COMPLEX NUMBERS AND FUNCTIONS - NIT Kurukshetra

Engineering Mathematics through Applications418

COMPLEX NUMBERS AND FUNCTIONS

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418

6.1 INTRODUCTION

Complex numbers find applications in electric circuits, mechanical vibrating systems, etc. A

number of the form (a + ib), where a and b are real numbers and 1,i = − is called a complexnumber. In (a + ib), a is called the real part, written as R(a + ib) and b is called the imaginarypart, written as I(a + ib). A pair of complex numbers (a + ib) and (a – ib) are said to beconjugate to each other. For complex number z(= a + ib), complex conjugate is denoted by

( ).z a ib= −

Basic Properties of Complex Numbers

1. If z 1 = (a1 + ib1) and z2 = (a2 + ib2) are two complex numbers, then z1 = z2 implies a1 = a2and b1 = b2, i.e. the numbers being equal means their real and imaginary parts areseparately equal.

Note: However, the inequalities between complex numbers such as: z1 > z2 or z1 < z2 has no meaningsince the field of complex numbers cannot be ordered.

2. Sum, difference, product and quotient of any two complex numbers is a complexnumber.

(i) their sum z1 + z2 = (a1 + ib1) + (a2 + ib2) = (a1 + a2) + i(b1 + b2) = (A + iB), a complex number

(ii) their difference z1 – z2 = (a1 + ib1) – (a2 + ib2) = (a1 – a2) + i(b1 – b2) = (A – iB), a complex number

(iii) their product z1 · z2 = (a1 + ib1) (a2 + ib2) = (a1a2 – b1b2) + i(a1b2 + a2b1) = C + iD, a complex number

(iv) their quotient 1 1 1 1 1 2 2

2 2 2 2 2 2 2

( ) ( ) ( )( ) ( ) ( )

z a ib a ib a ibz a ib a ib a ib

+ + −= =+ + −

1 2 1 2 2 1 1 2

2 2 2 22 2 2 2

,a a b b a b a b

i C iDa b a b

+ −= + = ++ +

Complex Numbers and Functions 419

3. Every complex number is expressible in the form r (cosθ + isin θ) briefly written ascisθ. For z = (a + ib) = r (cosθ + isinθ), real and imaginary parts are

2 2 2

1

( ),( ) cos ,implying( ) sin tan

r a bR a ib rbI a ib ra

−

= + + = θ + = θ θ =

The above form of the complex number isknown as trigonometric form or polar form. Here

2 2r a b= + (taking positive sign only is calledthe modulus or absolute value or norm of thecomplex number z(= a + ib) denoted by |z| and θis called the argument or amplitude or phase ofz(= a + ib).

Observations

(i) Although z and z have the same moduli, and equal value of the argument but different in sign.(ii) As amplitude θ has an infinite number of values, the value θ which lies in between –π - π is called the

principal value of the argument. Unless otherwise stated, we take amplitude (z) to the mean principalvalue.

(iii) Amplitude θ, if directed along the positive X-axis, is positive in the anti-clockwise direction and isreckoned as negative in the clockwise direction.

6.2 (1) GEOMETRICAL REPRESENTATION OF IMAGINARY NUMBERS

Kuhn of Denzig was the 1st Mathematician who proposed the geometric representation ofimaginary number i.

Let OP be a real number represented by x along the positive direction of X-axis and OP'by –x along OX'.

Since –x = i2x = i(ix)It means that multiplication of real number x by i twice results

in rotation of OP through two right angles to trace OP' whichnaturally follows that the multiplication of real number by i isequivalent to the rotation of OP through one right angle to theposition OP".

Here if Y'OY be a line perpendicular to the real axis X'OX,then all the imaginary numbers are represented by points onthe Y'OY, called imaginary axis with positive imaginary numbersalong OY and negative imaginary along OY'.

6.2 (2) GEOMETRICAL REPRESENTATIONS OF COMPLEX NUMBERS (ARGAND

DIAGRAM)

Geometrical representation of complex numbers came into being through the memories ofJean Robert Argand, Paris 1806.

Consider the axis X'OX and Y'OY perpendicular to each other with all the positive realnumbers along OX and negative real axis along X'O.

o o

o

X' P' O P X

Y'

Y

P"

Fig. 6.2

Fig. 6.1

M iy( ) P z x iy( = + )

XX' O

Y'

Y

x

x a =

θ

y = b

L x( )

r


As the multiplication of the real number by i is equivalent to the rotation of its directionthrough 90°. Therefore, all the positive imaginary number along OY and negative imaginarynumber along OY'. Let the point M on OY represent the imaginary number iy (Fig. 6.1)

Complete the rectangle OLPM so that the cartesian co-ordinate (x, y) uniquely representsthe complex number z = x + iy on the complex plane Z. This diagram which geometricallyrepresents the complex number Z is called Argand’s diagram.

Example 1: Express 3 i− + into modulus amplitude form and find its principle argument.

Solution: Given 3 with 3, 1.a ib i a b+ = − + = − =

So that 2 2 3 1 2r a b= + = + =

Here, 3 1cos 3 , i.e. cos and sin 1, i.e. sin

2 2r r−θ = − θ = θ = θ =

Here cosθ is negative and sin θ is positive, collectively implies θ should lie in the 2nd

quadrant with value 52 2 ;6 6

n nπ π θ = π − ± π = ± π n = 0, 1, 2, …

So modulus amplitude form of 3 i− + is 5 262

i ne

π ± π with the principle argument lying

between –π ≤ θ ≤ π and with value 5 .6π−

Example 2: Show that the multiplication of a complex number by i results in a clockwiserotation of the corresponding vector through right angle.

Solution: Let z = a + ib = r cisθ = re iθ

So that 22( ) ( )ii iiz i re e re re

π π θ+ θ θ= = =

Clearly the argument of iz is 2π θ + which is 2

π more

than the argument of z.Hence OZ is rotated through a right angle in

anticlockwise direction (Fig. 6.3)

Example 3: Reduce 1 + sin ααααα + icosααααα in modulusargument form.

Solution: Here, cos 1 sin 1 cos ,2

r π θ = + α = + − α

sin cos sin2

r π θ = α = − α

So that 2 2

1 cos sin2 2

r π π = + − α + − α

θ + π2

iz

Y

z

XOθ

Fig. 6.3


2 21 cos 2cos sin2 2 2π π π = + − α + − α + − α

22 1 cos 2 2cos 2cos2 4 2 4 2

π π α π α = + − α = − = −

and 2

sin 2sin cos2 4 2 4 2tan

1 cos 2cos2 4 2

π π α π α − α − − θ = =π π α + − α −

or tan tan4 2π α θ = −

Thus, (1 sin ) cos cis 2 cos cos sin4 2 4 2 4 2

i r i π α π α π α + α + α = θ = − − + −

Example 4: Convert 12 ∠∠∠∠∠–60° to the rectangular form.

Solution: Take 12 ∠–60° = a + ib = r (cosθ + isinθ)

So that 1cos 12cos( 60 ) 12 62

r θ = − ° = × =

and 3sin 12 sin( 60 ) 12 6 3

2r θ = − ° = × − = −

Resulting in, θ + θ = + = ∠ − ° = −(cos sin ) 12 60 6 6 3r i a ib i

Example 5: Find the smallest positive integer n for which 1 11

nii

+ = −

Solution:1 11

nii

+ = − ⇒1 1 11 1

ni ii i

+ + × = − +

or1 1 2 1

1 1

ni− + = + ⇒ in = 1 = (i)4 ⇒ n = 4

Example 6: If a2 + b2 + c2 = 1, b + ic = (1 + a)z, prove that 11 1a ib iz

c iz+ +=+ −

Solution: Given that b + ic = (1 + a)z or1b icz

a+=+

Now taking right hand side,

1 (1 ) (1 )1 11 (1 ) (1 )1

1

b ici a ib c a ib ciz ab iciz a ib c a c ibi

a

++ + + − + + −+ += = =+− + − + + + −−+


[ ][ ]

[ ][ ]

(1 ) (1 )(1 ) (1 )

a ib c a c iba c ib a c ib

+ + − + + += ×

+ + − + + +On rewriting above equation,

[ ][ ]

[ ][ ]

(1 ) (1 )11 (1 ) (1 )

a ib c a ib ciziz a c ib a c ib

+ + − + + ++ = ×− + + − + + +

2 2 2

2 2 2

(1 2 ) 2 (1 )1 ( ) (2 2 2 )

a b c a ib aa b c a c ac

+ − − + + +=+ + + + + +

22( )

2(1 )a a ib iab

a c ac+ + +=+ + + , using a2 + b2 + c2 = 1

(1 )( ) ( )(1 )(1 ) (1 )

a a ib a iba c c

+ + += =+ + +

ASSIGNMENT 1

1. Express the following in the form a + ib

(i)2 3

1i

i−

+ (ii) 2 21 1

(2 ) (2 )i i−

+ −

2. Express (1 – cosα + sinα) in modulus amplitude form.

3. If 1 1 1;x iy u iy

+ =+ +

x, y, u, v being real quantities, express v in terms of x and y.

4. If x and y are real numbers, solve the equation 3 4

1 3iy y i

ix x y+−

+ +

5. Find what curve (1 ) (1 ) 0zz i z i z+ + + − = represent?

6. If z = 1 + i, find

(i) z2, (ii)1 ;z and plot them on the Argand diagram.

7. In the Argand diagram, show that 9 + i, 4 + 3i, –8 + 8 i and –3 – 4 i form a square.8. If |z1| = |z2| and amp (z1) + amp(z2) = 0, then show that z1 and z2 are complex conjugate

to each other.

6.2(3) SOME GEOMETRICAL RESULTS

(A) Geometric Representation of z1 + z2

Let P, Q represent the complex numbers z1 = x1 + i y1 and z2 = x2 + iy2. Complete theparallelogram OPRQ. Draw PL, QM and RN ⊥ s to OX. Also draw PK ⊥ RN (See Fig. 6.4).

