- be the path:

z(t)=a+re t1rl , -1:::;t:::;O; z(t)=a+(1-2t)r, O:::;t:::;1.

Find the values of integrals

1 f f(u) du 2n lc+ u - z '

a) For z inside oJ the path C+.

b) For z inside oJ the path C-.

1 f f(u) du. 2n lc- u - z

3) Using 2) prove that the Junction f is analytic on the whole set A. 00

4) Let A : Izl < r, and J(z) = L akzk. Denote k=O

00

MI(r,J) = L lakl Tk, M(r,J) = sup IJ(z)l·

k=O Izl=r

Prove that Jor 0 < r < r + 8 < R :

r+8 M(r, J) :::; MI (r, J) :::; -8- M(r + 8, J).

5) Using 4J prove

Solution. 1) Let a E A and Im a = o. Since A is an open subset of C it is a neighborhood of each its points and so also of the point a. Therefore there exists a


disc D(a, r) which lies in A, D(a,r) C A.

A..y I ___ .... ~

x

--~~--~----~.-. Q-r o o -~ o+r

Figure 8.9

2) The paths C+ and C- given by

C+:z(t)=a+(2t+1)r, -l$t$O; z(t)=a+ret1l"', O$t$l.

C- : z(t) = a + (1 - 2t)r, 1 $ t $ 1; z(t) = a + ret1l"', -1 $ t $ 1

are shown on Figures 8.9 and 8.10, respectively.

""y ,

Q-r Q ..--

o

Figure 8.10

207


Consider first the integral on the path C+. The only singular point of the function

J(u) is the point u = z, which is a pole of first order. We have u-z

_1 f J(u) du = { Res (!~~) _ 21rZ Jc+ u - z u-z o

for z inside of the path C+

for z inside of the path C-,

since for z outside of path C+ the function J(u) is an analytic function inside of u-z

the path C+.

Since

Res (J(u)) = lim(u _ z) J(u) = J(z) u - z u-+z U - Z

u=z

(the function J is continuous everywhere on C+), the value of the integral is

_1 f J(u) du = { J(z) 21rZ Jc+ u - z 0

for z inside of the path C+

for z inside of the path C-.

We obtain in an analogous way for the integral on the path C-

_1 f J(u) du = { J(z) 2n Jc- u - z 0

for z inside of the path C-

for z inside of the path C+.

3) We shall prove that the function J is analytic on the whole set A. By the supposition it is analytic on An 1+ and An /-. We have to prove that it is analytic on the set {z I Imz = 0, z E C} n A, on the path of the x-axe which lies in A.

By 1) for each real point a E A always there exists a disc D( a, r) with the center in a which lies in A. We shall prove the analicity of the function J on the part of x-axis which lies in D (a, r). Since J is continuous on C+ and C- we have that

_1 f J(u) 2n Jc- u - z

define analytic functions for z E C+ and z E C- , they are analytic functions on the part of x-axis which is in D(a,r). Then we have

_1 (1 J(u) du + 1 J(u) dU) = _1 f f(u) du, 2n c+ u - z c- u - z 2n JK u - Z

where K is the boundary of the disc D( a, r).

By 2) we have

1 1 f(u) - -- du = f(z) for z E D(a,r), 27rl K u - Z

(8.1 )

without x-axis. The integral on the left side of (8.1) defines an analytic function for all z E D(a, r), and it is an analytic extension of f on real axis in D(a, r). By the uniqueness of the analytic continuation it follows that (8.1) holds on the part of real axis in D(a,r). Since this holds for every point a E A from real axis we obtain that f is analytic function on the whole set A.

4) We shall prove first that M(r,l):S; M1 (r,l). Namely, since

If(z)1 = I ~ akzkl :s; ~ lakllzlk = ~ lakl rk

for every z for which 0 < Izl = r < R, we have

00

M(r,J) = sup If(z)l:S; L lakl rk = M1 (r, f). Izl=r k=O

We shall prove now the second part of the inequality,

r+8 M1(r,J) :s; -8- M(r + 8,1).

We have 00

M1 (r, I) = L lakl rk k=O

= ~ 1_1 1 f(z) d I k L.J 2 zk+l Z r k=O 7rl Izl=r

= f 1_1 r f(z) dzl rk k=O 27rl 1Izl=r+5 zk+l '

by the equivalence of integrals on paths Izl = rand Izl = r + 8 (0 < r < r + 8< R).

Since

1_1 r f(z) dzl < 27rl 1lzl=r zk+l -

1 r If(z)1 d 2n 1lzl=r+o Izlk+1 z

sup If(z)1 Izl=r+o r27r r + 8

< 2 10 (r + 8)k+1 dt

sup If(z)1 Izl=r+5 (r + 8)k+l '

we obtain

M(r,f) 00 sup If(z)1

< E Izl=r+6 rk

(r + 6)k k=O

00 r k

= Izr~!~)f(z)1 ~ (r + 6) 1

sup If(z)1 1 r Izl=r+6 - r+6

= r+6 = -1:- sup If(z)l,

u Izl=r+6

since ~ < 1, which implies the inequality and the convergence of series. So we r+u

have proved that

r+6 M(r,f)::; M1(r,f)::; -6- M(r + 8,f) for 0< r < r + 6< R.

5) Taking

n 00 1 1 1 r(z) 1 k M 1(r,j ) = E - ~ dz r , k=O 21ft Izl=r z

we can easily prove in an analogous way as in 4) that

Putting 6 = .!.. and taking the n-th root we obtain n

since

1 < Y'1 + nr < Y'n 2 + n2 = Y'2n 2 -+ 1 as n -+ 00. - -Therefore by the inequality (8.2) using the comparison criterion for sequences we obtain

Example 8.23 Let

lim r!M1(r,r) = M(r,f). n-+oo

1 1 F(z) = -. - - -.

smz z


a) Find the singular points ofthe function F(z). Is the function

analytic at the point z = O?

{ F(z) for z '" 0

f(z) = 0 for z = O.

b )Find the residues of the Junction J in all poles.

c) Show

/ 1 /2 2

sin z = cosh 2y - cos 2x .

211

d) Show that there exists a natural number no such that the Junction J is bounded independently of n ~ no on the boundary of a the square K n given by

z = ±(n + 1/2)7r + ~t, -(n + 1/2)7r ~ t ~ (n + 1/2)7r;

z = t ± t(n + 1/2)7r, -(n + 1/2)7r ~ t ~ (n + 1/2)7r.

e) For an arbitrary but fixed n E N find the integral

In = _1 { f(u) du. 27ft lKn u(u - z)

f) Prove that liIIln-+oo In = 0 and the equality

00 (_l)n F(z) = 2 E z2 _ (n7r)2'

g) Find Jor the preceding series:

(i) the domain oJ the convergence,.

(ii) the domain oJ the uniform convergence;

(iii) the domain oJ the absolute convergence.

Solution. a) The singular points of the function J are z = k7r for k = ±1, ±2, ... We shall examine the point z = O. We have

lim J(z) z-+o

1. z - sin z 1m .

z-+O Z Sin Z

1 - cosz = lim -c.:-----

z-+O sm z + z cos z smz

lim -------z-+O cos Z + cos Z - Z cos z

= 0,

limz-+oj(z) = j(O). Hence the function j is continuous at the point z = O. The function F has the same singular points as the function j, and at z = 0 it has removable singularity.

Now we shall show that the function j is analytic at the point z = O.

r j{O + ~z) - j(O) z~ ~z

r ßz - sin~z z~ ~z2sin~z r 1- cos~z

= z~ 2ßz sin ~z + ~Z2 cos ~z r cos~z

= z~ 2 sin ~z + 2ßz cos ~z + 2~z cos ~z - ß Z2 sin ~z r cos~z z~ 2 cos ~z + 4 cos ~z - -4~z sin ~z - 2~z sin ~z - ~z2 cos ßz 1 - < 00. 6

Since the preceding limit exists we conclude that the function j has a derivative at z = 0, it is analytic at the point z = O.

b) The function j has at z = 0, k = ±1, ±2, ... poles offirst order. We calculate the residues of the function f at z = k1r

Res{f (z)} z=k1r = (--:._Z_-_sI_n_Z_) Sin z + z cos Z z=k",

k1r _ (_l)k k1r cos k1r - .

c) U sing the equality sin z = elZ 2z e -1% we obtain

1

2 2z 4

= (e'X-Y _ e-,x+Y)(e'x+y - e-'x- y) 2

2 2 2

= cosh 2y - cos 2x .

d) The path K n is given in Figure 8.11.

(n#-jJ1i!

Kn

-"''''JIN" 0 {n#-IPi ~ X

, -(n-#T]1i"f.

Figure 8.11

We shall show that there exists a natural number no such that the function f is bounded independently of n ;::: no on the boundary of a square Kn • We have on the sides of K n

z = ±(n + 1/2)7r + tt, -(n + 1/2)7r $ t $ (n + 1/2)7r;

the following inequalities

If(z)1 = I ±(n + ~)7r + tt - sin(±(n + ~)7r + tt) I (±(n + ~)7r + tt) sin(±(n + ~)7r + tt)

< I ± (n + ~)7r + ttl + I sin(±(n + ~)7r + ti)1 7r(n + ~)I sin(±(n + ~)7r + tt)1

< (n + ~)7r + Itl + cosh Itl (n + ~)7r1 sin(±(n + ~)7r + tt)1

< 2(n + ~)7r + cosh Itl 2 (n + ~)7r cosh2t - cos2(±(n + ~)7r)

< 2(n + ~)7r + cosh(n + ~)7r 2 (n+~)7r 1+1

< M

for a natural number no, fixed M > 0 and n ~ no. In a quite analogous way we can prove the boundedness of f on the paths

e) The function f for a fixed n is analytic on K n • Its singular points (poles of the first order) inside the square are at the points u = k7r, k = ±1, ±2, ... , ±n. The function under the integral in In additionally has a pole of first order at the point u = z when the point z is inside the square K n (taking n enough big we can always manage this), and at the point u = 0 we have a removable singular point. Hence

I - 1 r f(u) d - t R ( f(U)) R ( f(u) ) n - 271"~ lKn u(u - z) u - k=-n es u(u - z) u=k1r + es u(u - z) u=z·

Since

and

R ( f(u) ) es u(u - z) u=k1r

R ( u - sinu ) es u2(u - z) sin u z=k1r

h (h)2(h - z) cos h

( _l)k

h(h-z)

R ( f(U)) f(z) es u(u - z) u=z z

we finally obtain (the term in the sum with the index 0 is equal 0)

(8.3)

f) We shall show that limn _ ao In = O. We have by d) If(u)1 < M for u E K n and n ~ no. Hence

IInl < 1 1 If(u)1 d 271" K n lu(u - z)11 ul

< M 1 Idul 271" K n lullu - zl

< M i du 271" K n lul(lul-lzl)·

Since we have minuEKn lul = (n + D1I" and lul- Izl 2: (n + !)11" -Izl we obtain

< M [ Idul 211" JKn (n + ~)1I"((n + ~11" -Izl)

= M [ Idul. 211"(n + !)1I"((n + })11" -Izl) lKn

Using that one side of the square K n is equal 2(n + ~)11" we obtain

111 M8(n+ t)1I" _ 4M n < 211"(n + ~)1I"((n + ~)71" - jzl) - (n + !)11" - JzJ·

Taking n ---t 00 we obtain that the last member in the preceding inequality tends to zero, what implies limn->oo In = 0.

To prove the desired equality we shall take n ---t 00 in (8.3) from e)

After some transformations we obtain

z

Finally we have

-1 (_l)k 00 (-l)k

k~OO h(z - h) + .t; h(z - h) 00 (_l)k 00 (_l)k

-E +E . k=1 h(z + h) k==1 h(z - h)

00 (-l)k -z+h+z-h E ----;;;-. (z + h)(z - h) 00 (_1)k

2E 2_(k F· k==1 Z 71"

This implies the desired equality. g) The series

00 (-1 )kz Ez2_(h)2

converges absolutely in the disc D(O, 71"), and uniformly in every closed region in D(O, 11"). The absolute convergence of the series follows by the inequality

I (-1)k z I R Z2 - (h)2 :::; (hF - R2

for jzl :::; Rand k 2: ~, since J( n1l")2 - Z2J 2: j( n1l"? - JzJ 21 holds.

216 CHAPTER 8. RES1DUES

z+u Example 8.24 I) Let W = J(z) = k -- Jor k 1= o.

z+v Z· +u

1) We denote by Wi = k -'--, i = 1, 2, 3. Prave that Zi+ W

---=------

Prave that iJ there are given three different points Zl, Z2, and Z3 in the z-plane and three different points Wb W2, W3 in the w-plane then there exist a unique bilinear transformation J wh ich map Zi on Wi, i = 1,2,3, respectively. Find the bilinear transformation J which maps the unit disc Izl < 1 on the half-plane Im W > o.

2) Find the image in w-plane by the Junction J{z) = 2 z + 1 oJ the circle from Z

z- plane, wh ich does not contain the point (0,0)7 For which points is this Junction a conformal mapping?

1"" a-l II) 1) For 0 < a < 1 find the value oJ V P -IX dx. We are taking Jor za-l

o -x the bmnch which is real for real z.

2) 1f 0 < a < 1 and 0 < b < 1, then

1"" xa-l - X b- 1 ---- dx = 7l'(cot a7l' - cot b7r)

o 1- x

Solution. I) 1) See Example 4.8. Since the bilinear transformation is uniquely determined by the given images of three points we shall find the desired transformation by three points.

Taking O - 1 1 h Zl = ,Z2 - 2' z3 = :4' we ave

u _ k 1 + 2u 1 + 4u Wl = k ;-, W2 - 1 + 2v' W3 = k 1 + 4v '

respectively.

The condition Im Wi ~ 0, i = 1,2,3, implies

ku v-ku v Im Wl = Ivl2 2z ~ 0, Im (k u v) ~ 0, (8.4)

Im - 2ku(I + 2v) - 2ku(I + 2v) 0 I (k) > 21 (k-) W2 - 11 + 2vl2 2z ~ , m u _ - m uv (8.5)

and 4ku(1 + 4v) - 4ku(1 + 4v)

Imw3 = ~ 0, 11 + 4vl 2 2z (8.6)

Im(k u) ~ -4Im(kuv).

Since (8.6) is a consequence of (8.4) and (8.5) we can omit it. Therefore the desired bilinear transformation is

w_k Z +u ki=0 - z+v' ,

where Im(kuv) ~ ° and Im (ku) ~ -21m (kuv), see also Example 4.5,

-1

Figure 8.12

where the bilinear transformation w' = ~ - Z maps the unit disc Izi < 1 on the ~+ z

half-plane Re w' > 0, see Figures 8.12 and 8.13. If we rotate for the angle 7r /2, w = e7r•j2 w, we obtain the desired transform (see Example 4.18 )

7r.j2 ~ - z w=e --. ~+z


Figure 8.13

z+l 2) For fez) = 2 -2- we have

Therefore we obtain the desired image in the following way. The transformation

[~ ~] maps the circle azz + bz + cz + d = a which does not cross (0,0) from the

z-plane onto the circle dWIWI + CWI + bW1 + a = 0 in the w-plane.

Applying now the transformation [~ ~] on dWIWl +CWl +bWl +a = 0: we first

translate the circle by 2 and enlarge the radius by 2. Write down the equation of the obtained circle in the w-plane. For every point z E ewe have J'(z) #- O. This is not true only in the extended complex open at z = 00. The bilinear transformation

fez) = 2 z + Z is a conformal mapping at all points of the complex plane except at z

z = 0 and z = 00. za-l

II 1) The function -- has the branching points 0 and 00. Therefore the cutting l-z

za-l [0,00) divide the branches of --. This function has a pole of first order at z = l.

l-z a-l

Therefore we shall integrate the function fez) = _z __ on the path L which is l-z

given at Figure 8.14.

-R

We have

Figure 8.14

a-1 f _z_dz = 0, JL 1- z

since the function f is analytic in the region bounded by L.

We have (see Figure 8.14)

1 za-1 --dz

L 1- z 11- E xa-1 1 za-1 JR x a- 1

--dx+ --dz+ --dx r 1 - X kl 1 - Z 1+E 1 - x

+ _z __ dz + _x __ e2(a-1)1I"' dx 1 a-1 J,1+E a-1

k(R) 1 - Z R 1 - X

a-1 r a-1 + f _z_dz + f _x __ e2(a-1)1I"' dx + f

lk2 1 - z lt+< 1 - x Jk3(r)

za-1 --dz. 1-z

219

a-1

Since 0 < a < 1, we have lim z _z __ = O. Therefore the value of the fourth integral z-+oo 1- z

a-1 is zero as R --+ 00. Since lim z _z __ = 0, the value of the last integral is also zero

z-+o 1 - z . as r --+ O. On the other side, since z = 1 is a pole of first order the values of second and sixth integrals are

1 za-1 a-1 -- dz = -1l"Z Res (-lZ) = 1l"Z,

kl 1 - z - Z z=1+.0

{ za-1 dz = -'In Res ( za-1 ) = 'In e2"a. 1k2 1 - z 1 - Z z=1-.0

(since 1 + z = eO., and 1 - zO = e2"') as c --t O. Letting R --t 00, r --t 0 and c --t 0 we obtain

Hence

(1- e2(a-1)1r,)VP {OO x a-

1 dx = -7l"z(l + e2,..a,). 10 1- x

100 xa-1 e2,..a. + 1 V P -- dx = 7l"Z 2 = 7l" cot a7l" for 0 < a < 1. ° 1 - x e "a. - 1

2) By 1) we have for 0 < a < 1, 0< b< 1,

x q- 1 _ X b- 1

1- x dx VP f'X> x q- 1 dx _ VP {OO

10 1 - x 10 = 7l"(cot a7l" - cot b7l").