Since ON = OL + LN = OL + OM = x1 + x2, [³ LN = PK = OM]


NR = NK + KR = LP + MQ = y1 + y2

The co-ordinates of R are (x1 + x2, y1 + y2) and it representsthe complex number

z = (x1 + x2) + i(y1 + y2) = (x1 + iy1) + (x2 + iy2) = z1 + z2

Thus, the point R which is extremity of the diagonal ofthe parallelogram having OP and OQ as adjacent sides,represents the sum of the complex numbers P(z1) and Q (z2)such that |z1 + z2| = OR and amp (z1 + z2) = ∠XOR

(B) Geometric Representation of z1 � z2.

Let P, Q represent the complex numbers z1 = x1 + iy1 and z2 =x2 + iy2. Then the subtraction of z2 from z1 may be taken asaddition of z1 to – z2.

Produce QO backwards to R such that OR = OQ. Then the co-ordinates of R are evidently (–x2, –y2) and so it corresponds to thecomplex number (–x2 –iy2) = –z2.

Complete the parallelogram ORSP, then the sum of z1 and

–z2 is represented by OS, i.e. 1 2z z OS QP− = =uuur uuur

. (See Fig. 6.5.)

Hence the complex number z1 – z2 is represented by thevector QP.

(C) Geometric Representation of z1z2

Let P, Q represent the complex numbersz1 = x1 + iy1 = r1(cosθ 1 + isinθ 1)

and z2 = x2 + iy2 = r2(cosθ 2 + isinθ 2)Measure OA = 1 unit along OX. Construct ∆OQR on OQ

directly similar to ∆OAP, so that

OQOR

OP OA= , i.e. OR = OP·OQ = r1r2

and ∠AOR = ∠AOQ + ∠QOR = ∠AOQ + ∠AOP = θ 2 + θ 1

∴ P represents the number r1r2[cos(θ 1 + θ 2) + isin(θ 1 + θ 2)].(See Fig. 6.6.)

Hence the product of two complex numbers z1 and z2 isrepresented by the point R, such that

(i) |z1z2| = |z1| . |z2| (ii) amp(z1z2) = amp(z1) + amp(z2)

(D) Geometric Representation of 1

2

zz

Let P, Q represent the complex numbersz1 = x1 + i y1 = r1(cosθ 1 + isinθ 1)

and z2 = x2 + i y2 = r2(cosθ 2 + isinθ 2)

X

R

Q

P

z z1 2+ z2

K

z1

z2

M L N

Y

O

Fig. 6.6

rr

12

r2

r1

θ2θ1

θ1

R

Q

P

XO A

Y

Fig. 6.5

Y

Q

P

S

R

O

z2z1

z z1 2–

–z2

X' X

Y'

Fig. 6.4


Measure OA = 1 unit and construct triangle OAR on OAdirectly similar to the triangle OQP, so that

1

2

, i.e.rOR OP OPOR

OA OQ OQ r= = =

and ∠XOR = ∠QOP = ∠AOP – ∠AOQ = θ 1 – θ 2.∴ R represents the number

[ ] θ − θ + θ − θ

11 2 1 2

2

cos( ) sin( ) .r

ir (See Fig. 6.7)

Hence the complex number z1/z2 is represented by thepoint R such that

(i) 1 1

2 2

z zz z

= (ii) 11 2

2

amp amp( ) amp( )z

z zz

= −

Example 7(i) The modulus of the sum of two complex numbers is always less than or at the most

equal to their moduli.or

If z1 and z2 be any two non–zero complex numbers, prove that |z1 + z2| ≤≤≤≤≤ |z1| + |z2|,

(ii) The modulus of the difference of two complex numbers is greater or equal to thedifference of their moduli.

orIf z1 and z2 be two complex numbers, prove that

|z1 – z2| ≥≥≥≥≥ |z1| – |z2|

Geometrical Proof: (i)

Let P and Q represent the two complex numbers z1 and z2respectively in the Argand plane. Complete theparallelogram OPRQ.

Here OP = |z1|, OQ = PR = |z2|, so that OR = |z1 + z2|Now from the ∆OPR, OP ≤ OP + PQ or OC ≤ OP + OQ∴ |z1 + z2| ≤ |z1| + |z2|

Alternately: Let z1 = r1(cosθ1 + i sinθ1), z2 = r2(cosθ2 + isinθ2)

∴ |z1 + z2| = |r1(cosθ1 + isin θ1) + r2(cosθ2 + isinθ2)|

= |(r1cosθ1 + r2cos θ2) + i (r1sinθ1 + r2sinθ2)|

2 21 1 2 2 1 1 2 2( cos cos ) ( sin sin )r r r r= θ + θ + θ + θ

2 21 2 1 2 1 2 1 22 (cos cos sin sin )r r rr= + + θ θ + θ θ

2 21 2 1 2 1 22 cos( )r r r r= + + θ − θ

rr

12

r2

θ2

θ1

P

Q

R

XO A

Y

r r1 2/

Fig. 6.7

Fig. 6.8

X

R z z( + )1 2

Q(Z )2

P Z( )1

Y

O


2 21 2 1 2 1 2 1 2 1 22 ( )z z r r r r r r z z+ ≤ + + = + = + [³ cos(θ1 – θ2) ≤ 1]

Cor. |z1 + z2 + … + zn| ≤ |z1| + |z2| + … + |zn|7(ii) Left for readers

Example 8: If pair opposite vertices of a square be represented by the complex numbers1 + 2i, 3 + 4i, show that the other pair is represented by 3 + 2i, 1 + 4i.

Solution: Take the square ABCD with vertices A and C as 1 + 2i and 3 + 4i, respectively.Let the other two opposite vertices be (a + ib) and (c + id)

∴ ( ) (1 2 ) ( 1) ( 2)AB OB OA a ib i a i b= − = + − + = − + −uuur uuur uuuur … (1)

also (3 4 ) ( ) (3 ) (4 )BC OC OB i a ib a i b= − = + − + = − + −uuur uuur uuur … (2)Now, whenever i is multiplied to a complex number, the number changes direction by

π/2, which implies that iAB BC=uuur uuur … (3)

or i [(a – 1) + i (b – 2)] = [(3 – a) + i (4 – b)]

[(2 – b) + i (a – 1)] = [(3 – a) + i (4 – b)]When two complex numbers are equal, their real and

imaginary parts are separately equal.

Thus, }32 3 521 4 1

ab a a bba b a b

=− = − + = ⇒ ⇒ =− = − − = And whence the vertex B is (3 + 2i)Let the diagonals AC and BD intersects at F, then F is the common middle point of AC and

BD.

∴(1 2 ) (3 4 ) (3 2 ) ( )

2 2i i i c id+ + + + + += or (4 + 6 i) = (3 + c) + i (2 + d) … (5)

On equating real and imaginary parts } }3 4 12 6 4

c cd d

+ = =⇒+ = = …(6)

∴ Vertex D represents (1 + 4 i).

Example 9: Find the locus of z given by(i) |3z – 1| = |z – 3| (ii) |z| = |z – 2|

(iii) is real2

z iz

++ (iv) is purely imaginary

2z iz

−−

Solution: Let z = (x + iy), so that }( 3) ( 3)(3 1) (3 1) 3

z x iyz x iy

− = − +− = − + ... (1)

Now,

3 33 1 31 13

z APz zBPz

−− = − ⇒ = =

− ... (2)

Also, |z – 3| = |3z – 1| implies 2 2 2 2( 3) (3 1) (3 )x y x y− + = − +x2 + 9 – 6x + y2 = 9x2 + 1 – 6x + 9y2 or x2 + y2 = 1 ... (3)

Fig. 6.9

D c d( , ) C(3, 4)

A(1, 2) B a b( , )

F

Y

X'O

Y'

X


Thus, from (2) and (3), we see that locus of P(z) is a circle with centre (0, 0) and radius 1unit, and P moves in such a way that its distance from two fixed points, say, A(3, 0) and

1 ,03

B , always bears a ratio 3. (Fig. 6.10)

(0, 0) 13

, 0

O B

P z( )

Y1

Y

(3, 0)A

XX1

Fig. 6.10

(ii) Given |z| = |z – 2| ⇒0

12

z APz BP

− = =−

In other words, P moves in such a way that its ratio of the distances from two pointsA(0, 0) and B(2, 0) is always 1, i.e. it is the right bisector of AB.

(iii) Here( 1)( )

( 2) 2 ( 2)x iy i x i yz i

z x iy x iy + + + ++ = = + + + + +

( 1) ( 2)

( 2) ( 2)x i y x iy

x iy x iy

+ + + − =+ + + −

2 2 2

( ( 2) ( 1)) (( 2)( 1) )( 2)

x x y y i x y xy

x i y

+ + + + + + − =+ −

2 2 2 2

( 2) ( 1) ( 2)( 1)( 2) ( 2)

x x y y x y xyi

x y x y+ + + + + −= +

+ + + +

Now 2

z iz

+ +

is real means its imaginary part is zero, i.e.

(x + 2)(y + 1) – xy = 0 or x + 2y + 2 = 0Whence the locus of z is a straight line.

(iv) Here ( 1)

2 ( 2)x i yz i

z x iy+ −− = − − +

( 1) ( 2)

( 2) ( 2)x i y x iyx iy x iy

+ − − − = ×− + − −


2 2 2

[ ( 2) ( 1)] [( 2)( 1)( 2)

x x y y i x y xy

x i y

− + − + − − − =− −

2 2 2 2

( 2) ( 1) ( 2)( 1)( 2) ( 2)

x x y y i x y xyx y x y− + − − − −= +

− + − +

Now, 2z iz

− − is purely imaginary implies its real parts is zero.

i.e. x (x – 2) + y (y – 1) = 0

x2 + y2 – 2x – y = 0

2

2 1 5( 1)2 4

x y − + − =

Clearly, the locus of z is a circle with centre 11, 2

and radius 5

2 units.