100 xa-1 - X b- 1 Then V P 0 ---- dx

I-x

X b- 1

I-x

= {01 x a- 1 - X b- 1 ],C x a- 1 - X b- 1 {OO x a- 1 _ X b- 1 1n 1 - x dx + 1 1 _ x dx + 1c 1 _ x dx.

Examine the integral also at the point o. First and second integrals are not improper at x = 1, since at x = 1

1. xa-1_xb-1 . (a-1)xa- 2 -(b-1)xb- 1 1m ----- = hm = -a + b.

",-+1 1 - X ",-+1 -1

The last integral is absolutely convergent since 0 < a < 1 and 0 < b < l. Then

X - x dx = V P x - x dx. 100 a-1 b-1 100 a-l b-1

o I-x 0 I-x

Example 8.25 Suppose that J is an analytic Junction in the region D which conta ins the point a. Let

F(z) = z - a - q J(z),

where q is a complex pammeter.

1) Let K C D be a circle with the center at the point a on wh ich the Junction J is zero. Prove that the Junction F has one and only one zero z = w on the closed disc (K) whose boundary is the circle K, iJ

. Iz - al Iql < ~W J(z)1 .

8.2. COMPOS1TE EXAMPLES 221

2) Let G be an analytic function on the disc (K) together with the boundary. Using the theorem on residues prove that

where w is the zero jrom 1).

G(w) = _1 ( G(z) dz F'(w) 2?ri lK F(z) ,

. 1 3) 1f z E K, prove that the functlOn F(z) can be represented as a convergent

series with respect to q :

4) Using 3) and 2) prove

G(w) 00 qn dn n

F'(w) = G(a) +; n! dan (G(a) j (a)).

5) Prove that ij G is oj the form

G(z) = H(z) (1 - q !,(z)),

where H is an analytic junction on (K), then

H(w) = H(a) + f ~~ ::n-~l (H'(a) j(n)(a)). n=l

Solution. 1) We shall apply the Rouche theorem. Let

F(z) = z - a - q j(z) = cp(z) + '!fJ(z),

where cp(z) = z - a and '!fJ(z) = -q f(z). We shall prove that:

(i) cp(z) and '!fJ(z) are analytic functions for all z E (K)j

(ii) cp(z) f 0 for z E Kj

(iii) Icp(z)1 > 1'!fJ(z) I for z E I<.

(i) is obvious since cp is a polynomial of first order and so an analytic function in the whole complex plane and the function f is analytic by the supposition in the region D which contains (I(). Hence -qf is also analytic function in D.

(ii) follows because it can not be z = a on the path K, since K is the circle with the center at a.

(iii) will be proved using the fact

. Iz - al Iql < ~f{ If(z)l·

Iz -al Namely, Iql < If(z)1 for z E K. Hence

11/J(z) I = Iqllf(z)1 < Iz - al = Icp(z)l· The conditions (i) , (ii) and (iii) by Rouche theorem imply that the number of zeroes of the function F = cp + 1/J and cp are equal. Since cp( z) = z - a is a polynomial of first order it has only one zero z = a, we obtain that the function F has also one and only one zero z = w on the closed disc (I<).

2) Since the function F has a zero of first order at z = w, the function ~ has a

pole of first order at z = wand also the function ~, since G is an analytic function

in the disc (I<).

We have by Residue Theorem

r G(z) (G(z)) JK F(z) dz = 2n Res F(z) z=w·

Since ~ has pole of first order at z = w

( G(Z)) G(w) Res F(z) z=w = F'(w) ,

and so r G(z) G(w) JK F(z) dz = 27rZ F'(w)'

which gives us the desired equality.

3) Since F(z) = z - a - q fez), we have

1 1 1 1 F(z) = z-a-qf(z) z - a 1 _ qj(z) .

z-a

Since

/qf(Z) / < 1, z-a

for z E I< (this inequality is equivalent to Iql < 1~(z~I, which holds by the condi

tion in 1)) we can represent the second factor in the last equality as an absolutely convergent series

1 1 00 (qf(z))n F( ) = - L ( ) for z E I<,

z z - a n=O Z - a n

8.2. OOMPOSITE EXAMPLES

which implies the desired equality.

4) By 2) we have

Using 3) we have

G(w) F'(w)

1 [ G(z) d 27r~ JK F(z) Z.

G(w) 1 [ ~ (q f(z))n d F'(w) = 27r~ JK G(z) ::a (z _ a)n+l z.

223

By the uniform convergence of the series for z E I<, we can exchange the order of the integration and sumo Therefore

G(w)

F'(w) = _1 f [ G(z) qn r(a) dz

27r~ n=O JK (z - a)n+l

= ~ qn _1 [G(z)r(z) dz. L..J 2n JK (z - a)n+l n=O

Then by the Cauchy integral formula we have

Therefore G(w) 00 qn n

F'(w) = G(a) + ~ n! (G(a) f (a)).

5) By G(z) = H(z) F'(z) and 4) we easily obtain the desired equality.

Example 8.26 Let f be a meromorphic function in the whole complex plane for which there exists an increasing sequence { rn } which tends to +00 as n ~ 00 such that

(8.7)

where M > 0 is a constant independent of n and (). We denote by u a point in the complex plane which is not a pole of the function fand On is the circle Izl = r n ,

positively oriented.

I) 1. Prove that the integral

converges to zero as n -t 00.


I) 2. Prove that

f(u) - f(-u) = -2u L Res ( !(z) 2) , aEP Z - U z=a

where P is the set of all poles of the function f. II) 1. a) Prove that for the complex number a > 0 there exists A> 0 such that

z = x + zy and Iy I ~ a imply 11 + e2•z 1 ~ A.

II) 1. b) Prove that for the complex numbers z which satisfy the inequality

there exists a number B > 0 such that

II) 2. Prove that the function tan z satisfies (8.7) taking rn = mr (n E N).

II) 3. Using I) 2. and II) 2. prove that

00 -2u tanu=L 2 ((2 )")2·

n=l U - n + 1 2"

II) 4. Examine the uniform convergence of the senes from II) 3.

Solution. I) 1. For r n ~ lul we have

Letting n --+ +00 we obtain that the integral tends to zero.

I) 2. By the theorem of residues we have

f ~dz = 2n "Res ( f(z) ) Je Z2 - u2 L..J Z2 - u2

n aEGn z=a

where Cn is the closed region bounded by Cn. Letting n --+ 00, we obtain by I) 1.

(8.8)

. . J(z) where S IS the set of all poles of the functlOn 2 2 .

Z -u

For J(u) =I- 0 and u =I- 0, u is a pole of the first order for the function !(z) 2' z -u

and therefore R (JJ!L) = J(u)

es 2 2 2u z -u z=u

We have analogously for J( -u) =I- 0

Res C!~ ~2 ) z=-u = - J~~u) . Putting this in (8.8) we obtain

J(u) - J( -u) = -2u L Res ( !(z) 2) . aEP Z - U z=a

The previous equation is true also for u = 0, or J(u) = 0 or J( -u) = 0, what can be easily checked.

11) 1. a) Let z = x + zy. Then le2,z l = e-2y . For y ~ a we have

11 + e2,z l ~ 11 _le21Z 11 = 1 - e-2y ~ 1 _ e- 2a •

For y :::; -a we have

We take

11) 1. b) The function 11 + e2,z l is periodic with the period 'Ir. Therefore it is enough to consider the case 0 :::; x :::; 'Ir.

By 11) 1. a) we have for lyl > a that 11 + e21Z1 ~ A. For lyl :::; a, 0 :::; x :::; 'Ir, Iz'Ir/21 ~ a the function 11+e2, z l is different from zero. Therefore inf 11+e21Z 1 = m > 0, where z belongs to the given region. Finally, taking B = min (m, A) we obtain the desired inequality.

11) 2. By the definition of the function tan z we have

e2•z - 1 2z tan z = -z = -z + .

e2•z - 1 e21Z + 1

Then for z E Cn : z = n'lr, we obtain by 11) 1. b) 1 tanzi:::; l+i = M independently of n (0 < a < 'Ir /2).

11) 3. By 11) 2. the condition (8.7) is fulfilled for the function f(z) = tg z and r n = mr (n E ~). Therefore by I ) 2.

+00 ( tanz) tanu -tan(-u) = -2u L Res 2 2 .

n=-oo Z - U z=(2n+l)"./2

Since Res ( tan z ) _ 1

Z2 - u2 z=(2n+1)"./2 - u2 - ((2n + 1)11"/2)2

we obtain (tan u is an odd function)

+00 1 tanu = -2u L

n=O u2 - ((2n + 1)11"/2)2·

11) 4. Let F be an arbitrary closed bounded region of the complex plain without the points z = (2n + 1)11"/2 for n = 0, ±1, ±2, ....

F. We shall prove that the functional series from 11) 3. is uniformly convergent on

Let d = max lul. For u E Fand (2n + 1)11"/2 > d we have for n 2: no uEF

<

1 IIul2 - ((2n + 1)11"/2)21

1 u2 - ((2n + 1)11"/2)2 - d2

1

as n --+ 00.

Since the series +00 1

~ ((2n + 1)11"/2)2

is convergent, also the series

+00 1

~ ((2n+ 1)11"/2)2 -d2

is convergent.

Hence the series representing the function tg z is uniformly convergent on F.

Chapter 9

Analytic continuation

9.1 Continuation

9.1.1 Preliminaries

Definition 9.1 Let f be an analytie function in a region A. An analytie function 9 on a region Al that intersects A is an (direet) analytie eontin uation

of f from region A to region Al if f = 9 throughout Al n A.

Such analytic continuation is uniquely deterrnined.

Definition 9.2 Let f be analytie in a dise D and Zo E oD. f is regular at Zo if f ean be eontinued analitieally to a region A with Zo E A. Otherwise, we say that f has singularity at zoo

Theorem 9.3 If apower series E~=o an zn has a positive radius of eonvergenee R, then the function

00

n=O

has at least one singularity on the eirc1e \z\ = R. In partieular, if R < 00 and an ~ 0 (n E NU {O}), then f has a singularity at z = R.

Definition 9.4 If f( z) = E~=o anzn has a singularity at every point on its eirc1e of eonvergenee, then that eircle is a natural boundary of f.

Theorem 9.5 If 00

f(z) = L ankznk and k=o

227

1· . f nk+1 1 ImIn -- > , k_oo nk

228 CHAPTER 9. ANALYTIC CONTINUATION

then its circle oJ convergence is a natural boundary Jor f

Theorem 9.6 (Schwarz' reflection principle) Let A be a bounded region that is contained in either the upper or lower half-plane and whose boundary contains a segment L on the real axis. IJ f is analytic in A and continuous on AC, and f(z) is real for real z, then it can be defined an analytic extension g oJ J to the region AULUA*, where A* = {zlzE A}, by

{ J(z)

g(z) = _ f(z)

for

for

zEAUL

z E A*.

Let Do be a disc centered at a point zo0 Let z(t), t E [a, b) be a path beginning at Zo and whit the end point W. If

is a partition of the interval [a, b), the we denote by Di a disc containing z(ai)' We say that the sequence {Do, D h D 2 , ••• ,Dn } is connected by the path along the partition if the image z([ai' ai+l]) is contained in D i. Let J be analytic on Do. An analytic continuation of (J, Do) along a connected sequence (along a path C) [Do, . .. ,Dn ) is a sequence of pairs (canonical elements)

such that (Ji+1' Di+1) is a direct analytic continuation of (Ji, Di), i = 0,1, ... ,n-1.


Example 9.1 The Junction J is defined on the disc Izl < 1 with the power senes

00

L( -1t(2n + 1)zn n=O

and Jor other values Jrom the complex plane by analytic continuation. Find this continuation summing the power series.

Solution. The radius ofthe convergence ofthe given power series is 1. Therefore for z = w 2 with Iwl < 1 we have

00

J(w2 ) = L( -lt(2n + 1)w2n , n=O

9.1. CONTINUATION 229

00

= L(-1tw2n+1

=

n=O W

1 +w2 '

where we have exchanged the integral and the sumo We have by the analyticity of the function f( w2 ) in Iwl < 1

putting z = w 2 we obtain l-z

f(z) = (1 + Z)2'

The function obtained is the analytic continuation of the function f given by power series in Izl < 1, on the whole complex plane without the point z = -1, where the function f has a pole of the second order.

The desired analytic continuation can be obtained also by another method. We have for Izl < 1

Since

00

(obtained by differentiating the series L(-1)nzn), we obtain n=O

2z 1 1 - z f(z) = -(1+Z)2 + 1+z = {1+Z)2'

Example 9.2 The funclion f is defined on the disc Izl < 1 by the power series

Find its analytic continuation outside the disc Izl :5: 1.

Solution. Let x be real and -1 < x < 1. Then

(x f(x))' = f) -ltx3n = A· n=O + X

Hence

J dx 1 V x 2 - X + 1 v'3 2x - 1 x f(x) = -- = -- log + - arctan-- + C.

1+x3 3 x+1 3 v'3 We obtain the constant C by the condition f(O) = 1: C = 7rv'3/6. Therefore for -1 < x < 1 :

~ (_l)n 3n 1 ( 7rv'3 I vx2 -x+1;;; 2X-l) L...J -- x = - -- - og + V 3 arctan -- . n=O 3n + 1 3x 6 x + 1 v'3

By the uniqueness of the analytic continuation this equality holds also for Izl < 1.

Since on the right side is an analytic function in the whole complex plane cutted from the points -1, e"r/3, e-"r/3 in the radial direction to z = 00, Figure 9.1, so in this region this is the analytic continuation of the left side of the equality.

x

Figure 9.1

Example 9.3 The power series 00 zn L - and n=l n

00 (z 2)n Z7r + L(-lt -

n=l n

have no common region 01 convergence. Prove that they are analytic continuations of each other.

9.1. CONTINUATION 231

Solution. Both series define the same function J(z) = -log(1-z) which is analytic outside of the cut from the point 1 to 00.

Example 9.4 The Junction 1 is defined in the disc Izl < 1 by the power series

00 (_1)n-l L z2n. n=l n(2n - 1)

Find its analytic continuation outside 01 the disc Izl ~ 1.

Solution. For real x with -1 ~ x ~ 1 we have

I'(x) = 2 t (_1)n-l x2n-\

n=l 2n - 1

and

Hence by 1'(0) = 1(0) = 0 we have 2x arctan x - log(1 + x 2 ) the obtained formula holds also for Izl < 1, and this gives the desired analytic continuation in the whole complex plane without radial sections from the points z, -z till 00, Figure 9.2.

y

-- -/' "-I '\

I \

\ 0 I X , / "- ./ - ./

-I

Figure 9.2

Example 9.5 Do there exist functions which are analytic at z = 0, and which satisfy the condition

Solution. a) On the set

{~,nEN}, which has the accumulation point 0, the function f(z) = Z2 satisfies the given condition.

b) There is no function which satisfies the given condition since on the set

{~, nE N} it would be f(x) = f(-x) = x3 •

Exercise 9.6 The function j dejined on the unit disc Izl < 1, by the power series

00 ( l)n a) ~ - z2n.

n==2 n(n-1) ,

00 z4n b) ~ -I'

n==O n.

Find their analytic continuations outside of the disc Izl :::; 1.

Answers.

a) j(z) = (Z2 + 1) log(l + Z2) - Z2 in the complex plane with cuttings from points l, -l fix 00, Figure 9.2.

b) The radius of convergence of the given power series is 00. Therefore this power series extend analyticly itself on the whole complex plane. We have

00 z4n ~ -, = (coshz + cos z)/2 n==O n.

which easily follows by the fact that the given power series satisfies the following differential equation

j(iv)(Z) - f(z) = 0

and the initial conditions flll(O) = 1"(0) = 1'(0) = 0 and f(O) = 1.

Example 9.7 The function j is dejined in the disc Izl < 1 by

Prove that it can not be analytically extended outside the disc Izl :::; 1.

Solution. The function satisfies the functional equations

Then, since z = 1 is a singular point of the function I, the solutions of z2 = 1, Z4 = 1, z8 = 1, ... are also singular points of the function f. The set of all singular points is dense on the circle Izl = 1 and they form the natural boundary of the function I.

00

Exercise 9.8 Let I(z) = L unzn , with the radius 01 convergence R = 1. Putting

z = _z_ we have l+z

n=O

I(z) = I (_Z_) = F(z) = f vnzn .

1 + z n=O

Denote by p the radius 01 the convergence 01 the last power senes. Prove that

a) p ~ 1/2, where il z = -1 is a singular point 01 the [unction I we have p = 1/2.

b) 111/2 < P < 1, then the equality I(z) = F(z) = I (_Z_) allows the l-z

analytic continuation 01 the [unction I outside o[ disc Izl < 1, and inside o[ the

circle I-z-I = p. l-z c) II p = 1, then by the equality [rom b) the [unction F analytically extend I on

the half-plane Re z < 1/2, (see Example 9.11 )

Exercise 9.9 Let land 9 be arbitrary entire [unctions and let

Sz= ---00 (I-zn l_Zn-1 )

() ~ 1 + zn 1 + zn-l .

Prove that I(z) = (f(z) + g(z)) /2 + S(z) (f(z) - g(z)) /2

reduces to the function I in the regionlzl < 1 and to the function g in the region Izi > 1.


Example 9.10 1. Prove that lor Izl < 1

Iln(l + z)1 $ -ln(1 -Izl).

z 2. Find the singular points of the function In -- and the regions of analyticity

z-l for each bmnch.

S. The bmnch of the function Ln_z _ which is positive for z > 1 expand in a z-l

Laurent series in the annulus 1 < Izl < R.

a) Find the coefficients of this Laurent se ries by the formula

b) Find the coefficients by some other method.

4. The functions

and 12 = 7l'Z + f( -lt (z - 1t n=l n

analytically extend each other. Prove that.

Solution. 1. We have for Izl < 1

00 zn -In(l + z) = L( -lt -.

n=l n

Therefore

For 2. and 3. see Exampie 7.17.

4. The both functions f1 and 12, see Exampie 9.3, are expansions in power series of the functiün fez) = -ln(l-z), where first in the neighbürhood üf zero für Izl < 1,

and the second in the neighborhood of the point 2 for Iz - 21 < 1, Figure 9.3,

smce

-ln(l- z)

y

Figure 9.3

= -ln(l - 2 - (z - 2)) = -ln(-1)(1+(z-2))

= -lne-1r' + In(l + (z - 2)) 00 (z l)n

7rZ + E(-lt -n n=1

235

It is interesting to remark that the regions of functions 11 and 12 have no common points, but they analytically extend each other (since they have common representation -ln(l - z)).

Example 9.11 1. Let ~ > 0 lor -1 < x < 1, and a is a complex number. Prove that the integral

11 dx 1= ,

-1 (x-a)~

is given by: I = - 'Ir for a > 1,

";a2 -1

I = ± 'Ir e·(31r-2t)/4 for a = ±et • for 0 < t < 'Ir, ";2 sin t

I ~t . ~ = J'1I"":"":2 slgn ny lOr

vI + y-a = tY,

~

1=+ ~ for a < -1, a2 -1

and for -1 < a < 1 the prineipal value (VP) is I = O.

2. The dise Izl ::; 1 map with the function w = _z_. What is its image in the l-z

w-plane ?

Map by the inverse function the eireles Iwl ::; r, 0 < r < 00, from the w-plane into the z -plane. What are the images?

00

3. Let f(z) = L anzn be a eanonieal element of the eomplex analytie function n=O

W F with mdius R = 1. Put z = -- and let F(z) = G(w). Let g(w) = J(z) be a

I+w . 00

eanonieal element for G( w) of the form g( w) = L enwn and let r > 0 be its mdius n=O

of eonvergenee. Prove that r ~ 1/2. The analytie function g (_Z_) analytieally I-z

extends f from the region Izl < 1 to a region O. Find this region 0 for the eases 1/2 < r < 1, r = 1, r > 1.

Solution. 1. We start with the function

1 J(z) = y'l-Z2' a 7' ±1, (z - a) 1- Z2

for which VI - x 2 > 0 for -1 < x < 1.

The function f has pole of the first order at the point z = a (a 7' ±1) and algebraic branching points of the first order at z = 1 and z = -1. The cut is ( -1, 1), since z = 00 is an ordinary point. We shall first consider the case a = ty, a< -1,


taking the path L from Figure 9.4.

y

-R R

Figure 9.4

We have i J(z) dz = 27rZ Res(J(z))z=a. (9.1)

The integral on the path L can be written in the following form

[ dz lL (z - a)Vf=Z2 = 11- e dx j dz

-1+. (x - a)v'f=X2 + "11 (z - a)Vf=Z2 (9.2)

[R dx [dz + A+e (x - 2h!(1- x)(x + l)c"" + lK (z - a).Jf=Z2

[He dx j dz + lR (x - a).J(1- x)(x + l)e31r' + "12 (z - a)v'f"=Z2

r1+e dx j dz + A-. (x - a).J(1 - X)(X + l)e21r' + 1'3 (z - a)Vf=Z2·

Since e-1r• = e31r', the third and fifth integrals on the right side of (9.2) are zero. Analogously we obtain that the integrals on /1 and /2 give zero since the functions under integrals are two different branches with opposite signs.

Since

11 (z-a~nl < r21f IRehz dtl

10 Re-t - lall hjl - R2e2.tl 2'1rR 1

< IR -lall J(R - 1)(R + 1)

2'1rR < IR -lall(R -1)'

the integral on K tends zero as R -+ 00.

If in the preceding inequality we take the path /3 instead of K and e instead of R we obtain again a true inequality, which implies

1 f(z)dz-+O as e-+O. 'Y3

Therefore by (9.1) and (9.2) we obtain for R -+ 00 and e -+ 0,

11 dx 2 ~ = 2'1rz Res(f(z))z=a.

-1 (x-a) l-x2

Since

Res(f(z))z=a = (h) , we obtain for a < -1 that I = 'Ir va2 -1' 1 - Z2 z=a

smce

arg (1 + a) = 'Ir, arg (1 - a) = 0,

and so

~ = J(a2 - l)e1f••

In the case a = zy we have on the upper side of the cutting the positive branch and on the down side the negative branch of the function~, and so we obtain

I 'lrZ • = ~slgnny.

vI + y-

In the case a = et., 0 < t < 'Ir, Figure 9.5,

we have

1= 'Ir

VI - e2t'

y

Figure 9.5

/ 1 ;1

I I

eh / 2e-1r./4 V2 sin t 'Ir e,(31r-2t)/4.

V2 sin t

239

x

In the case a = _et • for 0 < t < 'Ir, we put in the preceding case a = e·(t+1r) and we obtain

I = _ 'Ir e'(311"-2t)/4.

V2 sin t


In the case a > 1 we take the path L as in Figure 9.6,

-R R x

Figure 9.6

and we obtain in an analogous way as earlier

1= 7r Res(J(z))z==a, a> 1.

Since for a > 1,

arg (1 + a) = 0, arg (1 - a) = -7r,

we have

1=- 7r . y'ä2=t

For -1 < a < 1, we take the path as in Figure 9.7,

-R

and we obtain

,"y I

Figure 9.7

vp {lI dx - 0 (x - a)Vf=X2 - ,

241

R x

since by the analyticity of the function j in the region bounded by the path L we have

f j(z)dz+ r j(z)dz=O. 1s1 1s2

Remark 1. The path L for the case a < -1 can also be taken as in Figure 9.8,

-R R x

Figure 9.8

and also analogously for a > 1.

Remark 2. The following general theorem holds:

Theorem. If:

(i) the function J analytic in the complex plane with singular points ZI, Z2,.·. ,Zn;

(ii) the function J is analytic for a < z < b;

(iii) lim ZT+s+IJ(Z) = A f- = (Irl < 1, Isl < 1), then z---+oo

l b(x - s)'(b - x)J(x) dx = si:'lrs (E Res(z - a)'(b - z)'J(Z))Z=Zk - A) . The condition (ii) can be relaxed to:

(ii)' If the function J has finitely many singular points aI, a2, ... ,am on a < Z < b,

then V P l (x - a)'{b - x)' J(x) dx

'Ir n m

= -.- L Res((z-a)'{b-zYJ(z))Z=Zk+Cot'lrsL Res((z-a)'(b-z)sJ(z)L=Uk· SIn 'Ir S k=I k=I

2. We can write z 1

w=--=-l+--. 1-z 1-z


Then starting from the z-plane we map the disc Izl ~ 1, Figure 9.9,

1

Figure 9.9

first by the function Wl = 1- z (rotate for 7r and translate for 1), and we obtain the disc IWl - 11::; 1, Figure 9.10.

v,

o

Figure 9.10

Further, we map by W2 = l/Wl and we obtain the desired region Rew2 ;::: 1/2, where W2 : 0 I-t 00, 2 I-t 1/2, and the border li ne have to be normal on the real


axis, see Figure 9.11.

01

Figure 9.11

Finally, we take w = -1 + W2, translation for -1, and so the desired region is the half-plane

Rew;::: -1/2,

Figure 9.12.

Figure 9.12

The inverse function is given by

w Z=--.

1 +w

245

We have z = -1 w = 1 - -1 1 . So we ean in an analogous way as for the previous +w +w mapping find the images of dis es Iwl :$ r, 0 < r < 00.

Second method. Sinee I-z-I = Iwl :$ r, we have for z = x + zy and so 1-z

We have for r < 1

and for r > 1 :

(x + 1 ~2r2) 2 + y2 2: (1 ~2r2)2. We have for r = 1 the image x = 1/2.

1 2

Figure 9.13

00

3. Since J(z) = L anzn is the canonical element of the analytic function F with n=O

the radius of convergence R = 1,and we have by 2. that the unit disc Izl < 1 is

mapped by the function w = _z_ on the region Rew > -1/2, Figure 9.13, we find 1-z

that the dosest positive singular point for the canonical element g( w) (= J(z)) for G(w) is the point w = -1/2. Therefore r ~ 1/2, z = -l.

Now we shall ex amine how the analytic function 9 (_Z_) analyticallyextend 1-z

the function J from the region Izl < l.

(i) Let 1/2 < r < 1. By 2. the disc Iwl < r is mapped by the function z = ~ 1 +w

on the disc

( x + 1 ~2 r 2 ) 2 + y2 < (1 ~2r2)2' which gives us the region of the analytic extension of the function J by the function

9 (_Z_) , Figure 9.14. 1-z

--~~~~~~~~---~ X

Figure 9.14

If the set of singularities of J( z) = g( w) are not dense on the new cirde we can continue with the analytic continuation.

(ii) Let r = 1. By 2 the disc Iwl < r is mapped by the function z = ~ on the l+w

region Re z < 1/2, which gives us the region of analytic continuation of the function

f by the function 9 (_Z_) , Figure 9.15. 1- z

Figure 9.15

247

1 x

If the set of all singular points of the function f (z) = 9 (w) is not dense on the circle obtained the procedure of the analytic continuation can be continued.

iii) Let r > 1. By 2. the disc Iwl < r is mapped by the function z = ~ on 1 +w

the region


which gives us the region of analytic continuation, Figure 9.16.

+y

_ .-e:. 1_ r'l

Figure 9.16

For further analytic continuation it holds the same as for the previous cases.

Example 9.12 1) Let f be a mtional function which has on the positive part of real axis only poles of first order ih, b2 , ••• , bm • If there exist other poles, denoted by ab a2, ... , an, they are all different from zero. Let p be areal number such that

limzPHf(z) = lim zP+lf(z) =0. %-+0 z--+oo

Prove that for p real which is not integer

VP 10'>0 xPf(x)dx = __ 'lr_ e-1rp•

sin 'lrp m

n

2: Res(zP f(z))z=ak k=l

-'Ir cohp 2: 14Res(f(Z))z=bk, k=l

where xP > 0 for x > o. Prove that for pinteger

[00 n m

VP 10 xPf(x)dx = - 2: Res(zPf(z)Ln z)z=ak - 2: 14(lnbk + n) Res(f(z))z=bk' o k=l k=l

where Ln z = In z + z arg z, 0 ~ arg z < 21l". 00

2) Let g(z) = :LAiZ;, where ;=1

k-l

Al = 1, Ak = :LA;Ak- l + 1, k = 2,3, ....

The funciion h is given by

;=1

Z h 2(z) - h(z) + -- = o.

1-z

249

Prove that one branch of the funciion h is an analytic extension of the funciion 9 and find from which region to which region.

S) Map the region -1l"/n < argz < 1l"/n, Izl < 1, by thefunciion

z w(z) = 2' nE N,w(z) > 0 for z> o.

(1 + zn)n

Solution. 1) i) Let p is areal number different of a integer. The function under the integral is f(z)zP and the path L as in Figure 9.17, where R is enough big that all poles of f ; ak, k = 1,2, ... , n and bk , k = 1,2, ... , m, will be inside. The points z = 0 and z = 00 are branching points of zP (p is not an integer) and we have a cut from 0 to 00.


Using the partition of the path L from Figure 9.17 we represent

y

-R R x

Figure 9.17

the integral on the path L in the following way

kzPJ(z)dz =

where fi is the upper semi-circle and "ti is the lower semi-circle around bio

Because of lim zpH J(z) = 0 we have z-+oo

liR

zP J(z) dzl S; RPH L21r IJ(Re9')1 dO -+ 0 as R -+ 00

(exchange of the limit and integral). We obtain in an analogous way

{ zP J( z )dz -+ 0, as c:-+ 0, lK,

since limz ..... o Zp+1 J( z) = O. Since the points bk, k = 1,2, ... , m, are poles of the first order we have for € -+ 0

1 zP J(z) dz = -7rlll,;Res(J(Z))z=bk' Ik

and on lower semi-circles

1 zP J(z) dz = -7rZ eP(lnb k+2,n)Res(J(z))z=bk Ik

By Residue Theorem we have

1, zP J(z) dz = 27rZ t Res(zP J(Z))z=ak' L k=l

Therefore we have for € -+ 0 and R -+ 00

(1 - e27r')V P 1000 J(x) dx = 27rZ t Res(zP J(Z))z=ak k=l

m

+7rz(l + e27rp,) L 14Res(J(z))z=bk' k=l

Then

v P 1000 xP J ( x) dx

m

-7r cot7r L lI,;Res(J(z))z=bk' k=l

where xP > 0 for x > O.

1) ii) pis an integer. We apply the integral on the function 'ljJ(z) = zP J(z)Lnz on the same path L, where Ln z = In Izl + zarg z, 0::; arg z < 27r. Using the partition of the path we obtain

i zPJ(z)Lnz dz = tl-~ xPJ(x)lnx dx

+f LzPJ(z)Lnzdz+ 'Y:1 l b'+I-<xPJ(x)lnxdx i=1 I. ;=1 b.+<

+ [R xPJ(x) lnxdx + [ zPJ(z)Lnz dz lbm+< lKR

+ l bm+< xP J(x)(ln x + 27rz) dx + f 1 zP J(z)Lnz dx R i=l "

+I:1 [bi-< xPJ(x)(lnx+27rz)dx+ [ zPJ(z)Lnzdz. i=1 lbi+I+< lK,


We have by supposition zP+1J(z) -t 0 as z -t 0 and also as z -t 00. Therefore also zP J(z)Ln z -t o. Hence the integrals on ](R and ](. tends to zero as R -t 00 and E -t 0, respectively.

We have at poles of first order bk , k = 1,2, ... , m, on the upper semi-circles as c-tO

1 zPf(z)Lnz dz = -7fZ bpnbkRes(J(z))z=bk' "Ik

and on the lower semi-circles

~ zP J(z)Ln z dz = (-7fZ b~ + 27fz)Res(J(z))z=bk • Jik

By Residue Theorem we have

1 zPf(z)Lnz dz = 27fZ tRes(zPf(z)Lnz)z=ak. L k=1

Letting R -t 00 and E -t 0 we obtain

n

-27fZVP lX> xPJ(x)dx = 27fZ LRes(zpJ(z)Lnz)z=ak k=1

m m

+27fZ L Res(J(Z))z=bk~ lnbk - 27f2 L ~Res(J(Z))z=bk' k=1 k=1

which gives us the desired equality. 00

2. Let h(z) = L aif Then i=1

implies

Hence

h2(z) _ h(z) + _z_ = 0 1-z

00 00

h(z)h(z) + Lzi = Laizi, i=1 ;=1

k-1 a1 = 1, ak = L ajak_j + l.

;=1

Therefore the power series of the functions h (one branch) and gare equal.

We find h solving (9.2)

(9.2)

The branching points of the function h are 1/5 and 1. Then the radius of convergence of the power series of h, and also for g, is 1/5, we have

00

h(z) = g(z) = E Aii for Izl < 1/5. ;=1

The function h from (9.2) is an analytic extension of 9 on the whole complex plane out of the cut from 1/5 to 1.

3) The given function can be written in the following form

Remark. If n is odd the function w = w( z) has n branching points of (n - 1)-th order in the points

z = (e,,·/n) k, k = 1,2, ... , n.

For n even, n = 2m, w(z) has the representation by two function

m wm _± _z __

1,2 - 1 + z2m

When w(z) > 0 for z > 0 we have the behavior of the maps in Figures 18 and 19.

y

•

o

Figure 9.18


o

Figure 9.19

Chapter 10

Integral transfornns

10.1 Analytic Functions Defined by Integrals

10.1.1 Preliminaries

Let 0 be a region in rc. We denote by A(O) the set of all analytic functions in O.

Definition 10.1 A sequence {fn} from A(O) converges in A(O) to f E A(O) if for every closed subsei F of 0 and every c > 0 there exists no E N such that

max Ifn(z) - f(z)1 < c zEF

for every n ~ no.

Theorem 10.2 If a sequence {fn} from A( 0) converges uniformlyon every closed and bounded subset of 0, then the limit function is also regular on O.