Example 10: If the argument 1 ,1 4

zz

− π = + prove that the point representing z on the

argand diagram lies on the fixed circle with its centre at the point i.

Solution: Given 11 4

zz

− π = +, i.e. the difference between arguments of (z – 1) and (z + 1) is

π/4.

We have }− = − +

+ = + +1 ( 1) ,1 ( 1)

z x iyz x iy

so that

−

−

θ = − θ =

+

11

12

tan ,1

tan1

yx

yx

…(1)

Now given, arg 1 211 4

zz

− π = θ − θ = +

1 1tan tan1 1 4

y yx x

− − π− =− +

1 1 1tan41

1 1

y yx x

y yx x

−

− π− + = +

− +

[ ]2

( 1) ( 1)tan 1

( 1)( 1) 4y x xx x y

+ − − π= =− + +

2y = (x2 – 1) + y2 or x2 + y2 – 2y – 1 = 0

(x – 0)2 + (y – 1)2 = 2

which represents a circle with centre (0, 1) and radius 2 units.Hence the locus of z in the Argand diagram is the circle with its centre at the point i.

P z( )

Y

i

XO( , )O O

Fig. 6.11


Example 11: If z1, z2 be the two non-zero complex numbers, show that

( )2 2 2 21 2 1 2 1 22z z z z z z+ + − = +

Solution: Let 1 2,OA z OB z= =uuuur uuur be the two given complex numbers and OC z=uuur be anothernumber. Complete the parallelogram.

Now in ∆OAC by triangle law of forces, orOA AC OC OA OB OC+ = + =uuuur uuur uuur uuuur uuur uuur or z1 + z2 = z ... (1)

(since in the parallelogram, OBuuur || ACuuur )

In ∆OBA, OB BA OA+ =uuuruuur uuuur i.e. BA OA OB= −uuur uuuur uuur

i.e. BAuuur = position vector of point A – position vector of point B = z1 – z2 ... (2)

Clearly in the parallelogram OACBO, OCur and BAuuur are twodiagonals represented by complex numbers (z1 + z2) and (z1 – z2)respectively.

Now in any parallelogram, the sum of the squares of thediagonals is equal to twice the sum of the squares of the twosides, i.e.

OC2 + BA2 = 2{OA2 + OB2}

or { }2 2 2 22OC BA OA OB+ = +uuuruuur uuuur uuur

Mathematically, + + − = + 2 2 2 2

1 2 1 2 1 22z z z z z z

Hence the proof.

Alternately: If z1 = r1(cosθ1 + isinθ1) and z2 = r2(cosθ2 + isinθ2) then, |z1 + z2|

2 = |r1(cosθ1 + isinθ1) + r2(cosθ2 + isinθ2)|2

= |(r1cosθ1 + r2cos θ2) + i (r1sin θ1 + r2sin θ2)|2

= (r1cosθ1 + r2cosθ2)2 + (r1sinθ1 + r2sin θ2)2

= r12 + r2

2 + 2r1r2cosθ1cos θ2 + 2r1r2sin θ1sinθ2

= r12 + r2

2 + 2r1r2cos (θ1 – θ2)Similarly,

|z1 – z2|2 = |r1(cosθ1 + i sin θ2) – r2(cosθ2 + isinθ2)|

2

= |(r1cosθ1 – r2cos θ2) + i (r1sinθ1 – r2sinθ2)|2

= (r1cosθ1 – r2cos θ2)2 + (r1sinθ1 – r2 sinθ2)2

= r12 + r2

2 – 2r1r2cosθ1cosθ2 – 2r1r2sinθ1sinθ2

= r12 + r2

2 – 2r1r2cos(θ1 – θ2)On adding the two, we get|z1 + z2|2 + |z1 – z2|

2 = 2r12 + 2r2

2

{ }2 2 2 21 2 1 22 2 2z z z z= + = +

Hence the proof.

B z( )2A z( )1

C z( )

D

OFig. 6.12


Example 12: If |z1 + z2| = |z1 – z2|, prove that the difference of amplitudes of z1 and z2 is πππππ/2.

Solution: Let z1 = x1 + iy1 and z2 = x2 + iy2, so that

2 2 2 21 1 1 2 2 2

1 21 11 2

1 2

andtan tan

r x y r x y

y yx x

− −

= + = +

θ = θ =

... (1)

Further, (z1 + z2) = (x1 + i y1) + (x2 + iy2) = (x1 + x2) + i (y1 + y2), ... (2)

(z1 – z2) = (x1 + iy1) – (x2 + iy2) = (x1 – x2) + i (y1 – y2) ... (3)

Also given that |z1 + z2| = |z1 – z2|

⇒ 2 2 2 21 2 1 2 1 2 1 2( ) ( ) ( ) ( )x x y y x x y y+ + + = − + −

⇒ x12 + x2

2 + 2x1x2 + y12 + y2

2 + 2y1y2 = x12 + x2

2 – 2x1x2 + y12 + y2

2 – 2y1y2

⇒ x1x2 + y1y2 = 0 ... (4)which shows that the two vectors are perpendicular to each other.

Now, we need to prove that the difference of amplitude (the phase difference) between z1and z2 is π/2.

Alternately: For z1 = r1cisθ1 and z2 = r2cisθ2

2 2 21 2 1 2 1 2 1 2

2 2 21 2 1 2 1 2 1 2

2 cos( ),

2 cos( )

z z r r r r

z z r r r r

+ = + + θ − θ

− = + + θ − θ [See Example no. 7(i, ii) Page 424]

Given, |z1 + z2|2 = |z1 – z2|2

⇒ 4r1 r2cos(θ1 – θ2) = 0

⇒ cos(θ1 – θ2) = 0 (Since r1 and r2 are not zero).

⇒ 1 2 2πθ − θ =

Example13: A rectangle is constructed in a complexplane and its sides are parallel to the axes and its centreis situated at the origin. If one of the vertices of the

rectangle is 1 3,i+ find the complex numbersrepresenting the other three vertices of the rectangle.Find also the area of the rectangle.

Solution: Let ABCD be the given rectangle with its sidesparallel to the axis XOX/ and YOY/ with centre at theorigin O(0, 0) as shown in the Fig. 6.13.

Here position of one of the edge, say, A(z1) is given

1 1 1 11 3 (cos sin )z i r i= + = θ + θY'

Y

( )z B2 A(1 3)

X'

( )z C3 D z( )4

XO

60°

120° 60°

60°

Fig. 6.13


⇒ 1 1 1 1cos 1, sin 3r rθ = θ = so that r1 = 2 and θ1 = 60°

Hence OAuuuur , i.e. z1 is such that it makes an angle of 60° with the initial axis and possessesmodulus of 2 units.

Now for the given rectangle, it is clear that each OA, OB, OC and OD are 2 units and OBuuur

is obtained by rotating OAuuuur through 60° further, and then OCuuur is ordained by rotating OBuuur

through 120° and then ODuuuur by rotating OCuuur through 60° further.

1

21 2 3 4

3

4

6060 60 120 ( 60 )

and 2 units120 120 240 ( 60 )240 60 300 (2 60 )

r r r r r

θ = ° θ = ° + ° = ° = π − ° = = = = =θ = ° + ° = ° = π + ° θ = ° + ° = ° = π − °

Therefore, z2 = rcisθ2 = 2[cosθ2 + isinθ2] = 2[cos(π – 60°) + isin(π – 60°)]

1 32 1 32 2

= − + = − + z3 = rcisθ3 = 2[cos(π + 60°) + isin(π + 60°)]

1 32 1 32 2

i i = − − = − − z4 = rcisθ4 = 2[cos(2π – 60°) + isin(2π – 60°)]

1 32 1 32 2

i i = − = − Now, the area of the rectangle ABCD = BC × AB.

2 2( 2) 0 0 (2 3) 2 2 3 4 3= − + × + = × =

Example 14: An equilateral triangle constructed in the complex plane has its one vertex atthe point 1 3,i+ find the complex numbers representing the other vertices. Find alsothe area of the triangle.

Solution: Let ABC be the given triangle so that∠ABC = ∠BCA = ∠CAB = 60° or in other words ifO be the circumentre of the triangle,

∠AOC = ∠AOB = ∠BOC = 120°Hence clearly, first we need to locate the position

of 1st edge or the angle made by OAuuuur with the

initial axis and then rotate OAuuuur through 120° to

obtain OBuuur (i.e. 2nd edge B) and further, through

120° to obtain OCuuur (i.e. 3rd edge C). Fig. 6.14

B(– 2, 0)

O120° θ1 = 60

A(1, 3)

C(1, – 3)

120°

Y1

X

Y

X1


Here given 1st edge A i.e. 1 11 1 3 cisz r= + = θ

So that 1 1cos 1 and sin 3r rθ = θ = ⇒ r = 2 and θ1 = 60°

Clearly, θ2 = θ1 + 120° = 180° θ3 = θ2 + 120° = 180° + 120° = 300°

Whence edge B, i.e. z2 = rcosθ2 = 2[cosπ + isinπ] = –2

and the edge C, i.e. z3 = rcisθ3 = 2[cos300° + isin300°]

= 2[cos(2π – 60°) + i sin(2π – 60°)]

1 32 1 32 2

i i = − = −

Now area of the ( )( )( ), where 2

a b cs s a s b s c s + +∆ = − − − =

1 1 1or sin sin sin2 2 2

bc A ca B ab C ∆ = = =

Here 9 3 2 3,a BC= = + =

20 (2 2) 2 3b CA= = + =

2 2( 3) ( 3) 2 3,c AB= = − + = otherwise also sides of equitatral are equal

∴ 2 3 2 3 2 3 3 3

2s + += =

Whence, { } { } { }3 3 3 3 2 3 3 3 2 3 3 3 2 3 3 3 3 3 3 3 3∆ = − × − × − = =

Example15: Show that the equation of the ellipse having foci at z1, z2 and major axis 2a, is|z – z1| + |z – z2| = 2a. Also find its eccentricity. Further, find the locus given by|z – 1| + |z + 1| = 3

Solution: Let P(z) be any point on the given ellipse with foci at S1(z1) and S2(z2)

So that 1S P =uuur Position vector of P – Position vector of S1 or S1P = |z – z1| ... (1)

Similarly 2S P =uuur Position vector of P – Position vector of S2 or S2P = |z – z2| ... (2)

We know that the sum of distances of the two foci from some general point on the ellipseis equal to the major axis (see Fig. 6.15).i.e. S1P + S2P = 2a or |z – z1| + |z – z2| = 2a ... (3)which is the desired equation of the ellipse.