Theorem 10.3 If a family {fq}qEQ, Q C lW. from A(O) converges in A(O) to f E A(O) as q ~ qo, then the family {f~}qEQ converges to f' in A(O).

Theorem 10.4 Let 0 be a simple connected region in C and 9 = g( z, t), 9 : 0 X L ~ C, for L : Re z ~ a, Im z = 0 satisfies the following conditions: (i) 9 is a continuous function of t on for fixed z E 0; (ii) 9 is an analytic function of z E 0 for fixed t; (iii) the family {f: g(z, t) dth~a converges in A(O). Then the function

f(z) = t g(z, t) dt

255

256 CHAPTER 10. INTEGRAL TRANSFORMS

is analytic in 0 and we have

J'(z) = l b g(z, t) dt. a Bt

Theorem 10.5 Let A be a region in C and 9 = g(z, t),g : A x L -+ C, be a continuous function of t onto L : a ::; t ::; b, for fixed z and an analytic function of z E A for fixed t. Then the function

f(z) = t g(z, t) dt

is analytic in A and we have

J'(z) = t g(~/) dt.


Example 10.1 Prove that the gamma function

r(z) = 1'XJ e-t tz - 1 dt

for Re z > 0 is an analytic function.

Solution. The given integral is improper because of the point 0 and the infinite interval. Therefore we shall write

100 e-tt z - 1 dt = 11 e-tt z - 1 dt + 100 e-tt z - 1 dt.

The integral 11 e-tt Z - 1dt converges absolutely for Re z > 0 and uniformlyon every

bounded and closed region G in Re z > o. Namely, if mi[l x = 8 for z = x + zy, we zEO

have !e- ttZ - 1 ! = e-ttRez-l ::; to- 1 for z E G.

Since the integral 11 tO- 1dt converges, then by Weiersrtass' criterion follows the uni

form convergence of the integral on G. The integral 11 e-te- 1dt is then an analytic

function for Re z > o. On the other hand 100 e- te-1 dt converges uniformly and

absolutelyon every bounded region G. Namely, if Gis in the disc Izl ::; M, then for z E G,

! -ttZ-1! < -t M+1 e _ e e ,

10.1. ANALYTIC FUNCTIONS DEFINED BY INTEGRALS 257

and the integral 100 e-ttM-1dt converges, and therefore by Weierstrass' criterion the

integral 100 e-te-1dt is an analytic function in the whole complex plane.

So it follows that the gamma function r is a regular function in half-plane Re z > o.

Example 10.2 Prove that the gamma function r(z) satisfies

a) r(z + 1) = z r(z) for Rez > 0;

b) r(n + 1) = n! for nE N.

Then extend analytically the function r outside of Re z > o.

Solution. Applying the partial integration on r(z) = 100 e-te-1dt for Rez > 0

we obtain

r(z) 100 e-ttz- 1 dt

= e_t~ICO + ~ (CO e-t t(z+1)-l dt z 0 z Jo

= r(z + 1)

z

b) Putting z = n and applying a) we obtain

r(n + 1) = n(n - l)(n - 2)···3·2· r(I).

Since r(l) = 100 e-tdt = 1, we obtain r(n + 1) = n! for n E N.

We can find the analytic extension of r by the equality in a). Namely, we have

for -1 < Rez < 0 that r(z) = r(z + 1) is an analytic function, since Re (z + 1) > z

O,Rez> -1. Further, we have for -2< Rez ~ -1 without -1

r(z) = r(z + 1) = r(z + 1). z z(z+l)

So for the general case -( n + 1) < Re z ~ -n without -n we have

r(z)= r(z+(n-l)) . z(z + 1) ... (z + n)

In this way we obtain the analytic continuation of r on the whole complex plane without the poles at z = 0, -1, -2, ... For those points we have

Remark. The gamma function is a special case of the Mellin transform. This transform is defined for complex functions of real variable <p(t) which satisfy the condition

1<p(t)1 ~ cr'" for 0< t < 1; 1<p(t)1 ~ c1Cß for t> 1, and a < ß.

Then the Mellin integral transform for the function <p(t) is given by

4>(z) = 1'''' <p(t)tZ-

1 for a< Rez < ß·

Example 10.3 Prove that the beta funciion

B(p, q) = laoo tP- 1(1 - t)q-l dt

/or Re p > 0 and Re q > 0 satisfies

a) B(p, q) = B(q,p); b)B(p, q) = r(p)f(q). f(p + q)

Solution. a) Putting t = 1 - u we obtain

1 1

B(p, q) = la tP- 1(1 - t)q-l dt = J uq-1(1 - U)p-l du = B(p, q). o

b) We take in f(z) = laoo e-te-1dt the substitution t = x2 (and y2). Therefore

for z = p and z = q we obtain

(OO 2 f(p) = 2 10 e-X X 2p- 1 dx,

(OO 2 f(q) = 2 10 e-Y y2q-l dy,

respectively.

Taking the polar coordinates:

x=pcos<p, y=psin<p, O<p<oo, O<<p<1f/2,

8(x,y) . and so 8(p, <p) = p, we obtam

f(p)f(q) = 4100 laoo e-(x2+y2)x2p-ly2q-l dxdy

= 4 100 la7r/2 e-p2 p2P+2q-2p( cos <p ?p-l(sin <p )2q-l d<pdp

= 2 100 e- p2 p2P+2Q-l dp· 2la7r/2 (cos <p )2P-l(sin <p )2q-l d<p

= f(p + q)B(p, q),

10.1. ANALYTIC FUNCTIONS DEFINED BY INTEGRALS

where we have taken t = cos2 cp in

B(p, q) = la1 tp-1(I - t)q-l dt,

B(p, q) = 2la f (cos cp?p-l(sin cp)2q-l dcp.

Example 10.4 Solve the Bessel differential equation 0/ n-th order

Z2W" + zw' + (Z2 - n2)w = 0,

when n is not an integer.

259

Solution. We shall solve the differential equation in the reals and then using analytic continuation we shall obtain the complex solution. So we shall find the solution of

x 2y" + xy' + (x 2 - n 2 )y = 0

in the following power series form

(10.1 )

Putting this series in (10.1) and taking the coefficients of the same power of xequal to zero

(2n + I)al = 0, (2n + k)ak + ak-2 = 0, k = 2,3, ....

Hence the coefficients by odd powers are zero and even coefficients are given by

(-I)kao a2k = 22kk!(n + I)(n + 2) ... (n + k) for k = 1,2, ...

Putting ao = 2nr(~ + 1)' we obtain a particular solution of the equation (10.1) for

n E ~ (n can be also integer):

x n 00 1 (X)2k Jn(x) = (2) E(-I)k k!r(n + k + 1) 2

Extending the obtained solution on the whole complex plane (the series has radius of convergence 00) we obtain the general solution of the Bessel differential equation

Z2W" + zw' + (Z2 - n2)w = 0 (n is not an integer).

It is given by

260

for A,B E C.

The function

CHAPTER 10. INTEGRAL TRANSFORMS

( z)n 00 k 1 (Z)2k Jn(z) = 2 {;( -1) k!r(n + k + 1) 2

is the Bessel function of first order with index n. Why is the general solution not a solution for n integer?

Example 10.5 Prave that

00

a) ez(t-l/t)/2 = E Jn(z)tnj n=-oo

Solution. We shall find the Laurent series of ez (t-l/t)/2 in the neighborhood of w=O:

1 1 ez(t-l/t)/2 with U n = - dt.

27rZ Itl=l tn +1 n=-oo

P . . h I' I 2u b' uttmg m t e ast mtegra t = - we 0 tam z

U n = 2~Z (~) n jUI=tlzl eu - z2 /4u u:+l du for n = 0, 1,2, . .. .

By the theorem on residues have

To find the preceding residue we shall start from the following expansion

The essential singularity of the above expansion of eu-z2/4U at point u = 0 will give us the coefficient by u-1 (the residues). Therefore

( z)n ~ k 1 (Z)2k un = 2 f:'o(-l) k!r(n+k+1) 2 =Jn(z), n=0,1,2, ...

It is easy to check that

Ln(z) = (-ltJn (z) for n=0,1,2, ....


b) By the preceding a)

00

ez (t-l/t)/2 = L Jn(z)tn. n=-oo

Then

ez (t-l/t)/2 (~ (t + ~) )

= n=-oo

Hence 00 00 00

n=-oo n=-oo n=-oo

Remark. The function ez (t-l/t)/2 is the generator of the Bessel function Jn(z).

Exercise 10.6 Let us consider the Wright function in the integral form

where u-a is the principal branch 0: > -1, and the path C is a on Figure 10.1

y

x

Figure 10.1

(starting at -00 on the real axis, going around (0,0) in the positive direction and coming back to -00).


Prove that: If 0: > 0, ß > 0, then the Wright function has the following representation in every finite part of the complex plane

1 lXo+,oo -Q

cI>(ß, a, z) = - w-ß eW +z .w dw for 27rZ Xo-'OO

Xo > 0,

where l XO+,y lxo+.oo lim = .

y--+oo xo-"Y xo-too

Hints.Use the path C' from Figure 10.2.

y ___ .0-----.. E

w

A.

~ F

Figure 10.2

Remark. The Wright function has the following power series representation

00 zk

cI>(ß, 0:, z) = E f(k + l)f(ß + ak)'

where ß is a complex number and 0: > -1 (see E.M. Wright: On the coefficients of power series having exponential singularities, J. Lond. Math. Soc. S., 1953, 71-79).

Exercise 10.7 Prove that the gamma function f can be extended over the whole region of the existence by the equality

(( _wY-1 = e(z-l)ln(-w)), where C is the path given on Figure 10.3.

y

c

x

Figure 10.3

Example 10.8 Let 0 and a E ~ and <pet) = 0, for t < O. The Laplace transform of 'P is given by

4>(z) = 100 <p(t)e-zt dt,

(if it exists, generally in the sense of the Lebesgue integral) which is an analytic function in the half-plane Re z > Q.

Find the Laplace transforms of the following functions

a) <pet) = 1 for t;::: 0; b)<p(t) = ekt for t;::: 0; c) <pet) = tn - 1 for t;::: 0, n > 0; d)<p(t) = 1 - ate-at, t > -a, a E ~;

t;::: -a, a E~. e) <pet) = 1 - (1 + at + a2t2 /2)e-at for

Solution. As usual, we write 4>(z) = L(<p(t)).

a) 4>(x) = {OO e-xt dt = ~ for x> O. We obtain by the analytic extension Jo x

100 1 4>(z) = e-zt dt = - for Re z > 0,

o z

L(I) = l/z.

b) 4>(x) = ['>0 e-xtektdt = {OO e(k-x)tdt = _1_ for x > k. Jo Jo x - k

With the analytic continuation we obtain L(ekt ) = ~k for Rez > k. z-

c) In an analogous way we obtain

a2

cl) We have L(cp) = ( )2 for Rez > -a zz+a

a3

e) We have L(cp) = ( )3 for Rez > -a. zz+a

Example 10.9 Prave that if for the function cp there exist cp(n)(t) and l cp(t)dt then.

a) L( cp(n)(t)) = zn L( cp(t)) - zn-lcp(O) _ zn-2cp'(0) _ ... _ cp(n-l)(O).

b) L (l cp(t) dt) = L(:(t)).

Solution. The proof goes by incluction. Let n = 1. Then

L(cp'(t)) 10'00 cp'(t)e-zt dt

cp(t)e-ztl: +z 100 cp(t)e-ztdt

zL(cp(t)) - cp(+O).

Supposing now that a) is true for n = k we can prove a) in an analogous way for n=k+1.

b) Using a) we obtain

L ((l CP(t)dt),) = L(cp(t)) = zL (l cp(t)dt) ,

L (l cp(t) dt) = L(:(t)).

Example 10.10 Let cp(t) = L-1(<P(z)) be the inverse Laplace transform. Find with it the solution of the following differential equation

w'(z) + aw(z) = a(l - aze-az _ e-az ),

for w(O) = 0, a E llt.


Solution. We shall solve first the equation over the reals and then by analytic extension we shall obtain the its solution in the complex plane. We start from the problem

y'(t) + a y(t) = a(l - at e-at - e-at ), y(O) = O.

Applying the Laplace transform and using Example 10.9 and Example 10.8 d)

L(y' + a; y) = aL(l - ate-at _ e-at ),

L(y') + aL(y) = aL(1 - ate-at _ e-at ).

Further , we have a3

xL(y) - y(O) + aL(y) = ( )2' xx+a

Hence a3

(x + a)L(y) = ( )2' xx+a

a3

L(y) = x(x + a)3

Applying now the inverse Laplace transform L-1 we obtain

Since by Example 10.8 e)

we have y(t) = 1 - (1 + a + at + a2t2/2)e-at .

With the analytic extension we obtain the complex solution

w(z) = 1 - (1 + az + a2z 2 /2)e- az for Rez> -a.


F(n)(z) = L((-ttcp(t)),

for n E N. Using this result find a particular solution of the Bessel differential equation

Z2W " + zw' + (Z2 - n 2 )w = 0,

with w(O) = 0, for n E NU {O}.


Solution. By the uniform convergence of the integral we can exchange the derivative and the integral

F'(z) = 1000 e-zt ( -tcp(t)) dt = L(-tcp(t)).

Then the proof follows by the induction.

We shall find now a particular solution of the differential equation in real

We put u(t) = tny(t), and then we obtain

tu"(t) + (1 - 2n)u'(t) + tu(t) = 0, u(O) = o.

Applying the Laplace transform on the preceding equation, taking

we obtain

dF F(x) = L(u(t)), L(tu(t)) = - dx' L(u'(t)) = xF(x),

L(u"(t)) = x2 F(x) - u'(O), L(tu"(t)) = _~(x2 F(x) - u'(O)), dx

(x 2 + I)F'(x) + (2n + l)xF(x) = O.

The general solution with respect to F is given by

( 1 ) -(n+I/2)

F(x) = C(x2 + 1)-n-1/2 = CX2n- 1 1 + x2 '

where C is an arbitrary real constant.

By the power series expansion we have

L(u(t)) = F(x) = n!C ~ (-I)k (2n + 2k)! (2n)! t:o 22kk!(n + k)! X2n+2k+I .

Applying the inverse Laplace transform we obtain

_L-1(n!C 00 (_I)k (2n+2k)!) u(t) - (2n)!Ek!(n+k)! x2nHk+1

n!C 00 (_I)k -1 ((2n+2k)!) = (2n)! E 22k k!(n + k)!L X2n+2k+1'

Since L-1 ((2n + 2k)!) _ 2n+2k . h () m!

X2n+2k+1 - t ,wIt L tm = Xn+1'


we obtain {or n!C 1 (2n)! = 2n' {or C = 1 :

(t) _ ~ ( -1 )k t2n+2k u - f:'o 2"k!(n + k)! .

Since it was u(t) = tny(t), we have

00 ( -1 )k (t) n+2k y(t) = E k!(n + k)! 2" .

With the analytic continuation we obtain the complex solution

00 (_l)k (Z)n+2k w(z) = L klr( k ) -2 = Jn(z) {or Rez > O.

k=o·n++1

Exercise 10.12 Let f be a function defined by 1000 e-ztet sin etdt.

a) Find the region where f is an analytic function.

b) Extend analytically the function f on the whole half-plane Re z > -1.

Solution. a) Re z > O.

b) Putting et = x the given integral reduces to f(z) = [00 sinx dx {or X Z = ezlnx . J1 XZ

Applying the partial integration we obtain

[00 cos x f(z) = cos 1 - A x z+1 dx.

The last integral converges {or Re z > -1.

Exercise 10.13 The Riemann zeta fundion is given by

00 1 00

((z) = L --; = L e-zlnn • n=1 n n=1

a) Expand the function (z) in a Taylor series in a neighborhood of z = 2 and find the corresponding radius of convergence.

b) Prove that for Re z > 1

1 [00 wz - 1

((z) = r(z) Jo ew - 1 dw.

c) Extend analytically the fundion ((z) on the whole complex plane without the point z = 1.

Answers.

and 00 1 11'2

UO=2:-=-. k=l k2 6

The radius of the convergence is R = 1.

b) and c) For the analytic extension consider the integral

1 (_WY-l ~-'--dw,

c eW -1

where C is the path given on Figure 10.3.


Example 10.14 1. Find lor the function

its singular points and their nature lor different values 01 the complex parameter a. For a = 2 expand the given lunction in apower senes in the neighborhood of z = 1. Find the disc 01 convergence.

2. The lunction f is given by

Prove: 2 a) f can be represented by two senes

00 00

fez) = 2: Un(Z) + 2: vn(z) for Rez > 0, n=l n=l

where

2b) Both senes converges uniformly in every closed region 01 the form

0< Xo:::; Rez:::; Xl< 00,

and r is an analytie function at the right half - plane.

3) Starting from the integral

where L is the path in Figure 10.4,

~y I

I ,

I

Figure 10.4

represent the following two integrals by r -sine and -eosine functions given by

1= ca cos t dt, 1= ca sin t dt for 0 < a < 1,

respeetively, and prove that they exist.

Solution. 1. (i) a E M. Then the point Zl = 0 is a pole of order a.

(ii) a is a negative integer. The point Z2 = 00 is an essential singularity.

269

(iii) ais a rational number, i.e., a = E., where (p, q) = 1. Then points Zl = 0 and q

Z2 = 00 are branching points of order q.

(iv) ais a rational number or a complex number al + W2 (a2 :j:. 0).

Then the points Zl = 0 and Z2 = 00 are branching points of infinite order.

For a = 2 the given function is J(z) = -i-. The expansion in the neighborhood at Z eZ


z = u (u i= 0) in power series we obtain in the following way. Let w = z - u, z = w+u. Then

On the other side we have

1 1, 1 1 00 (-1 )nwn A = Z2 = (w + U)2 = U (w); u(w) = - w + u = -;: E un '

and so

Then for u = 1

1 = (E(-l t (n + l)(z -lt ) (e-1 E (_l)n~ _l)n)

00 n ( 1)" e-1 L L T,(-l)n-k(n - k + l)(z _l)n

n=O k=O .

00 n 1 = e-1L L k,(-lt(z- l t(n-k+1).

n=O k=O .

The radius of convergence is Rand

R ~ min[R(A), R(B)] = 1.

2. a) We have for Rez > 0

r(z) fooo e- ttz- 1 dt

= fo1 e-t tz- 1 dt + 100 e-te-1 dt

100 e- t tz- 1 dt + 100 e-~u(z+1) du,

1 where we have taken u = t'

Then

and

So we obtain 00 00

f(z) = L un(z) + L vn(z) for Rez > o. n=O n=O

2. b) We shall prove that both series from 2. a) converges uniformly in every closed region: 0 < Xo :::; Re z :::; Xl < 00.

We have

Since the series Ee-n(n+1Y' _nX1

n=O Xl

00

is convergent we have by Weierstrass criterion that L Un(Z) is absolutely convergent n=l

for 0 < Re z :::; Xl. what implies its uniform convergence on the given closed regions. We obtain in an analogous way

Since the series

Ivn(z)1 Iln+1 e-l/tr(z+l) dtl

< e- l /(n+1) l n+1 C(xo+l) dt

= e- l /(n+1) (n + 1)-"'0 - n-XO

-xo

E e-l/(n+l) (n + l)-XO - nXO

n=l -x 00

is convergent we have by Weierstrass' criterion that the series L vn(z) is abso-n=l

lutely convergent for 0 < Xo :::; Rez, which implies the uniform convergence in the previously mentioned closed region.

By Theorem 10.2 we have {(vn } C A(O), n E N) and Ek=l Vk(Z) converges uniformlyon every closed and bounded subset of 0, then E:=l Vn E A(O); and «un ) c A(O), n E N) and Ek=l Uk(Z) converges uniformlyon every closed and bounded subset of 0, then E:=l u" E A(O). Then by 2. a) we have

(Eun E A(O) and ~vn E A(O))::::} r = Eun + EVn E A(O), i.e. ,

r is an analytic function on 0 = {z I Rez > O}.

3. See next Example 10.15.

e'" Example 10.15 I) Let F(z) = -. zP

1. Find the zeros, singular points and periods of F for different values of the parameter p. For p = 2, find the real and the imaginary parts of F.

2. Let 0 r > 0. These points make a closed path P which consists of straight parts AB,BC,CD and a part DA on a circle with the center (0,0), Figure 10.5.

A B x Figure 10.5

Using the path P find the integrals

100 cos x 100 sin x --dx and --dx, o xP 0 xP

i ~nowing that 1000 e- IIyP-1dy = r(p) (gamma function).


3. Using 2. find also the following integrals

and prove that they exist.

II.) Let fand 9 be two given complex functions such that f(a) =f 0 and 9 has a zero of second order at z = a.

/1.1) Find the residues of L at the point z = a with the values of these functions 9

and their derivatives at z = a.

II.2) Find the coefficients with negative indices in the Laurent series ofthe func

tion L using the values of fand 9 at the point z = a. 9

Z2 elZ

II.3) Apply the results from 1) and 2) on the function (2 )2' Z + 1

Solution. eiz

1. (i) For pE N the function F(z) = - has a pole of the order p at the point zp z = 0, since

e'z 1 00 znzn 00 1 -;; = zp ~ --;:J = n~p (n + p)!zn+pzn.

It has no zeroes since e'Z f. 0 for all z E C and for z = 00 it has an essential singularity.

1. (ii) For p = -k (k E N) the function F has a zero of the order k at z = 0, and for z = 00 it has an essential singularity.

1. (iii) For p = 0 the function F(z) = elZ has an essential singularity at z = 00.

It is periodic with the period 271".

1. (iv) For p = ~, where k and n are integers different of zero without common n

divisors, then z = 0 is an algebraic branching point of the order n.

1. (v) For p an irrational number the point z = 0 is a transcendental branching point.

1. (vi) For p = Pt + ZP2 (P2 f. 0) be a complex number the point z = 0 is an infinite branching point of F.

For p = 2 we have for z = x + zy

e'Z e'("'+'Y) e-Ye''''(x2 - y2 - 2zxy) F(z) = -z2 = (x + .y)2 = (2 2)2 4 2 2 • x-y +xy

Hence -y

Re F(z) = (x2: y2)2 (x 2 - y2) cos X + 2xy sin x),

12. The integral ~ dz for 0 < p < 1, can be written in the following form i .z

p zP

i eU i R e'X 11 e,(,Rt+(I-t)R) (zR - R)dt -dz= -dx-;/-

pzP T xP 0 (zRt+(I-t)R)p

iT e-Y 10 e'T e9'zre9'dO +z --dy+ . R (zy)p f rPe,9p (10.2)

The function e'Z has no singularities in the region bounded by the path P and p

therefore by Cauchy theorem we have

i e'Z -dz = O.

p zP

Letting R -+ 00 the integral on the path Be tends to zero, since

I [1 e,(,Rt+(I-t)R) (zR - R) dt I < [1 e-Rt ..,j2Rdt Jo (zRt + (1 - t)R)p Jo (zRt + (1 - t)R)p

as R -+ 00, where we have used that

< [1 e-Rt .j2Rdt Jo RP:iiB.

2

= 2R-P 11 e-Rt dt

-+ 0

IzRt + (1 - t)RI > min IzRt + (1 - t)RI = .j222R -09:9

(10.3)

_ e'z for 0 :::; t :::; 1. The integral on the path DA tends to zero as r -+ 0, since z zP -+ 0

as z -+ 0 for 0 < P < 1.

Therefore by 10.2 and 10.3 we have for R -+ 00 and r -+ 0

VP100 cosx + zsinx d VP1°O e-Y d x- z- y=O. o xP 0 zPyP

10.2. GOMPOSITE EXAMPLES 275

Since the integrals in the previous equality converge absolutely we can ornit the symbol V P. Since

we obtain

100 cos x + z sin x d ( 7fp . 7fP )r(1 ) x = z cos - - z sm - - p . o xP 2 2

Therefore 100 cos x . 7rp -- dx = sm _. r(l- p)

o x P 2 and 100 sin x 7rp

-- dx = cos - . r(1 - p), o xP 2

for 0 Z

Remark. By the analyticity of the function - outside of the point z = 0, it zP

was possible to interchange the path BG by one on circle ReB• (0 ~ () ~ 27r), and then to prove that the integral an BG tends to zero as R ---7 00.

3. We easily obtain the desired integrals putting p = ! in 2.

100 cos x 100 7r ( 1 ) r,:;; dx = 2 cos e dt = sin - . r - , o yX 0 4 2

where we have taken x = t 2 • Therefore

100 y'2 ,j2i COS x 2 dx == -r(1/2) = --

o 4 4

and

The convergence of the first integral follows by

1"" 00 jV(k+ 1)1f/2 1 00 1(k+1)1f/2 COS x cosx2 dx = L cosx 2 dx = - L r:::: dx.

o k=O Vkrr/2 2 k=O krr/2 Y X

Since this series converges by Leibnitz criteria since the coefficient

1(k+ 1 )1f/2 cos x --dx

krr/2 yX

converges monotonically zero as k ---7 00 and the sign is alternatively changed~

We can prove analogously the existence of the integral 100 sin x 2 dx.


11.1) The function F(z) = :~;j has at point z = a a pole of second order, since

9 has a zero of second order at z = a and J(a) i= O.

Therefore

R (F( » _ = ( _ )2J(Z»)' = (f'(Z)h(Z) - J(Z)h'(Z») es z z_a Z a (z) h2(z) ,

9 z=a z=a

where g(z) = (z - a)2h(z), h(a) i= 0, since 9 has zero and order u at z = a. We find by the last equality

Therefore

g'(z) 2(z - a)h(z) + (z - a?h'(z), g"(z) 2h(z) + 4(z - a)h'(z) + (z - a?h"(z), g"'(z) 6h'(z) + 6(z - a)h"(z) + (z - a)2h"'(z).

h(a) = g"~z), h'(a) _ g"'(z) - 6 .

Putting the obtained results in residues of F we obtain

Res(F(z»z-a = 2f'(a) _ 2J(a)g"'(z). - g"(a) 3(g"(a»2

11 2. Since F has a pole of second order at z = a, the Laurent series of this function has the form

00

F(z) = L un(z - at· n=-2

We shall find U-l and U-2.

We have that 11.1. implies

2f'(a) 2J(a)g"'(a) U-l = Res(F(z»z=a = g"(a) - 3(g"(a»2 .

Further integrating through circle J( with the center at the point a, taking J( inside of the annulus of the analicity of the function F, we have

= 1 [ F(z) dz 2n lK (z - a)-l 1 [ J(z)dz

2n lK (z - a)h(z) 27rz R ( J(z) ) 27rz es (z-a)h(z) z=a


( f(z) ) h(z) + (z - a)h'(z) Z=(l

f(a) h(a) 2f(a) g"(a) ,

277

where we have used the procedure for finding the residue of the fraction of two functions from which the function in denominator has a pole of first order.

2f(a) Hence U_2 = -(-) .

g" a Z2eiz

11.3. For F(z) = (2 )2 we take z + 1

The points Zl = ~, Z2 = -~ are zeros of second order for the function g.

Since f'(z) = ze'Z(2 + zz), g'(z) = 4z(z2 + 1),

and g"(z) = 12z2 + 4, g"'(z) = 24z,

we obtain by 11.1. that

By the same formula we obtain

2f(~) 1 By II.2 we have for z = Z, U-2 = -(-) = - and for z = -Z, U-l = ~e/2, U-2 = e/4. g" Z 4e

Exercise 10.16 Let a function F : IC X IC --+ IC be defined by

w F(z,w) = --.

eW - z

a) Find the branching points and on fixing one variable find the analyticity domain with respect to the other variable.

b) Let q(z) = 10'''' F(z, t) dt. Find the analyticity region of q.

00 n

c) Let u(z) = E ( z )2' Prove that q is an analytic continuation of u(z). n=O n + 1

Specify from which region to which region.

d) Let Cn : (2n + l)11"et., 0::; t ::; 211". Find

X(z) = r F(l,w) dw JCn w(w - z)

(z is inside the region bounded Cn). Find lim X(n). n_oo

e) Prove that w 00 2w2

F(1,w)=1--2 +E 24 22 ' n=l W + n 11"

For which w does the preceding equality hold?

Hints.a) Use the analyticity regions of the elementary functions involvecl.

b) Use Theorem 10.4.

c) Write the function F(z, t) for a fixecl t as power series with respect to z.

cl) Apply Residue Theorem.

Example 10.17 I Let the function f be represented by apower senes f(z) = 00

E anzn with the radius of convergence R = 1, and let the function g be given n=O by

I a) Prove that g is an entire function.

I b) Prove that for Izl < 1

100 e-tg(zt) dt = f(z),

(apply the partial integration n times and then let n _ +00).

II Let the region G be obtained in the following way: through each singular point v of the function f take a normal straight line on Ov. G is the convex region which contains Izl < 1, and with the boundary consisting of described straight lines.

II a) Prove that for z E G, t ;::: 0 :

g(zt) = -21 r f(u)ezt/u du, nJc. u

where Cz is the circle u = z/2 + (lzl/2 + b)e,q, 0:5 q< 211" (b> 0 enough smalI).

II b) For z E G the integral 100 e-tg(zt) dt converges.

III For

00 zn 00 Z _ Z n

f(z) = L - and fl(Z) = 1I"z/4 -ln2/2 + L (-) /4, n=l n n=l 1 - z

prove that they analytically extend each other. Specify from which region to which region. Give the function which cover all analytic extensions of the function J.

00

Solution. I a) Since f(z) = L anzn is apower series with radius of convergence n=O

R = 1 we have 1

RJ = -1 lim lan+11- . n~oo an

00

Since g(z) = L a~ zn we obtain by R J = 1 n=O n.

111 Rg = ----;;---- = ----,..,.-------,.---.....-- = 00.

an+l l' I an+11 1 1m -- lim ---lim (n + I)! n-+oo an n-+oo (n + 1) n~oo an

n!

Hence the function 9 is analytic in the whole complex plane, 9 is an entire function.

I b) To prove that for Izl < 1 we have fooo e-tg(zt) dt = J(z) we introduce a function F by

F(z) = 100 e-tg(zt) dt.

Applying the partial integration on F we obtain

F(z) = -e-tg(zt)l: + z l'XJ e-tg'(zt) dt.

Applying again the partial integration we obtain

F(z) = -e-tg(zt)l: + z( - e-tg'(zt)l: + z 100 g"(zt)e- t dt).

We can conclude that after applying the partial integration n times we shall obtain

Prove this by mathematical induction.

Since

(k) 00 a p(p - 1) ... (p - k + 1) -k 9 (z) = I: p , zp

p=k p.

we obtain for z = 0

(k)(O) _ akk! _ 9 - k! - ak·

Therefore

We shall prove that the second summand on the right hand side tends to zero as n -+ 00. N amely, we have

Since

from laHm I < M for m E N we obtain

Since Izl < 1, the right hand side of the preceding inequality tends to zero as n -+ 00.

Therefore for n -+ 00 we obtain F = f. II a) The circle Cz : u = z/2 + (lzl/2 + b) eq·, 0 ~ q < 27r, Z E G, and small

enough b > 0, completely belongs to convex region G since its center is at the point


z/2 with a radius Izl/2 + b, Figure 10.6.

Figure 10.6

Hence the function J is analytic in the disc bounded by C •. So we have

00

J(z) = E anzn n=O

. 1 1 J(u) wüh an = -2 ~ du.

1I"Z GUn

Therefore

where we have interchanged the order of integration and the series, since the series

(z E G, t 2::: 0) uniformly converges on Cz •

11 b) We shall prove that 100 e-tg(zt)dt converges for z E G and t 2::: o. Since

(z) Izl max Re - = -I -I -b = S < 1, uEe. U Z +

we obtain by 11 a)

Ig(zt)1 < ~ f /f(u)ezt/u/ldul < Aest , 271" Je. lul

where A is areal constant.

Remark. The function F(z) = 1000 e-tg(zt)dt analytically extends the function

f on the whole region G. This is the so called Borel method of extension. So we can apply this method on the following power series a) E:=ozn; b) E:=oz2n; c) E:=oz4n j (given regions Gon Figures 10.7, 10.8 and 10.9, respectively). Find the corresponding functions 9 and F.

Figure 10.7


-1

Figure 10.8

y

Figure 10.9

III 00 n

For the function 1 given by I(z) = E ~ the radius of convergence is R = 1. n=1 n

We shall find the function represented by this power series.

Differentiating the given power series and using its uniform convergence for Izl < 1, we obtain

j'(z) = f nzn-1 = f: zk = _1_. n=1 n k=O 1 - z

Integrating we obtain

I(z) = -ln(l - z).

We represent now the function z f--t In(l - z) by its power series at u = z

-ln(l-z) -ln(l-z-(z-z)) z-z = -ln(l-z)(l-l_z)

-ln(l - z) -ln (1 _ z - Z) 1 - z

= -ln V2e-,n/4 + t .!.(z - zr n=1 n 1 - z

7rZ ~ 1 (z - Z) n - - In 2/2 + L.J - --4 n=1 n 1- z

11 (z).

The new power series converges for

Iz - zl < 11 - zl = V2,

in the disc with center at z, and radius .J2, Figure 10.10.

y

x

Figure 10.10

We have in the region

0= {z Ilzl < 1, Iz - zl < 11 - zl}

that fez) = ft(z) = -ln(1 - z).

Hence the functions fand ft analytically extend each other.

All possible analytic extensions of the function f from the region Izl < 1 on the whole complex plane out of the cut from 1 to 00 are given by the function z I--t -ln(! - z).

Example 10.18 Let C be a straight Une Re z = a > 0 and cp is an analytic function on the half - plane Re z :::; a, except at finite number of poles and essential singularities ab a2, ... , an which are in Re z < a. If cp( z) --. 0 as z --. 00 in the region Re z :::; a, prove that


where C is oriented from below to above. z-a

2. Let F(z, t) = ezt In --b' t > O. z-

2. a) Find the singular points of the function F.

2. b) Find the value of the integral

J zt I z + 1 d e n-- z c z -1 '

t> 0,

where C is the straight line Re z = a > 1, which is oriented from below to above. z-a

2. c) Find the Laurent series in the neighborhood of 00 of In --b and specify z-

the region where this representation is meaningful.

Solution. 1. The straight line C is exchanged by the path Cl given by Ree., 11" /2 :::; () :::; 31r /2, part of the straight line Re z for Re z = a for - R :::; Im z :::; Rand the part of the straight line from x + Rz to x - Rz (0 :::; x :::; a) (Figure 10.11).