Also we know that S1S2 = 2ae, where e being the eccentricity.

|OS2 – OS1| = 2ae, i.e. |z2 – z1| = 2ae implying 2 1

2z z

ea

−= …(4)


BA S2

S1

P z( )

O

Fig. 6.15

On comparing |z – 1| + |z + 1| = 3 with the equation (3), we see that locus of Z in thisequation is an ellipse with foci at z = ±1 and major axis 3.

Example 16:(i) P1(z1), P2(z2) and P3(z3) be any three points then

3 21 2 3

1 2

ampz z

P P Pz z

−∠ = −

(ii) Also prove that the points P1, P2, P3 are collinear, if 3 2

1 2

z zz z

− −

is wholly real.

(iii) Line P2P3 ⊥⊥⊥⊥⊥ P2P1, if 3 2

1 2

z zz z

− −

is wholly imaginary.

Solution: Let P1(z1), P2 (z2) and P3 (z3) be the threegiven points in the complex plane with reference pointO the origin as shown in the geometry.

Join OP1, OP2 and OP3. Extend P2P1 and P2P3backward meeting the axis XOX' in such a way thatthey make angles θ1 and θ2 with it respectively.

Now we have 2 1 1 2P P z z= −uuuur and P2P3 = z3 – z2

∴ P1P2P3 = θ2 – θ1 = The difference for the external angle and the 2nd internal angle = amplitude of P2P3 – amplitude of P2P1

= amplitude (z3 – z2) – amplitude (z1 – z2)

= amplitude 3 2

1 2

z zz z

− −

(ii) If P1, P2, P3 are collinear then they all lie on a straight line or their angle with initialaxis is zero or π.

3 2

1 2

(say) (cos sin )z z

z r iz z

− = = θ + θ−

, (θ being 0° or π)

Fig. 6.16

P z3 3( )

X1 X

θ2

P z1 1( )P z2 2( )

O

θ1


⇒ , for 0

which is wholly a real number., forr

z rθ == − θ = π

(iii) if P2P3 ⊥ P2P1, then

3 2

1 2 31 2

amp ,2

z zPP P

z z − π∠ = = −

Say, 3 2

1 2

, then clearly cis2

z zz z r ri

z z− π= = =−

which is wholly imaginary

Example 17: If z1, z2, z3 be the three vertices of an isosceles triangle right-angled at z2,prove that z1

2 + z32 + 2z2

2 = 2z2(z1 + z3).

Solution: Let ABC be the desired triangle with vertices A(z1), B(z2), C(z3) right-angled atB(z2) as shown in Fig. 6.13.

The triangle being given an isosceles one, therefore, its two sides, say, BA = BCi.e. |z1 – z2| = |z3 – z2|Secondly it is right-angled at B(z2), i.e. ∠ABC = 90°

Now, 1 2

3 2

cis ,z z

z rz z

−= = θ−

where 1 2 1 2

3 2 3 2

1 and arg . 90z z z zBAr zz z BC z z

− −= = = = θ = = ° − −

Therefore, 1 2

3 2

cisz z

z rz z

−= = θ−

1 2

3 2

1(cos90 sin 90 )z z

i iz z

− = ° + ° =−

or (z1 – z2)2 = i2(z3 – z2)

2

z12 + z2

2 – 2z1z2 = – (z32 + z2

2 – 2z3z2)

z12 + z3

2 + 2z22 = 2z2(z1 + z3)

ASSIGNMENT 2

1. The centre of a regular hexagon is at the origin

and one vertex is given by 3 i+ on the Argand diagram. Determine the other vertices.

2. Find the locus of P(z), when(i) |z – a| = k; (ii) amp(z – a) = α, where k and α are constants.

3. What domain of z-plane is represented by(i) 2 ≤ |z + 3| < 4 (ii) I(z) > 2

A z( )1C z( )3

B z( )2

Fig. 6.17


(iii) amp( )3 2

zπ π< < (iv) |z + 2| + |z – 2| < 2

4. Find the locus given by |z – 1| + |z + 1| = 35. If z1, z2, z3 be the three vertices of an equilateral triangle, prove that

z12 + z2

2 + z32 = z1z2 + z2z3 + z3z1

6. What are the locii given by |z + 3| = k|z + 1| for k = 1 and 2.7. Find the locus of the point z, where

(i)z a kz b

− =− (ii) amp , where and are constants.z a k

z b− = α α −

6.3 DE MOIVRE�S THEOREM

Statement: If n be (i) an integer, positive or negative [PTU, 2005] (cosθ + isinθ)n = cosnθ + isinnθ.

(ii) a fraction, positive or negative,one of the values of (cos θ + isinθ)n is (cosnθ + isinnθ)Proof: Case 1: when n is a positive integer.

By actual multiplication cis θ1cisθ2 = (cosθ1cosθ2 – sinθ1sinθ2) + i (sinθ1cosθ2 + cosθ1sinθ2)

= cos (θ1 + θ2) + isin(θ1 + θ2)

= cis (θ1 + θ2)

Similarly, cis θ1cisθ2cosθ3 = cos(θ1 + θ2 + θ3)Proceeding in this way,

cisθ1.cisθ2. cisθ3 …………… cisθn = cis (θ1 + θ2 + θ3 + ……………… + θn)

Now putting θ1 = θ2 = θ3 = ………………… = θn = θ, we obtain

(cisθ)n = cis nθ

Case 2: when n is a negative integerLet n = –m, where m is a positive integer.

∴ 1 1(c i s ) (cis )

(cis ) cisn m

m m−θ = θ = =

θ θ (By case I)

cos sin

(cos sin ) (cos sin )m i m

m i m m i mθ − θ=

θ + θ × θ − θ

[Multiplying the num. and deno. by (cos mθ – isinmθ)]

2 2cos sin cos sincos sin

m i m m i mm m

θ − θ= = θ − θθ + θ

= cos(–mθ) + i sin(–mθ) = cis(–mθ) = cisnθ [³ –m = n]


Case 3: when n is a fraction, positive or negative.Let n = p/q, where q is a positive integer and p is any integer +ve or –ve.

Now cis cis . cisq

qq q

θ θ= = θ ∴ Taking qth root of both sides, cis(θ/q) is one of the q values of (cisθ)1/q, i.e. one of the

values of 1/(cis ) cis .q

q

θθ =

Raise both sides to power p, then one of the values of /(cis ) cis cisp

p q pq q

θθ = = θ i.e. one of the values of (cisθ)n = cisnθ. (By case 1 and 2)

Thus the theorem is completely established for all rational values of n.

Cor. 1. cisθ1.cisθ2.cisθ3 …………… cisθn = cis(θ1 + θ2 + θ3 + ………… + θn)2. (cosθ – i sin θ)n = cosnθ – i sinnθ = (cosθ + i sinθ)–n

3. (cismθ)n = cismnθ = (cis nθ)m

Example 18: Prove that 2 3

9 5[(cos5 sin5 ) (cos7 sin7 ) ] 1[(cos4 sin 4 ) (cos sin ) ]

i ii i

θ − θ θ − θ =θ − θ θ − θ

Solution: Using DM Theorem, (cos5θ – isin5θ)2 = (cos10θ – isin10θ) = (cosθ + isinθ)–10

(cos7 θ – isin7θ)3 = (cos21θ – isin21θ) = (cosθ + sinθ)–21

(cos4θ – i sin4θ)9 = (cos 36θ – isin36θ) = (cosθ + isinθ)–36

Left Hand Side 10 21 31

36 5 31

(cos sin ) (cos sin ) (cos sin ) 1(cos sin ) (cos sin ) (cos sin )

i i ii i i

− − −

− −θ − θ θ + θ θ + θ= = =θ − θ θ + θ θ + θ

Problem 19: Find the modulus and principal value of the argument of 13

11(1 3)( 3 )

ii

+−

Solution: Say, 11 1 1 1

31 3 cis , where 2, tan 601 3

i r r − π+ = θ = θ = = ° =

Identically, 12 2 2 2

13 cis , where 2, tan 303 6

i r r − π− = θ = θ = − = − ° = −

Now

π π ⋅ + = = ⋅π − π ⋅−

13131 1313

11111 11

2112

cis cis 13(1 3) 3 3( 3 ) cis 11cis 66

rri

i rr

√

13

112 13 11cis cis ,2 3 6

π π= −

= = θ − θ θ 1

1 1Because ciscis( ) (cis )


2 13 112 cis3 6

= + π

2 22 cis 2 2 2 cis3 6 6π π π = π + + π − =

Hence the modulus is 4 and principal argument 6π

Example 20: Prove that the general value of θθθθθ which satisfy the equation

2 4(cos sin )(cos sin ) 1 is ( 1)

mi in n

πθ + θ θ + θ …… =+

where m is any integer.

Solution: (cosθ + isinθ)(cosθ + isinθ)2 … = (cosθ + isinθ)1 + 2 + 3 + … + n

( 1)

2(cos sin )n n

i+

= θ + θ

( 1) ( 1)

cos sin ,2 2

n n n ni+ + = θ + θ …(1)

= 1 = (cos2mπ + isin2mπ), Given …(2)

∴ ( 1) 42 or

2 ( 1)n n mm

n n+ πθ = π θ =

+

Example 21: If (a1 + ib1) . (a2 + ib2) … … … (an + ibn) = A + iBProve that

(i)1 21 1 1 1

1 2

tan tan tan tann

n

bb b Ba a a A

− − − −+ + … + =

(ii) (a12 + b1

2)(a22 + b2

2) + … + (an2 + bn

2) = A2 + B2

Solution: Let a1 = r1cosα1, b1 = r1sinα1 with 111

1

tanba

−α = ;

a2 = r2cosα2, b2 = r2sinα2 with 212

2

tanba

−α = ;

.................. ..............................................