Figure 10.11

Then by the theorem on residues (for enough big R) we have

We have to prove now only that the integral on KR : Res. ('Ir /2 :::; () :::; 3 'Ir /2) tends to zero as R -+ 00, as weH as the integral on AB and CD. By

and the fact that <p( z) -+ 0 as z -+ 00 in the region Re z :::; a, easily follows the convergence of the above integral to zero. Analogously there follows the convergence to zero of the integrals on AB and CD.

2. a) We shall find the singular points of the function

z-a F(z, t) = ezt In --b' t > o.

z-

The points z = a and z = bare the branching points of infinite order. The point z = 00 is an essential singularity.

2. b) Using 1. and 2. a) and the path C' from Figure 10.12

y

c' t o~ oe x

Figure 10.12

we obtain for R -+ 00

1 1 zt 1 z + 1 d R (zt 1 z + 1) - e n -- z = es e n -- . 27rz c' Z - 1 z - 1 z=<Xl

Letting R -+ 00, and using

Res eztln __ ( z + 1) z - 1 z=<Xl

e t - e- t 2. h --- = -sln t, t t

we obtain

- ezt In -- dz = - ezt In -- dz = - sinh t 1 1 z + 1 1 1<>+·00 z + 1 2 27rZ c' Z - 1 27rz 0-'00 Z - 1 t '

for t > O.

2. c) We shall expand in the neighborhood of the point z z-a P' 1 b' In --b in a Laurent series. uttmg z = - we 0 tam z- w

z-a In-

z-b In(z - a) -ln(z - b)

In(l - aw) -ln(l - bw) <Xl (aw)n 00 (bwt

-2:-+2:-, n=l n n=l n

00 the branch

where we have used the power se ries representation of In(l - z) at z = 0 (jzj < 1).

Taking w = ~ we obtain z

Since jawj < 1 and jbwj < 1, we have jzj > max(jaj, jbl).

Remark. The integral of the function J on the straight line Re z = a

l a+.b la+.oo i~~ a-.b J(z) dz = a-.oo J(z) dz

is the so called Brounwich-Wagner integral. The considered integral

(given by some additional suppositions on <p) is the inverse Laplace transform of

<po So we have obtained in 2. b) for the function <p( z) = In z + 1 that its inverse z-l

Laplace transform is J(t) = 2 sinh t/t, t > O.

Example 10.19 Let

[CO dt a > 0 Ja ts(t - x)' - ,

where X is a real parameter, s a complex number and t S = es/nt.

1. a) Investigate the convergence of the given integral with respect to a, X and s.

1. b) For a > 0, x < a prove that the given integral defines an analytic function in region 0 and find this region O.

2. Let

'Ir co 1 (a)k G s = +a-s ---() Ixl s sin 'lrS E k - s k '

a> 0, x< -a.

Find the region of the analicity of the function G(s).

3. Prove that for s > 0 different from integer (for parameters a > 0 and x < -a) we have F(s) = G(s).

Hints. Use the integral [ dz

Je zs(z - x)'

where the path C is given on the Figure 10.13 and let R --+ 00.

y

Figure 10.13

4. Using the principle of analytic continuation find the region where F = G and the region from wh ich the function G extends the function F. Find the region of this extension.

5. Prove that for lxi< a, a > 0, and sE 0

= 1 x k F(s)=a- s 2:-k -(-).

k=O + S a

Solution. 1. a) We shall consider the following cases. Ia > O.

I (i) x < a, and s = m + zn. Then

and

I [= dt I [= dt Ja ts(t - x) ~ Ja tmlt - xl

1 1 --,----,- '" -- as t ~ 00 tmlt - xl t mH '

and so the integral converges for m + 1 > 1, i.e., Re s = m > O.

I (ii) x = a. In this case the integral diverges, since

[= dt [b [= Ja ts (t - a) = Ja + A '

where the first integral on the right side diverges.

I (iii) x > a. In that case we have

[= dt [b [= Ja ts(t - x) = Ja + Jb '

where a < x < b, and therefore the first integral on the right side diverges for every sEC.

II a = O.

II (i) x < O. We have

1= dt 1b 1= ) = + ,b > 0, o ts(t - x 0 b

The first integral on the right side converges for Re s < 1 and the second integral for Res> 0 (as in I d)). Therefore the integral converges for 0< Res< 1.

II (ii) x = O. Then the integral

[00 dt [b [= Jo tsH = Jo + Jb ' b > 0,

diverges since we can not find such sEC for which both integrals on the right side would converge.

II (iii) x > o. Then

{CO dt {b t (CO Jo ts(t - x) = Jo + A + Je '

where for b < x < c, diverges for every sEC.

1. b) Let a > 0, x < a and F(s) = {CO (dt ). To find the region 0 of the Jo ts t - x analicity of the function F we shall prove that the function under integral g(s, t) =

(1 ) has the following properties. t s t - x

(i) g(s, t) is continuous with respect to for every sEC, since t- S = e-slnt . The function g(s,t) is continuous for every tE [a,oo) since x < a ~ t.

(ii) The function g(s, to) for to E [a,oo) is a analytic function for every sEC, since toS = e- slnto is analytic.

(iii) The family Fq(s) = r (dt ) uniformly converges on every closed and Ja ts t - x bounded subset A of the half-plane Re s > 0, since for s = m + zn

IFq(s)I ~ [ tm(td~ x) ~ J txO(:~ x)'

where Xo = min Re s. The function under integral on the right side has the following seA

behavior 1 1

txo(t-x) '" txo+l (q -+ 00).

Hence Res ~ Xo > o. Then (i), (ii), (iii) by Theorem 10.4 implies that the function F is analytic for

Res> o. 2. We shall find the region of the analyticity of the function

'Ir -8 CO 1 (a) k G(s) = I I· + a E -k- - for a > 0, x< -a.

x SSln'lrS k=l - S X

The first summand on the right side is an analytic function in the whole complex plane without integer points s = 0, ±l, ±2, ... for which sin 'Ir s = O. The second summand is an analytic function on C \ {I, 2, ... }, since on every closed and bounded subset A of the given region by

the series f: _l_(~)k, k=l k - s x

for a > 0, x< -a, uniformly converges, since 1;1 < l.

3. We have

[dz [R dt [ dz Je zs(z - x) = Ja ts(t - x) + JKR zs(z - x)

By Residue Theorem

(10.4)

1 dz = 27rzRes ( 1) 27rZ C zs(z - x) Z8(Z - x) z=", - lxises", ' smce x< -a, a> O.

Letting R -+ 00 we obtain (where x < -a, a > 0)

I [2" Re·9dO I Jo Rs e,s9 (Re·9 - x)

< 1 r" dO Rs-l Jo R - lxi

2 (for s > 0). RS-l(R + x)

Therefore by (10.4) letting R -+ 00 we obtain

(10.5)

Then

[ dz JJzJ=a zs(z - x)

[ dz JJzJ=a zs( -x)(l - z/x)

tZJ=a C8(~X) E (;)) dz

11 00 zk-. -- E-dz

x JzJ=a k=O x k

~ 1 1 k-Sd = - L....ti k+l Z Z, k=O x JzJ=a

where the interchange of the integral and series is allowed because of the uniform convergence of the series for Izl < lxi.

Therefore

r dz 11z1=" zs(z - x)

Putting this in (10.5) we have

F(s)(l - e-2'11"") = ~ + (1- e-2'11"'")a-s f: _l_(~)k. Ixl s e''1I"s k=l k - s k

Hence by e'l7rB _ e- I lI"S

Sin 7rS = 2z

we obtain 7r 00 1 (a)k F s = + a-s -- - •

() x S sin 7rS E k - s X k=l

4. The function F(s) is an analytic function on the whole half-plane Res> 0 (see 1. b)) for a > 0 and x < a. The function G (s) is analytic on the w hole complex plane without integers. The equality F(s) = G(s) holds in the region

1= {s I s > 0, s =I- 0,±1,±2, ... }

Therefore the analytic function G( s) analytically extends the function F( s) through the region I on the whole complex plane without integers for a > 0 and x < -a.

5. We shall start from the same integral and path as in 3. Therefore (10.4) holds. Since lxi< a, a > 0 in the region bounded by C there are no singular points of the function under the integral. Hence by Cauchy's theorem

1 dz CZS(z-x) =0.

Letting R -t 00, the integral on the path KR tends to zero (as in 3.) and we obtain

F(s)(l - e-2'11"1S) = r dz (s E 0). 1Iz l=" zs(z - x)

We have for the integral on the right side

r dz r dz 1Izl=" zs(z - x) = 1Izl=" zS+I(l -;)

where we have exchanged the order of the integration and series using the uniform . . lxi 1 convergence senes smce - < .

a Finally we have

00 1 x k F(s)=a-'L-k -(-) ,

k=O +s a

for lxi< 1, a> 0 and s E O.


00 n 1.../3/2 x cot 7rX dx 7r ( t;:;) = + - cot 7r 2 - v 3 . :; (n 2 - 3)V4n2 - 3 0 (3 - x 2 h/3 - 4x2 6

2. Let

and 00 (l)n 100

J(z) = L ,- + e-te-1 dt. n=O n.(z + n) 0

Prove that J analytically extend rand find Jrom which region on wh ich region. Find the singular points oJ the Junction J and the corresponding residues.

7r 3. Find the image oJ the region 0 ~ Re z ~ a, by the Junction w( z) = tan 2 4a z.

Solution. 1. To prove the desired equality we shall take the integral of the function

7rzcot7rZ J(z) = (3 _ z2)v'3 _ 4z2


on the path L given on Figure 10.14.

y

-R -2

-Ri

I Figure 10.14

The function J has the branching points of second order at V3/2 and V3/2, and we take the cut [-V3/2, V3/2]. The function J has poles of first order at points ±V3,±1,±2, ... We have

iJ(z)dz = j V3/2- E l 7rX cot 7rX dx i -;:-----;:~~~ + J( z) dz

-V3/2+E2 (3 - x2)V3 - 4x2 Kl

+ jO 7rX cot 7rX dx + [ J(z) dz + [ J(z) dz V3/2-El (3 - x2)e"'v'3 - 4x2 JI1 JK

1 la-V3/2+E2 7rxctg 7rX dx + J(z)dz+

12 0 (3 - x2 )e"'V3 - 4x2

27rZ( :E Res(J(z))z=k + Res(J(z))z=V3 + Res(J(z))Z=_V3)' ~-~ .


It is easy to prove that for R ~ 00 we have fK J(z) dz ~ 0, for Cl ~ 0 we have

r J(z) dz ~ 0, and for C2 ~ 0 we have r J(z) dz ~ O. Integrals r J(z)dz and k ~ k r J(z)dz together give zero. Since J is an even function we have Jl2

00 00 n '" Res(J(z)) -k = -2z '" --;:-~-r=~=== n~oo n- ~ (n 2 - 3)J4n2 - 3'

and therefore

i V3/2 7rX cot 7rX (00 n 7r ) 4 J dx = 27rz - 2z 2: )J - 2z-6 cot(7rV3) ,

o (3-x2 ) 3-4x2 n=1(n2 -34n2 -3 smce

y

-R R x

Figure 10.15


Remark. We can obtain the same equality using the same function j and the path C from Figure 10.15. 2. See Examples 10.1 and 10.2. So it is easy to obtain that j(z) = f(z) for Rez > 0 and j(z) analytically extend f(z) on the whole complex plane without points 0, -1, -2, ... , which are poles of first order of the function f. The corresponding residues are given in the following way (k E N U {O} )

Res(f(z))z=-k = 00 (-l)n(z + k) 100

lim L: + lim(z+k) e-te- l dt z-+-k n =1 (n + z)n! z-+-k 1

3. We have

= 1. ~ (-l)n(z+k) (_l)k 1m L." '( ) + -k'-z-+-k n=I,njk n. n + z .

( _l)k kl'

(e1rzt/4a _ e-1rZl/4a) 2 2 7r 2 w(z)=tan -z=(-z) 2'

4a (e1rZ./4a _ e-1rZ./4a)

Taking the sequence of transformations

w =~zz=~e1r·/2z· w -eW1 • w _(w)2. 1 4a 4a ,2 - , 3 - 2,

1 W4 = W3 + 1j Ws = -j Ws = 2zwsj

W4

Figure 10.16


o o

Figure 10.17

we obtain the final image (Figures 10.16, 10.17, 10.18, 10.19, 10.20).

t !

/- ........ / I " I I

Figure 10.18


i I

t

o

-1.

w 7

299

Figure 10.19

~v

u

Figure 10.20

In this way the region 0 ::; Re z ::; a, Figure 10.16 (left), is mapped by the

function w(z) = tan2 :a z on the unit disc Iwl ::; 1, Figure 20 (right).

Example 10.21 The closed regions D n , n = 0,1, ... , are given by

D n = {z = x+zy I lxi::; n+ 1/2, Iyl::; n + 1/2}.

1. Prove that

1. a) I sin2 1l"(x + zy)1 = sin2 1l"x + sinh2 1l"yj

1. b) for lxi = n + 1/2 we have I sin 1l"ZI > 1/2 e1l"IYl j

1. c) for lyl = n + 1/2 we have I sin 1l"zl ~ 1/4e1l"(n+t).

Hints. Use that et > 2e- t for t ~ 1l" /2.

2. Let f be an entire function. Prove that for any arbitrary but fixed n E N

[ !(z) dz = 2z t (-ll f(k) JD SIn 1l"Z k=-n

(Dn is the border of D n ).

3. Let the entire function f satisfies the condition

e-aIY1Izf(z)1 < M (U)

for every z = x + zy and 0 S a < 1l", where M is areal constant. Prove that

n

lim L (-l)kf(k) = O. n--++oo

k=-n

4. Let g(t) be a complex function of real variable on the interval [-a, a] for which there exists

Prave that the function f(z) = iaa e,tzg(t) dt

is an entire function which satisfies (U).

Solution. 1. a) See Chapter 3.

1. b) By 1. a) we have for lxi = n + 1/2

I sin2 1l"(n + 1/2 + zy)1 = sin2 1l"(n + 1/2) + sinh2 1l"Y = 1 + sinh2 1l"Y = cosh2 1l"Y.

Since cosh 1l"Y > 1/2e1l"IYI, the preceding equality implies the desired inequality.

1. c) By 1. a) we have for lyl = n + 1/2

I sin 1l"zl ~ sinh 1l"(n + 1/2).

Since et

sinh t > 4' i.e., et > 2e- t

for t ~ 'Ir /2, we have sinh 'Ir(n + 1/2) ~ e,,(n+l/2) /4.

The preceding two inequalities imply the desired inequality.

2. The desired equality follows by Residue Theorem and

Res (f(Z)) = (-l)kf(k) Sin 'Ir z z=k 'Ir

for k = O,±1,±2, ....

301

3. The path Dn is divided on finite sides. On sides parallel to x - axis (one of them we denote by D~) we have lyl = n + 1/2 and Izl ~ n + 1/2, and therefore

I fv~ s{~;z dzl < fv~ I ~~;!lldzl < 4e-7r (n+!) [lf(z)lldzl

JD~ < 4e-,,(n+!)Mea(n+~) [ Idzl

lD~ Izi < 4Me(a-1r)(n + ~)2(n + ~)

n+! 2

where we have used 1. b) and (U). Letting n -+ +00 the right side of the preceding inequality tends to zero since a < 'Ir.

On sides parallel to y - axis (one of them we denote by D~) we have lxi = n+1/2 and Izl ~ n + 1/2, and therefore

I [ .ß:ldzl < lD~ sin 'lrZ

< ('Ir - a)(n + ~)'

where we have used 1. a) and (U). Letting n -+ +00 the right side of the preceding inequality tends to zero.

The preceding considerations imply

[ f(z) dz -+ 0 as n -+ +00. lDn sm 'lrZ

Therefore by the equality from 2. we obtain the desired result.

4. The function f is an entire function by Theorem 10.4.

We shall prove that the function J satisfies the condition (U). Applying the partial integration on the integral which defines J we obtain

and then IzJ(z)l::; eaIY1(lg(a)1 + Ig(-a)1 + iaa Ig'(t)ldt).

Taking

M = Ig(a)1 + Ig( -a)1 + iaa 19'(t)1 dt,

we obtain that the function J satisfies the condition (U).

Example 10.22 Let J(z)

R, R > 0, and let

00

L anzn be an analytic Junction in the disc Izi < n=O

F(z) = f a~ zn. n=O n.

I 1. Let M(r) = max IJ(z)l, 0< r < R. Izl=r

I 1. a) Prove that F is an entire Junction.

I 1. b) Prove that Jor every real number r, 0 < r < R,

IF(z)1 ::; M(r)e 1zl/r.

12. Let Kr be a circle Izl = r, 0 < r < R, oriented positively. Prove that

F(z) = _1 [ J(u)ez / u du. 27l'Z lKr U

00

II Let G(z) = L bnzn be an entire Junction which satisfies the inequality n=O

IG(z)1 ::; Bi1zl

(B and k are real constants).

II 1. Prove that for Izl = r


II 2. Using II 1. prove that the series

1 converyes jor k > Iz!-

II 3. Taking that jor n E N

prove that

1 jor Re - > k.

z

Solution. I 1. a) See Example 10.17 I a).

Second method. The Cauchy inequality implies

303

(10.6)

00 1 Therefore the series 2: ,anzn converges absolutely for every z and therefore the

n=O n. sum is an entire function.

11. b) Using (10.6) we obtain

lF(z)1 :s M(r) f ~(l:lr = M(r)e 1zl /r• n=O n. r

I 2. Residue Theorem implies

_1 r j(u)ez/u du = Res (! j(u)ez/u) . 2:n JKr U U u=o

The desired residue is the zero coefficient in the Laurent expansion of the function j(u)ez/u with respect to u

and this is just the expansion of the function F.

II 1. The desired inequality follows by the Cauehy inequality applyied to the function Fand eircle Kr.

II By II 1. we have Ibnl ~ Br-nekr = hn(r).

We want to find the minimum of hn(r) (the preceding inequality holds for 0 ~ r < +00). The minimum is at r = n/k and it is

Then Ibnl ~ min hn(r).

The series (by D'Alembert's eriterion

00

L n!n-nknenzn n=O

(10.7)

00

eonverges for Izl < l/k, and beeause of (10.7) the series L n!bnzn also eonverges n=O

for Izl < l/k.

II 3. By the uniform eonvergenee of the series whieh represents G on [0, 1] we have

Sinee

we obtain that the integral /'0 e-x/zG(x) dx

uniformly eonverges on the set w hieh satisfies Re (1/ z) ~ k + e for e > O.

The series

uniformly eonverges for Re (l/z) ~ k + e and e > O.

Therefore, letting L _ +00 we obtain

for Re (1/ z) > k, Le., on dise x2 + y2 < x/k whieh is eontained in the dise Izl < l/k.

Example 10.23 The function F is given by F(z) = {OO s~nhzt dt. 10 sm t a) Find the region where F is a regular analytic function.

305

b) For z real find the function F integrating the function elZU / sinh u on the path C given on Figure 10.21 and letting that Cl and C tend to zero, and R tends to infinity.

t x

Figure 10.21

c) Extend F maximally and find the singular points of this analytic extension U.

d) Map the region Rez > 0, 0< Imz < 1/2 by the function U.

Solution. a) We shall find the region 0 C C where the function

{OO sin zt d F( z) = 10 sinh t t

is analytic.

(i) The function s~nhzt is continuous on C x L, where L is the whole real line sm t

(Imz = 0). We have for t = 0

I· sin zt I· sin zt I· 1m -- = Im -.-- = Im t_O sinh t t_O sm zt t_O

z

sinzt z--~-z sin zt - . --zr-

(ii) The function sin zt is analytic with respect to z E C for every tEL, and

h r h f . sin zt. 1 I· t erelOre t e unctIon -:--h . IS a so ana ytIc. sm t

(iii) We shall show that the family { {q s~nhzt dt} converges in A(O) where 10 sm t q

0: -1 < Imz < 1. Namely, on a closed bounded sub set Fe 0 always exist

max Imz = Yo, minImz = Y1. zEF zEF

Then

ISin zt I sinh t

since e- t :<:; 1 for t ~ O.

<

I etz, - e-tz, I et _ e- t

letz'l + le-tz'l et - 1

Further we have for z = x + zy

I s~n zt I < smht

<

for -1 < YI :<:; Yo < 1.

!ext'e-ytl + le-xt'eytl

et - 1 e-Y1t + eyot

et -1

b) Let z be areal number. Then by the theorem on residues

-.-- du = 21l'zRes -.-- . i e'ZU ( e'ZU ) c smh u smh u U=1I"'

( e'ZU ) Since z = 7rZ is pole of first order we have Res -'-h-

SIn u u=1rt have

1 --. Further, we e1l"Z

i e'ZU --du

c sinh u j -q etzX 1 etzU

--dx+ --du -R sinh x K«,) sinh u

l R e'ZX 1211" e'Z(R + zy) 1< e,z(x+211"') + -.-- dx +. z dy +. dx

<I smhx 0 smh(R+zy) R smh(x+ 21l'z) [e lZu j-R etz (x+21r') [0 e,z(-R+,y)

+ JK«) sinh u du + -E sinh(x + 21l'z) dx + J21r sinh( -R + zy)

The residues at the poles z = 0 and z = 21l'Z of the first order give us

lim [ ~ du = -1l'zRes (~) = -1l'Z <1 .... 0 J K(.,) sinh u sinh u u=o '

lim [ ~ du = -1l'zRes (~) < .... 0 J K«) sinh u sinh u u=211"1 - e21rZ '

respectively.

Letting R ~ 0 we obtain

I [211" e,z(R+,y) I Ja sinh(R + zy) z dy I eR+,y 2 e-R-.y I

2e-zY eR e2R _ 1 dy

Z dy.

Analogously we have

1211" e,z(-R+,y) . z dy ~ 0 as R ~ 00.

o smh( -R + zy)

Putting the obtained results in the equality where we have represented the integral on C by the integral on parts of C and letting 15 ~ 0, Cl ~ 0 and R ~ 00, we obtain

100 eZ'" 1-00 e,z(x+2"') (1) 27l"Z V P -- dx + V p. dx = 7l"Z 1 + - --,

-00 sinh x 00 smh( x + 211"z) e211"Z e11"Z

( 1) 100 elZX (e11"Z _1)2 1 - - V P -- dx = 1I"Z ~---::----''-

e21rZ -00 sinh x e21rZ

Since e'zx = cos zx + Z sin zx, we have

100 sinzx e11"Z 1I"Z --dx = 11" --- = 11" tanh -2 .

-00 sinhx e'll"Z + 1

The last integral is without V P since we have proved in a) the existence of this integral for every real z.

Since the function under the integral is even we have

F(z) = [000 sinzx dx = ~1°O sinzx dx = ~ tanh 1I"Z

in sinhx 2 -00 sinhx 2 2

c) By a), the function F(z) = [00 s~nhzt dt is analytic for IImzl < 1. The 10 sm t function

11" 1I"Z U(z) = '2 tanh 2"""

is an analytic function on the whole complex plane except at the points

z = (1 + 2k)z, k = 0, ±1, ±2, ....

Since F(z) = U(z) for z real, functions Fand U are equal on real axis, the function U is an analytic extension of the function Fon the whole complex plane excluding the points z = (1 + 2k)z, k = 0, ±1, ±2, .... .

d) Now we shall apply the function

11" 1I"Z 11" e11"Z - 1 w = U(z) = '2 tanh 2""" = '2 ell"Z + 1

on the region Rez > 0, 0< Imz < 1/2, Figure 10.22.

z

Figure 10.22

Let Wl = e1fZ = e1fZ: e'1fY. Applying it on zo, where x> 0, 0< y < 1/2, we obtain in Wr plane the point with radius p = e1fX 2:: 1 and 0 < arg Wl = 7rY < 1/2, Figure 10.23.

Figure 10.23

We have 7r Wl - 1 7r ( 2)

W = "2 Wl + 1 ="2 1 - Wl + 1 .

We take that W2 = Wl + 1, and W3 = ~, then W4 = 1 - 2W3. Finally W = ~ W4. W2 2

Then the quarter of the circle: p = 1 and 0 < () < i, will be mapped onto the


following lines see Figures 10.24, 10.25, 10.26.

J

----------~/~/_/_/~~/--~--_\~~~---~------------~ -1' 0 ,I ~ \ / U,

" / ....... ./

Figure 10.24

---/' "-I \

I t \ \ Q \. _1-

'- Z ---

/ /

Figure 10.25

309

v

o o u

Figure 10.26

The part of real axis Ul > 1 is mapped on the part of real axis 0 1 is mapped in the w -plane on

'Ir ZVl - 1 'Ir vf - 1 t'lr 2Vl w--·-----·--+---- 2 ZVl + 1 - 2 v~ + 1 2 v~ + 1 .

Since w = U + zv we have

'Ir v? - 1 'lrVl

U = 2" v? + l' v = v~ + 1 .

Hence u2 + v2 = 'Ir 2 /4 for U > 0 and v > O. This means that the part oft he imaginary axis Vl > 1 is mapped on the quarter of the circle p = 'Ir /2, 0 < arg w < 'Ir /2.

Since the point Wl = 00 is mapped on the point w = 0, we conclude that the desired region in the w-plane is the quarter of the disc:

w = pes., 0 < P < 'Ir /2, 0 < () < 'Ir /2.


11" 2

o

v

311

w

u

Figure 10.27

Chapter 11

Miscellaneous Examples

Example 11.1 Give.an example of a continuous, closed curve which does not intersect itselj in a bounded region and whose length is not finite.

Solution. We start from a tri angle with equal sides ßABC, Figure 11.1 (left)

whose one side length is 1 and which lies in the circle Izl = V;.

E Figure 11.1

Dividing each side on three equal part we construct equal sided triangles (see Figures 11.1 and 11.2). Taking out the sides DF,FG,GJ,KM,MD, we obtain a

313


314 CHAPTER 11. MISCELLANEOUS EXAMPLES

closed curve.

Figure 11.2

Continuing this procedure ad infinitum we obtain a continuous clos~d, not intersecting itself in a bounded region, in the disc Izl $ R, where R is given by

R =

=

Since the length of this curve is given by

00 1 L = 3 + 1 + E 3 . 22n 3n '

n=l

and this series diverges, we conclude that the length of this curve is not finite.

Example 11.2 Find the region on which the function w(z) = Z2 maps the region bounded by straight lines x = 1, Y = 1 and x + y = 1, Figure 11.3.

315

z

x

Figure 11.3

Solution. Since w = z2 is equivalent with

u + w = (x + zy)2 = x2 _ y2 + 2zxy,

we obtain u = x2 - y2, V = 2xy. Then the following straight lines are mapped onto the corresponding parabolas

v2 X = 1 on u = 1 - y2, V = 2y, u = 1 - 4'

v2 Y = 1 on u = x 2 - 1, v = 2x, U = - - 1,

4

x + Y = 1 on u = x2 - (1 - x)2 = 2x - 1, v = 2x(1 - x) = 2x - x 2 , i.e.,

v = ~(1 - u2 ), Figure 11.4.

w

u

Figure 11.4

Remark that the map considered is conformal.

00

Example 11.3 If L laml < 00, find m=-oo

00 00

Solution. Since L lam 1 = (J < 00, there exists L am = S.

Let m=-oo m==-oo

1 00

Cn = 21 L lam- n + ... + am+nl· n + m=-oo

We shall prove that lim Cn = ISI. Let e > O. Then there exists a natural number n-+oo

M such that ~ laml < e. Then for n :2: M we have Iml>M

(2n + 1)Cn = L lam- n + ... + am+nl Iml>n+M

We have

and

+ n+M~lml>n-M

+ L lam- n + ... + am+nl· Iml~n-M

L lam - n + ... + am+nl < L (Iam-nl + ... + lam+nl) Iml>n+M Iml>n+M

< (2n+1) L laml Iml>M

< (2n + 1)5

(1 :::; 4M (1. n+M~lml>n-M n+M~lml>n-M

317

Since Iml:::; n-M implies m-n:::; -M:::; M:::; m+n, we havefor Iml :::; n-M

/Iam- n + ... +am+nl - ISI/:::; L laml < 5, Iml~M

and so

1 L lam - n + ... + am+nl- (2n - 2M + l)'SII

Iml~n-M

< L /Iam- n + ... + am+nl-ISII Iml~n-M

< (2n-2M+1)5.

Therefore

/(2n + l)Cn - (2n - 2M + l)ISI\ :::; (2n + 1)5 + 4M(1 + (2n - 2M + 1)5.

Dividing by (2n + 1) and taking n -t 00, we obtain

lim sup 1 Cn - ISII:::; c: + c: = 2c:. n-oo

Since c: > 0 was arbitrary we have lim Cn = ISI. n_oo

1 dz Exarnple 11.4 Find the integral,,; starting with the positive value

Izl=l 4z2 + 4z + 3 of the square raot at the point z = 1.

Solution. The zeros of the equation 4z2 + 4z + 3 = 0 are Zl,2 = -1/2 ± zV2/2, where IZl,21 < 1.

Taking a cut ZlZ2, we can take in the complex plane outside of this cutting an analytic part the function

1 J(z) = V4z2 + 4z + 3

1

2J(z + 1/2 - zV2/2) (z + 1/2 + zV2/2)'

The condition that we are starting with positive value of the square root at the point z = 1 means that J(l) > O.

Therefore

So we can take for z = 1 that

arg(z - zd = -(Y and arg(z - Z2) = (Y,

Figure 11.5. When the point z moves upto any position on the right part of the cut without crossing the cut arg(z - zd continuously changes to -7r /2, and arg(z - Z2)

to 7r/2.

y

Z1 . - J:;--------- ----

-1 1 0 ------f ..-------..-- l~_

Zt

Figure 11.5

Therefore on the right part of the cut arg((z - Zl)(Z - Z2)) = O.

We have (independence of the path from Figure 11.5)

f dz Jlzl=l J 4z2 + 4z + 3

where

I Z je; rel' dt I 2 -f vreh(zl - Z2 - reh)

1 ~ rdt

< 2"1!§ VrVl z1 - z21- r

7rVr =

~ 0, as r ~ O.

Analogously we can prove that

(/21 = I.: f~ rehdt 1 ~ 0 2Jf jreh(z2-z1+ret') ,

Taking r ~ 0 we finally obtain

as r ~ O.

f dz j:f. dy Jlzl=l J4z2 + 4z + 3 = Z -:f. j~ _ y2 = 1rl.

Example 11.5 Find the value 01 the integral I: e-s2/2elSZ ds.

Solution. We have to find the function

100 2 F(z) = -00 e-s /2 elSZ ds.

319

First we shall find the corresponding real function for z = t real, and then with an analytical continuation we shall obtain F for z E (C. We shall prove that

(11.1)

where L is the straight line u = s - zt (-00 < S < +00). From u = s - zt we have s = u + zt and so

Therefore

u2 + 2iut - t 2 =----

2 2 2

v P i e-t2/2-u2/2 du

e-t2 / 2V P i e- u2 / 2 du.

We can remove V P, since jetSt-s2 /21 = e- s2 /2, and the integral of this function absolutely converges. Since e-u2 / 2 is an entire function its integral on a closed path C is zero fc e-u2 / 2 du = o. We take the path C as in Figure 11.6.

o

Figure 11.6

We have

1s-.t 2 l S 2 e-U /2 du + e-U /2 du -s-~t S-tt

1s 2/ 1'-'t 2/ __ se-S 2ds+ -s e- U 2du

o.

We have

The last integral is bounded and therefore

lim 11s e-u2

/ 2 dul = 0, Isl .... oo s-d

the second and fourth integrals are zero as Isl --t 00. So we have

[ e-u / 2 du = 100 e-·/2 ds. JL -00

The last integral is 1: e-s/ 2 ds = ."fi;, and therefore by (11.1)

F(t) = ."fi;e-t2 / 2•

With an analytical continuation on the whole complex plane we obtain

F(z) = .j2;e-z2 / 2 (z E C).

Remark. The integral transform

cfl(z) = ~1°O f(t)e lZt dt v27r -00

321

is the Fourier transform of the functions (see [17]). The case f(t) = e- t2 / 2 is important in probability theory. In this way we obtain the characteristic function cp(t)

= ~F(t) for the normal distribution N(O, 1) : v 27r

za Exercise 11.6 Let w(z) = --, a E ~.

1+z a) At which points and what kinds of singularities does this function depending

on a?

b) For a = 1 prove that this function is a conformal mapping on the whole complex plane except at z = -1. What is the imageof the part of the unit disc from the first quadrant under the given function '?

c) For 0 < a < 1 expand one part of the given function in apower senes in a neighborhood of the point z = -2.

d) For -1< a < 0, using the function w(z), prove that

[00 x a 7r Ja 1 + x dx = - sin a7r .

Exercise 11.7 Let VI - Z2 be the branch of this function which is positive on the upper part of the cut [-1,1].

a) Prove that r dz r dz JK JI=Z2 = Je JI=Z2'

where K is the circle with the center at (0,0) and radius 2, and C is the path on Figure 11.7.

y

-1 o

Figure 11.7

b) Using a) prove that r ~ = -211". JK 1- z

Exercise 11.8 Let f(z) = v'ZLn 1 + ~ 1- z

a) Expand the function f with respect to powers of z for Izl < rand Izl > r. Find for which z this is possible.

b) Find the singular points of f. 1+v'Z. . 11 .

c) Let h(z) = In 1 _ v'Z' where y'Z zs the branch for whzch vI = 1. Map wzth h

the disc Izi < 1.

Exercise 11.9 Let

323

a) Find the region of analyticity of G.

b) Find the analytic form of G.

Example 11.10 The function f(z) = u(x,y) + w(x,y) is analytic at the point Zo = Xo + lYo and f(zo) = Co. Prove that

f(Z)=2U(z~zO, z;/o)_Co.

Solution. We have

Hence

u(X,y)

Putting

00

f(z) = L an(z - zo)n. n=O

Co+Co 00

-2- + L (an((x - xo) + l(Y - yo)t n=1

+an((x - xo) - l(Y - yoW)·

w - Zo x = xo+ --2-'

w-zo Y=Yo+-2-

and taking w enough dose to z we obtain

( w - Zo u xo+ --2-'

w - zo) Yo + -2l- = (Co + f(w))/2.