.................. ..............................................

an = rncosαn, bn = rnsinαn with 1tan nn

n

ba

−α = ;

A = rcosθ, B = rsinθ with 1tan ba

−θ =

(a1 + ib1) . (a2 + ib2) … (an + ibn) = A + iB

r1(cosα1 + i sinα1) r2(cosα2 + i sinα2) … rn(cosαn + i sin αn) = r(cosθ + isinθ)

r1r2 … rn[cos(α1 + α2 + … + αn) + i sin(α1 + α2 + … + αn)] = r(cosθ + i sin θ)


Therefore, we see that modulus and amplitude are

r1r2 … rn = r or (a12 + b1

2)(a22 + b2

2) + … (an2 + bn

2) = A2 + B2

and 1 21 1 1 1

1 2

tan tan tan tann

n

bb b ba a a a

− − − −+ + …… + =

Note: Alternately this problem can be taken under applications of log.

Example 22: Prove that 1 1

( ) ( )n na ib a ib+ + − has n real roots and find those of

{ } { }1 13 31 3 1 3 .i i+ + −

Solution: Let − = θ = θ = + θ = 2 2 1cos , sin so that and tan ba r b r r a b

a

∴ 1 1 1 1

( ) ( ) [ (cos sin )] [ (cos sin )]n n n na ib a ib r i r i+ + − = θ + θ + θ − θ

1 1 1

[cos(2 ) sin(2 )] [cos(2 ) sin(2 )]n n nr p i p p i p

= π + θ + π + θ + π + θ − π + θ

1 2 2 2 2

cos sin cos sinnp p p p

r i in n n n

π + θ π + θ π + θ π + θ = + + −

1 22cosn

pr

nπ + θ =

{ }1

2 2 1212 cos 2 tann ba b pn a

− = + π +

{ }1

2 2 112 cos 2 tann ba b pn a

− = + π +

which is real and gives n real values corresponding to p = 0, 1, 2 ,…, n – 1

Now if a = 1, 3, 3,b n= = then

1 1 1 1

3 3( ) ( ) (1 3) (1 3)n na ib a ib i i+ + − = + + −

1

2 3 12(1 3) cos 2 ;3 3

p× π = + π + p = 0, 1, 2

1

2 2 36

2(2 ) cos9

p×

π + π =


13

62.(2) cos

9pπ + π =

π π π =

4 4 43 3 37 132 cos , 2 cos , 2 cos

9 9 9

432 cos ;

9pπ = p = 1, 7, 13

Example 23: If p = cisθθθθθ, q = cisϕϕϕϕϕ, so that cot( ).p q

ip q

+ = − θ − ϕ−

Solution: We have p = cos θ + isin θ, q = cosϕ + i sin ϕIt implies p + q = (cos θ + cos ϕ) + i (sinθ + sinϕ)

and p – q = (cos θ – cosϕ) + i (sin θ – sinϕ)

θ + ϕ θ − ϕ θ + ϕ = cos cos 2cos cos

2 2

sin sin 2sin cos2 2

θ + ϕ θ − ϕ θ + ϕ =

cos cos 2sin sin 2sin sin2 2 2 2

θ + ϕ ϕ − θ θ + ϕ θ − ϕ θ − ϕ = = −

sin sin 2cos sin2 2

θ + ϕ θ − ϕ θ − ϕ = On sustituting these values,

θ + ϕ θ − ϕ θ + ϕ θ − ϕ + + =θ + ϕ θ − ϕ θ + ϕ θ − ϕ − − +

2cos cos 2 sin cos2 2 2 2

2sin sin 2 cos sin2 2 2 2

ip qp q i

2

2cos cos sin2 2 2

2sin sin cos2 2 2

i

i i

θ − ϕ θ + ϕ θ + ϕ + =θ − ϕ θ + ϕ θ + ϕ +

cot

2 cot2

ii

θ − ϕ θ − ϕ = = −

Example 24: If a = cis ααααα, b = cis βββββ, c = cis γγγγγ, then

( )( )( ) 8cos cos cos2 2 2

b c c a a babc

β − γ γ − α α − β+ + + =


Solution: Here a + b = (cosα + isinα) + (cosβ + isinβ) = (cosα + cosβ) + i (sinα + sinβ)

α + β α − β α + β α − β= +2cos cos 2 sin cos

2 2 2 2i

2cos cos sin2 2 2

iα − β α + β α + β = + … (1)

Similarly, 2cos cos sin2 2 2

b c iβ − γ β + γ β + γ + = +

… (2)

2cos cos sin2 2 2

c a iγ − α γ + α γ + α + = + … (3)

Also, abc = cis α cisβ cisγ = cos(α + β + γ) + isin(α + β + γ) … (4)On using (1), (2) and (3), we see that

α − β β − γ γ − α α + β α + β + + + = + ×

( )( )( ) 8cos cos cos cos sin2 2 2 2 2

a b b c c a i

cos sin cos sin2 2 2 2

i iβ + γ β + γ γ + α γ + α + × +

( )( )( ) 8cos cos cos2 2 2

a b b c c a α − β β − γ γ − α+ + + = ×

cos sin2 2 2 2 2 2

i α + β β + γ γ + α α + β β + γ γ + α + + + + +

( )( )( ) 8cos cos cos2 2 2

a b b c c a α − β β − γ γ − α+ + + =

or ( )( )( ) 8cos cos cos

2 2 2a b b c c a

abc+ + + α − β β − γ γ − α= (using (4))

Hence the result.

Example 25: Simplify[(cosααααα – cos βββββ) + i (sin ααααα – sin βββββ)]n + [(cos ααααα – cosβββββ) – i (sin ααααα – sin βββββ)]n

Solution: Let (cosα – cosβ) = cos θ and (sinα – sinβ) = sin θ … (1)so that r2 = (cos α – cosβ)2 + (sinα – sinβ)2

= cos2α + cos2β – 2cosαcosβ + sin2α + sin2β – 2sinαsinβ = 2 – [cos(α + β) + cos (α – β)] + [cos(α + β) – cos (α – β)]

= 2 – 2cos (α – β) = 2[1 – cos (α – β)]

2 ( )4sin2

α − β=

implying ( ) ( )2sin and 2 sin

2 2n n nr rα − β α − β= = … (2)


Further,

α + β α − βα − β α + β α + βθ π θ = = = = − = + θ α − β − α + β α − β

1 12cos ( ) sin ( )sin sinsin 2 2tan cot tan1 1cos cos cos 2 2 22sin ( ) sin ( )2 2

implying .2 2

α + βπ θ = + … (3)

Now [(cosα – cosβ) + i (sin α – sinβ)]n + [(cosα – cosβ) – i (sinα – sinβ)]n

= [r (cosθ + i sinθ)]n + [r (cosθ – isinθ)]n

= rn[(cosnθ + isin nθ)] + rn[(cosnθ – isinnθ)],

= 2rncosnθ, using (2) and (3)

Example 26: Prove that

(i) 1(1 sin cos ) (1 sin cos ) 2 cos .cos4 2 4 2

n n n n n ni i + π θ π θ + θ + θ + + θ − θ = − −

(ii)1 sin cos cos sin1 sin cos 2 2

ni n nn i ni

+ α + α π π = − α + − α + α − α

Solution:

(i) 1 sin cos 1 cos sin2 2

i iπ π + θ + θ = + − θ + − θ

22cos 2 sin cos4 2 4 2 4 2

iπ θ π θ π θ = − + − −

2 cos cos sin4 2 4 2 4 2

i π θ π θ π θ = − − + − …(1)

Similarly,

1 sin cos 2 cos cos sin4 2 4 2 4 2

i i π θ π θ π θ + θ − θ = − − − − …(2)

∴ (1 + sinθ + icosθ)n + (1 + sinθ – icosθ)n

2 cos cos sin4 2 4 2 4 2

n n n n n ni π θ π θ π θ = − − + −

2 cos cos sin4 2 4 2 4 2

n n n n n ni π θ π θ π θ + − − − −

2 cos .2cos4 2 4 2

n n n nπ θ π θ = − −

12 cos cos4 2 4 2

n n n n+ π θ π θ = − −


Identically, using (1) and (2), we have

1 sin cos 1 sin cos1 sin cos 1 sin cos

n ni ii i

+ α + α + α + α = + α − α + α − α

cos sin4 2 4 2

cos sin4 2 4 2

n

i

i

π α π α − + − = π α π α − − −

cos sin cos sin4 2 4 2 4 2 4 2

cos sin cos sin4 2 4 2 4 2 4 2

n

i i

i i

π α π α π α π α − + − − + − = × π α π α π α π α − − − − + −

2

2 2

cos sin4 2 4 2

cos sin4 2 4 2

n

i π α π α − + − =

π α π α − + −

2 2 2cos sin 2. .cos sin4 2 4 2 4 2 4 2

n

i i π α π α π α π α = − + − + − −

cos2 sin 2 ,4 2 4 2

n

i π α π α = − + − (using cos2θ – sin2θ cos2θ, 2sinθ cosθ = sin2θ)

π π = − α + − α cos sin .

2 2n nn i n

Hence the proof.

Example 27: If cosααααα + cosβββββ + cosγγγγγ = 0, then show that

(i) cos2 ααααα + cos2 βββββ + cos2 ααααα = 0 or cos2 0Σ α =and sin2 ααααα + sin2 βββββ + sin2 ααααα = 0 or Σ Σ Σ Σ Σ sin2 ααααα = 0

(ii) cos(ααααα + βββββ) + cos(βββββ + γγγγγ) + cos (γγγγγ + ααααα) = 0 or ΣΣΣΣΣcos (ααααα + βββββ) = 0

and sin(ααααα + βββββ) + sin(βββββ + γγγγγ) + sin(γγγγγ + ααααα) = 0 or ΣΣΣΣΣsin(ααααα + βββββ) = 0

(iii) 2 2 2 23 3cos cos cos or cos2 2

α + β + γ = α =∑

and 2 2 2 23 3sin sin sin or sin2 2

α + β + γ = α =∑

Solution: Let a = cisα, b = cisβ, c = cis γ … (1)Then a + b + c = cis α + cis β + cisγ


= (cosα + cosβ + cos γ) + i (sinα + sinβ + sinγ) = 0(using the given conditions) ... (2)

Now for proof of (i), (ii) and (iii), we need to prove a–1 + b–1 + c–1 = 0

By (1)

1 1

1 1

1 1

(cis ) cos sin(cis ) cos sin(cis ) cos sin

a ib ic i

− −

− −

− −

= α = α − α= β = β − β = γ = γ − γ

…(3)

On adding all, a–1 + b–1 + c–1 = (cosα + cosβ + cosγ) – i (sinα + sinβ + sin γ)

= 0 (using the given condition, a + b + c = 0) … (4)

Now, 1 1 10 bc ca aba b c abc

+ += + + =

0 bc ca ababc

+ += i.e. bc + ca + ab = 0. …(5)

where

cis cis cis( ),cis cis cis( ),cis cis cis( )

bccaab

= β × γ = β + γ = γ × α = γ + α = α × β = α + β

…(6)

On substituting (6) in (5), we get

cis(α + β) + cis (β + γ) + cis(γ + α) = 0

i.e. cos(α + β) + cos (β + γ) + cos(γ + α) = 0 or cos( ) 0α + β =∑

and sin(α + β) + sin(β + γ) + sin(γ + α) = 0 or sin( ) 0α + β =∑Now (a + b + c) = 0 (given)

∴ (a + b + c)2 = (a2 + b2 + c2) + 2(ab + bc + ca) = 0

a2 + b2 + c2 = 0 (on using (5)) …(8)

Now a2 = (cisα)2 = cis2α, b2 = (cisβ)2 = cis2β, c2 = (cis γ)2 = cis2γ …(9)

On substituting (9) in (8), we get cis2α + cis2β + cis2γ = 0

i.e. cos2α + cos2β + cos2γ = 0 or cos2 0α =∑

and sin2α + sin2β + sin2γ = 0 or sin 2 0α =∑On rewriting cos2α = 2cos2α – 1 in the above result, we get

(2cos2 α – 1) + (2cos2β – 1) + (2cis2γ – 1) = 0

or 2(cos2α + cos2β + cos2γ) = 3 or 2 3cos2

α =∑

Now in the above result, replacing cos2α by (1 – sin2α), we get

2 2 2 3(1 sin ) (1 sin ) (1 sin )2

− α + − β + − γ =


or 2 2 2 23 3 3sin sin sin 3 or sin2 2 2

α + β + β = − = α =∑

Hence 2 23cos sin2

α = = α∑ ∑

Example 28: If sin ααααα + 2sinβββββ + 3sinγγγγγ = 0, cosααααα + 2cosβββββ + 3cosγγγγγ = 0, then prove thatsin3 ααααα + 8sin3 βββββ + 27sin3γγγγγ = 18 sin(ααααα + βββββ + γγγγγ)

and cos3ααααα + 8cos3 βββββ + 27cos3 γγγγγ = 18cos(ααααα + βββββ + γγγγγ)

Solution: Let a = cisα; b = 2cisβ; c = 3cisγ … (1)∴ a + b + c = cisα + 2cisβ + 3cisγ

= (cosα + isin α) + 2(cosβ + isinβ) + 3(cos γ + i sinγ)

= (cosα + 2cosβ + 3cosγ) + i (sinα + 2sinβ + 3sinγ)a + b + c = 0 as given (cos α + 2cosβ + 3cosγ = 0 = sinα + 2sinβ + 3sinγ )) …(2)

But we know that (a + b + c)2 = a3 + b3 + c3 – 3abc∴ a3 + b3 + c3 = 3abc (on using (2)) …(3)where

3 3

2 3

3 3

(cis ) (cos3 sin 3 )(2cis ) 8(cos3 sin 3 )(3cis ) 27(cos3 sin 3 )

a ib ic i

= α = α + α = β = β + β = γ = γ + γ

…(4)

Further abc = cisα × 2cis β × 3cis γ = 6cis(α + β + γ) …(5)On using (4) and (5) in (3), we get

(cos3α + isin3α) + 8(cos3β + i sin3β) + 27(cos3γ + isin3γ)

= 18{cos(α + β + γ) + isin(α + β + γ)}

(cos3α + 8cos3β + 27cos3γ) + i (sin3α + 8sin3β + 27sin3γ)

= 18{cos(α + β + γ) + isin(α + β + γ)}Comparing real and imaginary parts on both sides, we get

cos 3α + 8cos3β + 27cos3γ = 18cos(α + β + γ)

and sin3α + 8sin3β + 27sin3γ = 18sin(α + β + γ)Hence the result.

Example 29: If ααααα, βββββ be the roots of x2 – 2x + 4 = 0, prove that cos .3

n n nπα + β =

Solution: The given equation is a quadratic in x. Hence its roots are given by

±± − × ± −= = =2 2 32 4 4 4 2 12

2 2 2i

x

Let 1 1 1 1 1 11 3 c is ; 2,3

i x iy r r πα = + = + = θ = θ =


2 2 2 2 2 21 3 c is , ; 2,3

i x iy r r πβ = − = + = θ = θ =

Now, αn + βn = (rcosθ + i r sin θ)n + (r cos θ – i rsinθ)n

= rn[cosnθ + i sinnθ + cosnθ – i sin nθ]

= rn[2cosnθ]

12 . 2.cos 2 cos3 3

n n nn +π π = =

Example 30: If x2 – 2xcosθθθθθ + 1 = 0, show that x2n – 2xncosnθθθθθ + 1 = 0.

Solution: We know that 2 4

2b b acx

a− ± −=

− − θ ± θ −

=2( 2 cos ) 4cos 4·1·1

2a

22 cos 4.( 1)(1 cos )

2θ ± − − θ

=

2cos 2 sin cos sin

2i iθ ± θ= = θ ± θ

Hence the two of the roots are x1 = cos θ + isinθ, x2 = cos θ – i sinθwith x1 + x2 = 2cosθ and x1x2 = (cos θ + isinθ)(cosθ – isin θ) = 1

Further, xn = cosnθ ± i sinnθNow the given equation is x2n – 2xn + 1 = 0

The root, 22cos 4cos 4 2cos 2 sin cos sin

2 2n n n n i nx n i nθ ± θ − θ ± θ= = = θ ± θ

Hence xn is the root of the given equation with sum of the root as x1 + x2 = 2cosnθ andproduct of the roots, x1x2 = 1.

Example 31: If cis ,2r r

x π= show that 1 2Lt 1nn

x x x→∞

…… = −

Proof: Here cis2r rx π = means for r = 1, 1 cis

2x π =

r = 2, 2 2cis

2x π =

r = n, cis2n nx π =


Thenπ π π……… = ………1 2 2

. . cis .cis .cis2 2 2n n

x x x

22 2 2ncis π π π = + + …… + (Using DM Theorem)

The bracket part is a G.P. with common ratio, 12

r = and the first term .2

a π=

∴ 1 2. . . cis1n

ax x xr

… = −

1 22Lt . cis 11

2

nn

x x x→∞

π …… =

−

→∞……… = π = π + π = −1 2Lt . cis cos sin 1n

nx x x i

Hence the solution.

ASSIGNMENT 3

1. Prove that (i) 4 5

3 4

(cos3 sin 3 ) (cos 4 sin 4 ) 1(cos 4 sin 4 ) (cos5 sin 5 )

i ii i −

θ + θ θ − θ =θ + θ θ + θ

(ii)4

cos sin cos 8 sin 8sin cos

i ii

θ + θ = θ + θ θ + θ

(iii)4

5

(cos sin ) sin(4 5 ) cos(4 5 )(sin cos )

i ii

α + α = α + β − α + ββ + β

2. If a = cis2α, b = cis2β, c = cis2γ and δ = cis2δ, prove that

(i) 2cos( )ab cc ab

+ = α + β − γ (ii) 2cos( )ab cdcd ab

+ = α + β − γ − δ

3. Find the general value of θ which satisfies the equation(cosθ + isinθ)(cos2θ + isin2θ) ……… (cosnθ + i sin nθ) = 1

4. Prove that 1 sin cos cos sin1 sin cos 2 2

ni n nn i n

a i+ α + α π π = − α + − α + − α

5. If 12cos xx

θ = + and 12cos ,yy

ϕ = + show that one of the value of

(i)1 is 2cos( )m n

m nx y m n

x y+ θ + ϕ [SVTU, 2007]


(ii) is 2 cos( )nm

n m

yx m ny x

+ θ − ϕ

6. If 12cos xx

θ = + prove that

(i) 12cos rr

r xx

θ = + , (ii)2

2 11 cos

cos( 1)

n

nx n

x x n−+ θ=+ − θ

7. If sinα + sinβ + sinγ = cosα + cosβ + cos γ = 0, prove that(i) sin2α + sin2β + sin2γ = 0

(ii) sin3α + sin3β + sin3γ = 3sin(α + β + γ)(iii) sin4α + sin4β + sin4γ = 2Σsin2(α + β)(iv) sin(α + β) + sin(β + γ) + sin(γ + α) = 0

6.4 ROOTS OF A COMPLEX NUMBER

Statement: There are q and only q distinct values of (cos θ + isinθ)1/q, q being an integer.Since cosθ = cos(2n π + θ) and sin θ = sin(2nπ + θ), where n is any integer.∴ cis θ = cis(2nπ + θ)

By De Moivre’s theorem, one of the values of

1/ 1/ (2 1)(cis ) [cis(2 )] cisq q nnq+ πθ = π + θ = … (1)

Replacing n by the values 0, 1, 2, ..., (q – 1) successively, we get the following q values of1

(cis )qθ ;

θ = π + θ =π + θ = ……………… ………… − π + θ = −

cis (for 0)

(2 )cis (for 1)

(4 )cis (for 2)

[2( 1) ]cis (for 1)

nq

nq

nq

qn q

q

…(2)

Putting n = q in (1), we get a value of (cisθ)1/q = cis(2π + θ/q) = cisθ/q, which is the same asthe value for n = 0.