Exchanging w with z we obtain the desired equality.

00

Exercise 11.11 Prove that for -'Ir < argPn ::::; 'Ir, the infinite product II Pn either

(Xl

converges or diverges simultaneously with the series L lnpn. n=1

Hints.The infinite product 00

PI . P2 ... Pn ... = II Pn n=1

converges if there exists a finite limit 00

lirn IIp; = p n ..... oo

;=1

different frorn zero.

n=1

(11.2)

If P = 0 and no factor Pn is zero, then we say that (11.2) diverges to zero Cn, in the opposite case we call it convergent to zero.


Exercise 11.12 Prove that a necessary and a sufficient condition lor the absolute 00 00

convergence 01 n (1 + Un ), i.e., absolute convergence 01 the series E ln(1 + u n ), is n=l n=l 00

that the series E un is absolutely convergent. n=l

00 00

Exercise 11.13 Suppose that n Un and n Vn converge. n=l n=l

Examine the convergence 01 the lollowing products:

Answer.

a) c)

a) Divergesj c) Converges;

00 n (un + vn)j b) n=l

00 n (un . vn)j n=l

b) Diverges (to zero); d) Converges.

Example 11.14 Let Pl>P2,'" ,Pn,'" be a sequence 01 all prime numbers and

00 1 (s) = E---;

n=l n

the Riemann zeta lunction (which is analytic in Re s > 1). Prove that

1 a) ((s) = 00 j

I1 (1 - p~") n=l

b) The Iunction (( s) has no zeros in the half-plane Re s > 1.

Solution. a) Substracting from the series

1 1 (s) = 1 + - + - + ...

2" 3"

the series 2-0 • (s), we obtain

(1- 2-0 )(s) = 1 + ~ + ~ + ~ + .... 3" 5" 7"

In the last sum there are missing the term displaystyle ;. with n which are divisible with 2.

Analogously we obtain

325

where in the last surn thcre are rnissing the membcrs ~ with n which are divisible n~

with 2 or 3. Generally we have

(1- pt")(l - P2"S) . '" • (1 - p;;'S)(s) = 1 + t~, n

where in the surn E' we are surnming through indices n (greater than one) which are not divisible with PIJP2, ... ,Pm' It is easy to prove that for Res> 1 +~, ~ >

1 0, E' - -+ 0 as m -+ 00, and therefore

n" 00

(s) n (1 - p;;'S) = 1. m=l

b) Since the series f 1.. converges for Re s > 1 + 5, 5 > 0 we have that the m=:I P

product 00 n (1 - p;;'S)

m=1

also converges. Therefore ( s) has DO zeroes in Re s > 1.

Example 11.15 Prove that the series f:~, where {Pn} is the sequence 01 alt prim n=1 Pn

numbers, diverges.

Solution. By Exarnple 11.14 we have for every 5 > 0

n°o (1 _ -(1+6» _ 1 n=1 Pn - (1 + 8)"

Therefore 00

lirn n (1 - p~(1+Ö» == O. 6-+0 n=1

Since (1- p;;l) < (1-p;(I+ö», the product E:=l (1-p;;l) diverges. Hence the series 00 1 L - also diverges .

n=1 Pn

Example 11.16 Let I be a bounded analytic lunction in the unit disc and 1(0) i= o. 11 {an} is the sequence 01 zeros 01 f in an open disc where the zeros appear in the sequence so many times as is their order, then

00 00 n lanl converges, i.e., L(1 - lanD< 00. n=1 n=1

Solution. For simplicity we suppose that I/(z)1 ~ 1, for Izl ~ 1.

In the function I has only finite number of zeros then the statement is trivially true. In the opposite I has countable many zeros al, a2, a3, . .. . Let Bn (z) be the finite product

n

Bn(z) = II t -_ak . k=l - ak z

The function Bn(z) is a rational function, analytic in Izl ~ 1 and IBn(eo')1 = 1, since

every function z - _ak has modulus one on the circle Izl = 1. Therefore BI((z)) is a 1 - akZ n Z

bounded analytic function in Izl < 1.

Since

Specially n

0< 1/(0)1 ~ IBn(O)1 = II lakl k=l

n

Since lakl < 1 for k = 1,2, ... and every partial product n lak I is not less than k=l

1/(0)1, the infinite product converges.

Remark. The analytic function

B(z)=zPll (an. an-=z)PR, n=l lanl 1 - anz

where:

(i) p, PI, P2, ... , are non-negative integers,

(ii) an are different non-zero numbers from the unit disc , 00

iii) the product n lanlPR converges, n=l

is the Blaschke product.

Example 11.17 Let f be a bounded analyticallunction in the unit disco Prove that I can be in a unique way represented as I(z) = B(z)g(z) where B is the Blaschke product, 9 a bounded analyticallunction without zeroes.

Solution. Since fez) "I- 0, we have I(z) = zPh(z), where h(O) "I- O. Let B be the product zP with the Blaschke product of zeros {an} of h.

Then g( z) = I (z) / B( z) (see the preceding example) is an analytic and bounded function in the disco The representation I(z) = B(z)g(z) is unique since the Blaschke product is uniquely determined by the zeros of the function.

327

Exercise 11.18 Find the integral Jn = ~zl=2 1 ~ z el/z dz, n = 0,1,2, ... , and find

A = lim J2n and B = lim J2n - I . n~~ n-+oo

Hints.The function 1 ~ z el/z in the disc Izl :::; 2 has a pole of first order at z = -1

and an essential singularity at z = O. Use Residue Theorem.

Solution.

(1 oe (_l)k) n (_l)k

Jn =2n(-lt --:L -,- = 27rx(-lt:L-,-, e k=n+1 k. k=Q k.

27rX lim J2n = - and 1· J 27rX

1m 2n = --. n-+oo e n-oo e

Exercise 11.19 Function F( z) is defined by the following integral

F(z) = (oe sinzt dt. 10 e27rt - 1

a) Find the region of analytieity of the function F.

eUU

b) For z real find F integrating the function 2 on the rectangle with ver-e 7rU_1

tiees 0, R, R + x, x, and letting R --+ 00.

e) Analytieally extend the function F as for as it is possible and examine the singularities of the analytieal extension .

d) Map the region Rez,O < Im < 7r/2 by the function w = (z) + 1/2z.

Hints.a) Prove that for Im z :::; 27r - ö,O < Ö < 27r, and big enough t :

I sinzt I< -M/ 27rt 1 - e 2. e -

Therefore F is analytic in the region IImzl < 27r.

b) Take the path in Figure 11.8 and apply Cauchy theorem.

y

R~I

~~------------------~

t o R x

Figure 11.8

Then let R --+ 00 and r --+ o. So we obtain

1 + e- Z 1 F(z) = 4(1 _ e-z ) - 2z·

1 + e-z 1 . c) The function (z) = ( ) - -2 analybcally extend F(z) on the whole

4 1- e-Z z complex plane without the points Zk = 2k7n, k = ±1, ±2, ... , where it has poles of first order and z = 00 is an essential singularity. d) The region Rez > 0, 0 < Imz< i, Figure 11.9 (left) is mapped by (z) + 1/2z onto the region in Figure 11.9

329

(right).

y

z w

Figure 11.9

Exercise 11.20 Find the real integral

("/2 I = 10 cosa x cos dx dx (b> a > -1).

Hints.Integrate the function J(z) = (Z2 + l)a zb-l-a along the path in Figure 11.10 ( section at [-1, 1]) and let r ---t O.

Answer.

f(a+ l)f(~) 1= sin~(b-a).

2f(1 + 1 +~) 2


L

o 1 x

-L

Figure 11.10

Answer. r(a + 1)r(~)

1= sin:::'(b-a). 2r(1 + 1 +~) 2

Exercise 11.21 Let

(Z2 + 3)(ea(1-Z)/2(I+z) - 1) F(z,a) = ,

(z - 1F{jz(z + 1F (a is a complex number).

a) Find the singularities oJ this multivalued Junction.

b) Choose the branch J(z,a) oJ the multivalued Junction F(z,a) which at the point z = 2 takes the value

and find the value J(z -l,a).

c) Let

7(e- a/ 6 - 1) J(2,a) = m

18

w(a) = [ J(z, a) dz, Jlz+1I=R

Find the region oJ analyticity oJ the Junction w( a).

R>2.

331

d) Prove that in the region Iz + 11 > 2 the function f has the following representation

00 a f(z,a) = ~ (z+n1)n'

What is the cut? Find the coefficient al.

e) Find the analytical form of w(a).

j) Map the region bounded by curves x = O,y = O,xy = f, with the junction w(a2 ).

Example 11.22 Let LI and L 2 be two closed paths with a common path L. Let j be a continuous junction on LI and L 2 and in regions bounded by LI and L 2 , where it is also analytical. Prove that then j is analytical also on L.

Solution. Let 0 be the union of regions bounded by LI and L 2 • Then we have

!cj(z)dz = 0

for any closed path contained in O. Namely, if C is completely in a region bounded either by LI or L 2 , then the preceding equality foliows by Cauchy's theorem 5.4. In the opposite case, Figure 11.1,

Figure 11.11

we have by the generalization of Cauchy's theorem 5.5 that

1 j(z)dz = 0 and c, 1 j(z) dz = 0

C2

Therefore

1 j(z)dz=l j(z)dz+l j(z)dz=O. c C, C2

Then by Morera's theorem f is analytic in O. Since L c 0 we obtain that j is analytic also on L.

Exercise 11.23 Find

where an are the maximum points 0/ the function sin x 2 / x 2 at (-00, +00).

Hint. Take for the path P the square of side 2 'Ir R centered at the origin and integrated on this path the function

zsinz (1 + z2)(sinz - zcosz)"

Then let R --+ 00.

Exercise 11.24 (Prime Number Theorem) Let 'Ir(N) denote the number 0/ primes less that 01' equal to N. P1'Ove that

1. 'Ir ( N) log N 1m N = 1, N_oo

i.e., 'Ir(N) rv N/logN.

Hint. Let ( 1 z - 2 )

g(z) = N Z :;; + R2 + 1 .

Integrate the function 1

L(z) = L -;/(z) pprime P

on the path P consisting of C the right semi-circle of radius R with center at z = 1; D the continuation of C for 1 - [j < Re z < 1; E the vertical segment Re z = 1 - [j from Im z = J R2 - [j2 to the x-axis; F the segments from 1 - [j to 1 and the circle around z = 1 clockwise; G : the refIection of E across the x-axis; H : the reflection of D across the x-axis. Use Cauchy's theorem and let R --+ 00. Prove that

1a L(z)/(z) dz = 2'1rz'Ir(N).

Use the estimations

" 1 < e log 4x r 1 L.J - - ( _ 1) x-I lor x > ;

p prime,p>N P" X N log N

and L ~ = 0 (Nl-x~Og(2 + lxI) + N(1-x)!210g N 10g(2 + lXI))

p prime,p<N P" (1 x) log N

for x < 1, where 0 is the small "0" limit sign.

Bibliography

[1] Ahlfors, L., Complex Analysis, McGraw-Hill Book Co., New York, 1966.

[2] Bak, J., Newman, D.J., Complex Analysis, Springer-Verlag, New York, 1997.

[3] Carathoodory, C., Theory of Functions of a Complex Variable, Vol II, Chelsea Publishing Co., New York, 1954.

[4] Conway, J., Functions of One Complex Variable, Vol. 1,11, Springer-Verlag, New York, 1995.

[5] Courant, R., Hilbert, D., Methods of Mathematical Physics I, II, Interscience Publishers, 1953, 1962.

[6] Evgrafov, M.A., Problem Book on Analytical Functions (in Russian), Moscow, 1973.

[7] Feyel, D., De la Pradelle, A., Exercices sur les fonctions analytiques, Arman Colin, Paris, 1973.

[8] Fuchs, W.H.J., Topics in the Theory of Functions of One Complex Variable, D. Van Nostrand Co., 1967.

[9] Hörmander, L., The Analysis of Linear Partial Differential Operators -I-IV, Springer-Verlag, Berlin, 1983-1985.

[10] John, F., Partial Differential Equations, Springer-Verlag, 1982.

[11] Ladyzenskaya, O. A., Uraltseva N. N., Linear and Quasilinear Elliptic Equations, Academic Press, New York, 1968.

[12] Lakshmikantham, V., Leela, S., Differential and Integral Inequalities,· Academic Press, New York, 1969.

[13] Lang, S., Complex Analysis, Springer-Verlag, New York, 1993.

[14] Lelong, J., Ferrand, Problemes d'analyse Cl, Paris, 1967.

333

334 BIBLIOGRAPHY

[15] Markushevich, A. 1., Theory 0/ Functions 0/ a Complex Variable, PrenticeHall, Englewood Cliffs, N.J., 1965.

[16] Polya, G., Szegö, G., Problems and Theorems in Analysis (2 vols), SpringerVerlag (vol. I, 1972; vol. 2, 1976).

(17] Pap, E., TakaCi, A., TakaCi, Dj., Partial Differential Equations through Exampies and Exercises, Kluwer Academic Publishers, Dordrecht, 1997.

[18] Rubinstein, Z., A Course in Ordinary and Partial Differential Equations, Academic Press, 1969.

[19] Rudin, W., Principles 0/ Mathematical Analysis, McGraw-Hill Book Co., New York, 1964.

[20] Rudin, W., Real and Complex Analysis, McGraw-Hill Book Co., New York, 1966.

[21] Schmeelk, J., TakaCi, Dj., TakaCi, A., Elementary Analysis through Examples and Exercises, Kluwer Academic Publishers, Dordrecht 1995.

[22] Sobolev, S. 1., The Equations 0/ the Mathematical Physics, Nauka, Moscow, 1966 (in Russian).

[23] Strang G., Introduction to Applied Mathematics, Wellesley-Cambridge Press, 1992.

[24] Taylor M.E., Partial Differential Equations I,II, III, Springer-Verlag, 1996.

[25] Titchmarsh, E.C., The Theory 0/ Functions, Oxford University Press, London, 1939.

[26] Tikhonov A., Samarski A. A., Equations 0/ Mathematical Physics, Pergamon Press, New York, 1963.

[27] Vvedensky, D., Partial Differential Equations with Mathematica, AddisonWesley, 1993.

[28] Vladimirov, V., S.,Equations 0/ Mathematical Physics, Nauka, Moscow, 1976.

[29] Volkovyskii, L.I., Lunts, G.L., Aramanovich, I.G., Problem Book on Functions 0/ a Complex Variable (in Russian), Moscow, 1961.

List of Symbols

z Rez Imz Izl argz AC XA n u \ ao Q liminf limsup Logz logz Res(f(z))z=zo

set of natural numbers set of integers set of real numbers n-dimensional real Euclidean space set of complex numbers .;=I imaginary unit conjugate of the complex number z real part of a complex number z imaginary part of a complex number z absolute value of the complex number z

argument of the complex number z complement of the set A characteristic function of the set A intersection of sets union of sets set difference border of the region 0 closure of the region Q limit inferior limit superior multivalued complex logarithm principal value of the complex logarithm residue of a function f at Zo

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Index

Absolute value 1 analytic continuation 227

along a path 228 direct 227

analytic function 54 anharmonic relation 87 argument 2 Argument Principle 192 aXlS

imaginary 1 real 1

Bernoulli numbers 141 Bessel function 260 beta function 258 bilinear transformation 75 Blaschke product 326 bounded set 32 branch of the logarithm 68 branching point 68

algebraic 68 transcedental 68

Brounwich-Wagner integral 288

Cauchy Generalized Integral Formula 192 inequality 130 integral formula 129 sequence 37 theorem 104 theorem generalization 104

Cauchy-Riemann equations 54,60 closed path

regular 192 closed set 32

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compact set 32 complex

number 1 conformal mapping 73 conjugate 1 connected set 32 curves

smoothlyequivalent 103 cutting 68

DeMoivre formula 2

entire function 54

Fibonacci numbers 145 function

analytic 54 conformal 73 continuous 53 differentiable 53 entire 54 meromorphic 172

Gamma function 256

Hadamard example 63 Three Circle Theorem 159

Integral Fresnel123 line 103

isolated singularity 171

Laplace equation 60

INDEX

transform 263 Laurent expansion 178 limit 53 Liouville

theorem 130 logarithm

principal value 68

Maximum Modulus Theorem 130 Mellin transform 258 rnerornorphic function 172 metric 32 Mittag-LefHer

theorem 179 Mittag-LefHer

theorem 189 Montel

theorem 160 Morera

theorem 130

Natural boundary 227

Open mapping theorem 130 open set 32

Path 53 Poisson

formula 154 power series 130 Prime N umber Theorem 332

Region 32 regular point 227 residue 191 Riemann

Mapping Theorem 73, 160 zeta function 267

Rouche Theorem 192

Schwarz lemma 130, 157 reßection principle 236

sequence accumulation point 37 bounded 37 Cauchy 37 convergent 37

series

set

absolutely convergent 45 convergent 44 power 130

accurnulation point 32 singularity 227

essential 171 pole 171 removable 171

stereographic projection 32

Tauber theorem 155

Taylor expansion for analytic function 130

theorem general residue 191 residue 192

Uniqueness theorem 130

Vitali theorem 159

Well-posed problem 63 Wright function 261

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Complex Analysis through Examples and Exercises

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