Similarly for n = q + 1, we get a value of (cisθ)1/q to be cis(2π + θ/q), which is the same asthe value for n = 1 and so on.

Thus the value of (cis θ)1/q for n = q, q + 1, q + 2, etc. are the mere repetition of the q valuesobtained in (2).

Moreover, the q values given by (2) are clearly distinct from each other, for no two of theangles involved therein are equal or differ by a multiple of 2π.

Hence, (cis θ)1/q has q and only q distinct values given by (2).


Example 32: Find the cube roots of unity and show that they form an equilateral trianglein the Argand diagram.

Solution: If x be the cube root of unity, then

1 1 1 13 3 3 3 2(1) (cos0 sin 0) (cos2 sin 2 ) (cis2 ) cis ,

3nx i n i n n π= = + = π + π = π =

where n = 0, 1, 2.For n = 0, x = cis 0 = 1

n = 1,2 2 2 1 3cis cos sin3 3 3 2 2

x i iπ π π= = + = − +

n = 2,4 4 4 1 3cis cos sin3 3 3 2 2

x i iπ π π= = + = − −

The three values of cube roots of unity are represented by the three points A, B, C in theArgand diagram such that OA = OB = OC and angle AOB = 120°, angle AOC = 240°.

So these three points lie on a circle with centre O and unit radius such that angle AOB =angle BOC = angle AOC = 120°. Hence A, B, C form an equilateral triangle.

Example 33: Find all the roots of x 12 – 1 = 0 and identify the roots which are also the rootsof x 4 – x 2 + 1 = 0.

Solution: Given x12 – 1 = 0 or x12 = 1 = cos0 + isin0 = cos2nπ + i sin 2nπ

1

12[cos2 sin 2 ]x n i n= π + π

2 2cos sin cos sin12 12 6 6n n n ni iπ π π π = + = + … (1)

Putting n = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, we get all the 12 roots as below:

n = 0, x1 = cos 0 + i sin0; n = 1, 2 cos sin ;6 6

x iπ π= +

n = 2, 32 2cos sin ;6 6

x iπ π= + n = 3, 43 3cos sin ;6 6

x iπ π= +

n = 4, 54 4cos sin ;6 6

x iπ π= + n = 5, 65 5cos sin ;6 6

x iπ π= +

n = 6, 76 6cos sin ;6 6

x iπ π= + n = 7, 87 7cos sin ;6 6

x iπ π= +

n = 8, 98 8cos sin ;6 6

x iπ π= + n = 9, 109 9cos sin ;6 6

x iπ π= +

n = 10, 1110 10cos sin ;

6 6x iπ π= + n = 11, 12

11 11cos sin ;6 6

x iπ π= + … (2)

By algebra, x12 = (x2 – 1)(x6 + 1)(x4 + x2 + 1).Means, the roots of (x4 + x2 + 1) = 0 are the roots of x12 – 1 = 0, excluding the roots of

(x2 – 1) = 0 and (x6 + 1) = 0.But the roots of (x2 – 1) = 0 are x2 = 1 or x = ± 1. … (3)


x = 1 implies x = cos 0 + isin0 and x = – 1 implies x = cosπ + isinπ.The roots of (x6 + 1) = 0 are

1 1 16 6 6( 1) (cos sin ) [cos(2 ) sin(2 )]x i n i n= − = π + π = π + π + π + π

cos(2 1) sin(2 1) ,6 6

x n i nπ π = + + + n = 0, 1, 2, 3, 4, 5

viz. π π+cos sin ;6 6

i 3 3cos sin ;6 6

iπ π+ 5 5cos sin ;6 6

iπ π+

7 7cos sin ;6 6

iπ π+ 9 9cos sin ;6 6

iπ π+ 11 11cos sin ;6 6

iπ π+ … (4)

On excluding two roots of (3) and six roots of (4) from twelve roots of (2), we get thedesired four roots of the equation (x4 + x2 + 1) = 0 as:

2 2 4 4 5 5cos sin ; cos sin ; cos sin ; cos sin ;

3 3 3 3 3 3 3 3i i i iπ π π π π π π π+ + + +

Example 34: Prove that the nth root of unity form a geometrical progression. Also showthat the sum of these n roots is zero and their continued product is (–1)n–1

Solution: Let nth root of unity be x, i.e. 1

1nx =

or1 1 2(cis 0) (cis 2 ) cis ,n n rx r

nπ = = π = r = 0, 1, 2, ……, n – 1 … (1)

Now roots for different values of r = 0, 1, 2, …, n – 1 from (1) are as follows:

r = 0, 02cis 0 1xnπ = × = ;

r = 1,1

12 2cis 1 cis (say)xn n

π π = × = = ρ ;

r = 2,2

22

2 2cis 2 cisxn nπ π = × = = ρ ;

………………………………………………………………………………………………………………………………

r = n–1, 1

11

2 2cis ( – 1)× cisn

nnx n

n n

−−

−π π = = = ρ

Therefore, the n roots x0, x1, x2, …, xn–1 are ρ0, ρ1, ρ2, …, ρn–1 and their sum Sn = ρ0 + ρ1 + ρ2 + … + ρn–1

0(1 )1

nρ − ρ=− ρ

2 21 cos sin

2 21 cos sin

n

in n

in n

π π − + =π π − +


1 1 02 21 cos sinin n

−= =π π − +

where 2 21 cos sin 0in nπ π − + ≠ for all non zero n.

Hence the sum of n roots of unity is zero.Further their continued product,

P = ρ0.ρ1.ρ2 ……… ρn–1 = ρ0+1+ 2+……+(n–1)

0 1 ( 1)

2 2n n nn + − − = ρ = ρ

1/22cisnn

n

− π =

−π = ×

12cis

2

nn

n(By DM Theorem)

= (cos π + isinπ)n–1 = (–1)n–1

Hence the continued product is (–1)n–1.

Example 35: Find the 7th root of unity and prove that the sum of their nth powers alwaysvanishes unless n be a multiple of 7, n being an integer, and then the sum is 7.

Solution: We know that 1 1 17 7 7 2[1] (cos 2 sin 2 ) (cis 2 ) cis

7

r

r i r r π = π + π = π = Putting r = 0, 1, 2, 3, 4, 5, 6, we get all the seven roots of unity and let these roots be

denoted by 1, ρ, ρ2, ρ3, ρ4, ρ5, ρ6, where 2cis7πρ = .

∴ Sum S of the nth powers of the seven roots = 1 + ρn + ρ2n + ρ3n + ρ4n + ρ5n + ρ6n

= 71 ,

1

n

n

− ρ− ρ

(G.P. with common ratio ρ)

In case when n is not a multiple of 7, ρ7n = (ρ7)n = (cis2π)n = 1i.e. 1 – ρ7n = 0 and 1 – ρn ≠ 0, as n is not a multiple of 7. Thus S = 0In case when n is a multiple of 7, say n = 7p, then

S = 1 + (ρ7)p + (ρ7)2p + (ρ7)3p + (ρ7)4p + (ρ7)5p + (ρ7)6p = 1 + 1 + 1 + 1 + 1 + 1 + 1 = 7

Example 36: Find the equation whose roots are 2 4 62 cos , 2 cos , 2cos .7 7 7π π π

Solution: Let y = cosθ + isinθ;πθ = = ……2 , 0, 1, 2, ,67

r r …(1)

Then y7 = (cosθ + i sin θ)7 ⇒ y7 = (cos7θ + isin7θ)

Implying y7 = 1 or y7 – 1 = 0


⇒ (y – 1)(y6 + y5 + y4 + y3 + y2 + y + 1) = 0 … (2)Leaving the factor (y – 1) which corresponds to θ = 0, we get

y6 + y5 + y4 + y3 + y2 + y + 1 = 0 … (3)

Its roots are y = cosθ, 2 4 6 8 10 12, , , , ,7 7 7 7 7 7π π π π π πθ =

Divide (3) by y3 throughout, we get

3 23 2

1 1 1 1 0y y yy y y

+ + + + + + =

3 21 1 1 13 2 1 0y y y yy y y y

+ − + + + − + + + =

or x3 + x2 – 2x – 1 = 0, where 1 2cosx yy

= + = θ

…(4)

Since

8 6 6cos cos 2 cos ,7 7 7

10 4 4cos cos 2 cos , are repeated roots.7 7 7

12 2 2cos cos 2 cos7 7 7

π π π = π − = π π π = π − =

π π π = π − =

Hence the roots of (4) are 2 4 62cos , 2cos , 2cos .7 7 7π π π

Example 37: If 2 4 3 5 62 2cos sin , and ,7 7

a i b a a a c a a aπ π= + = + + = + + show that b and c

are the roots of the equation x2 + x + 2 = 0

Solution: Given 2 2cos sin7 7

a iπ π= +

∴π π = + = π + π =

77 2 2cos sin cos 2 sin 2 1

7 7a i i …(1)

Now b + c = a + a2 + a3 + a4 + a5 + a6

= (1 + a + a2 + a3 + a4 + a5 + a6) – 1

71(1 ) 1

1aa

−= −−

(1 1) 1 11 a

−= − = −− (³ a7 = 1) …(2)


and bc = (a + a2 + a4)(a3 + a5 + a6)

= a4 + a6 + a7 + a5 + a7 + a8 + a7 + a9 + a10

= a4 + a6 + 1 + a5 + 1 + a + 1 + a2 + a3 (³ a7 = 1)

= (1 + a + a2 + a3 + a4 + a5 + a6) + 2

= 2 (³ 1 + a + a2 + a3 + a4 + a5 + a6 = 0) …(3)∴ The equation whose roots are b and c is

x2 – (b + c)x + bcx = 0, i.e. x2 + x + 2 = 0.

ASSIGNMENT 4

1. Mark the point on the Argand diagram, all the value of ( )151 3i+ √ and verify that they

form a pentagon.

2. Find all the values of (i) ( )141 i+ (ii) ( )

251 i− +

3. If ω is a complex cube root of unity, prove that 1 + ω + ω2 = 0.4. Use DM theorem to solve the equation x4 – x3 + x2 – x + 1 = 0

− + − + + = + = − = −

14 3 2 5 5Hint : ( 1)( 1) ( 1), find ( 1) , except 1x x x x x x x x

5. Solve the equation x6 + x5 + x4 + x3 + x2 + x + 1 = 0

+ + + + + + − = − = =

16 5 4 3 2 7 7Hint : ( 1)( 1) ( 1), find (1) , except 1x x x x x x x x x x

6. Find the roots common to the equations x4 + 1 = 0, x6 – i = 0.

7. Find the equation where roots are 3 52cos , 2cos , 2cos .7 7 7π π π

8. Show that the roots of the equation (x – 1)n = (x)n, n being a positive integer are

1 1 cot ,2

rinπ + r has the values 1, 2, 3, …, (n – 1).

6.5 (I) Expansion of sinnθθθθθ, cosnθθθθθ, tannθθθθθ in powers of sinθθθθθ, cosθθθθθ and tanθθθθθ respectively

(n being positive integer).

We have cosnθ + isinn θ = (cosθ + i sin θ)n (by De Moivre’s theorem)= cosnθ + nC1cos n–1θ(isinθ) + nC2 cosn–2θ (i sin θ)2 + nC3cosn–3θ(i sin θ)3 + ……

(by Binomial theorem)= (cosnθ – nC2cosn–2θsin2θ + ……) + i (nC1 cosn–1θ sin θ – nC3 cosn–3θsin3θ + …..) …(1)Equating real and imaginary parts from both sides, we getcosnθ = cosnθ – nC2cosn–2θsin2θ + nC4cosn–4θsin4θ – ….. …(2)


sinnθ = nC1 cosn–1θsinθ – nC3cosn–3θsin3θ + nC5cosn–5θsin5θ – …… …(3)Replacing every sin2θ by (1 – cos2θ) in (1) and every cos2θ by (1 – sin2θ) in (3), we get the

desired expansions of cos nθ and sin nθ.Dividing (3) by (2),

1 3 3 5 5

1 3 52 2 4 4

2 4

cos sin cos sin cos sintan

cos cos sin cos sin

n n n n n n

n n n n n

C C Cn

C C

− − −

− −θ θ − θ θ + θ θ − ……θ =θ − θ θ + θ θ − ……

and dividing numerator and denominator by cosnθ, we get

3 5

1 3 52 4

2 4

tan tan tantan

1 tan tan

n n n

n n

C C Cn

C Cθ − θ + θ − ……θ =

− θ + θ − …… …(4)

Example 38: Prove that 3 2 21 cos7 ( 2 1) , where 2cos .1 cos

x x x x+ θ = − − + = θ+ θ

Solution: L.H.S.

22 2

2

7 72cos cos cos71 cos7 2 2 , where1 cos cos 22cos cos

2 2

θ θ φ+ θ θ = = = = φ θ θ+ θ φ

… (1)

Now, cos7φ = 7C0 cos7 φ – 7C2 cos5 φsin2φ + 7C4 cos3φ sin4φ – 7C6 cosφ sin6φ = cos7φ – 21 cos5 φ sin2φ + 35 cos3 φ sin4 φ – 7 cos φ sin6φ

= cosφ[cos6φ – 21 cos4 φ sin2φ + 35 cos2φ sin4φ – 7 sin6φ]

Thus φ = φ − φ φ + φ φ − φφ

6 4 2 2 4 6cos7 cos 21cos sin 35cos sin 7 sincos …(2)

Given x = 2cosθ ⇒ 1 (1 cos )2x + = + θ

⇒ 2 21 2cos 2cos2 2x θ + = = φ …(3)

Similarly, 2 21 2cos 2sin2 2x θ − = = φ …(4)

On using (3) and (4), (2) becomes

3 2 2 3

1 1 1 1 1cos7 2 2 2 2 2 221 35 1 7cos 2 2 2 2 2 2

x x x x x x + + − + − φ = − + − − φ

3 2 2 31 2 2 2 2 2 221 35 7

8 2 2 2 2 2 2x x x x x x + + − + − − = + + +

= + + + + + − − + − − +3 2 3 2 3 21 ( 6 12 8) 21( 2 4 8) 35( 2 4 8)

64x x x x x x x x x

3 27( 6 12 8)x x x+ − + −


3 21 64 64 128 6464

x x x= − − +

= x3 – x2 – 2x + 1Therefore from (1),

φ+ θ = = − − + + θ φ

23 2 2cos71 cos7 ( 2 1) ,

1 cos cosx x x where 2

θ = φ

6.5 (II) Addition Formulae for any Number of Angles

We know cos(θ1 + θ2 + … + θn) + isin(θ1 + θ2 + … + θn)

= (cos θ1 + isinθ1)(cosθ2 + isinθ2) … (cos θn + isinθn) …(1)

Rewrite 1 1 1 1

2 2 2 2

(cos sin ) cos (1 tan ),(cos sin ) cos (1 tan )

and so on

i ii i

θ + θ = θ + θ θ + θ = θ + θ

…(2)

∴ cos(θ1 + θ2 + …… + θn) + isin(θ1 + θ2 + …… + θn)

= cosθ1cosθ2 ………… cosθn (1 + i tanθ1)(1 + i tan θ2) …… (1 + i tan θn)

= cosθ1 cosθ2 ……… cos θn[1 + i (tanθ1 + tanθ2 + ……… + tanθn)

+ i2(tanθ1tanθ2 + tanθ2tan θ3 + ……)

+ i3(tanθ1tan θ2tanθ3 + ……)] = cos θ1cosθ2 ……… cosθn(1 + i s1 – s2 – is3 + s4 + ……) …(3)

where s1 = tanθ1 + tanθ2 + ……… + tanθn,

s2 = Σtanθ1tanθ2,

s3 = Σ tan θ1tanθ2tanθ3, etc. …(4)

On equating real and imaginary parts in (3),

cos(θ1 + θ2 + ……… + θn) = cos θ1cosθ2 ……… cos θn(1 – s2 + s4 – ……)

sin(θ1 + θ2 + ……… + θn) = cosθ1cos θ2 ……… cosθn(s1 – s3 + s5 – ……)

Dividing the two,

1 3 5

1 22 4

tan( )1ns s s

s s− + − ……θ + θ + …… + θ =− + − ……

Example 39: If ααααα, βββββ, γγγγγ be the roots of equation x3 + px2 + qx + p = 0, prove thattan–1ααααα + tan–1 βββββ + tan–1 γγγγγ = nπππππ, except in one particular case.

Solution: Let α = tan θ1, β = tan θ2, γ = tanθ3 be the roots of the given equation.

Then, 111 1

0

tan ( 1)1pa

S pa

= θ = − = − = − = α∑ ∑

222 1 2

0

tan tan ( 1)a

S qa

= θ θ = − = = αβ∑ ∑


333 1 2 3

0

tan tan tan ( 1)a

S pa

= θ θ θ = − = − = αβγ∑ ∑

∴1 3

1 1 32

( ) ( )tan( ) 0,

1 1p pS S

S q− − −−θ + θ + θ = = =

− − except q ≠ 1

It implies 1 2 3tan( ) tannθ + θ + θ = π i.e. θ1 + θ2 + θ3 = nπ

Hence, tan–1α + tan–1β + tan–1γ = nπ, except one particular value q ≠ 1.

Example 40: If tan–1x + tan–1y + tan–1 ,2

z π= then show that xy + yz + zx = 1.

Solution: We know that tan–1x + tan–1y + tan–1z = 1 ( )tan

1 ( )x y z xyz

xy yz zx− + + −

− + +

It implies that 1 ( )tan

2 1 ( )x y z xyz

xy yz zx− + + −π =

− + +

( )

tan2 1 ( )

x y z xyzxy yz zx

+ + −π =− + +

π π− + + = + + − 1 ( ) sin cos [( ) ]2 2

xy yz zx x y z xyz

1 – (xy + yz + zx) = 0

i.e. xy + yz + zx = 1.

Example 41: Prove that the equation sin3 θθθθθ = asin θθθθθ + bcosθθθθθ + c has six roots and that thesum of six values of θθθθθ which satisfy it, is equal to an odd multiple of πππππ radians.

Solution: We know that sin3θ = 3sinθ – 4sin3θThen the given equation becomes,

3sinθ – 4sin3θ = asinθ + bcosθ + c … (1)Now we need to prove that equation (1) has six roots and their sum is equal to an odd

multiple of π radians.

Take

θ θ = = θ ++ θ=θ− − θ = =θ + +

22

22

22

2 tan 22sin11 tan

2 , where tan21 tan 12cos

11 tan2

tt

t

tt

…(2)

On substituting values of sinθ and cosθ in (1), we get

3 2

2 2 2 22 2 2 13 4

1 1 1 1t t t ta b ct t t t

− − = + + + + + +


or (b – c)t6 – 2(a – 3)t5 + (b – 3c)t4 – 4(a + 5)t3 – (b + 3c)t2 – 2(a – 3)t – (b + c) = 0 …(3)

which is a six degree equation in tan2

t θ = and has six roots.

Let its roots be t1, t2, t3, t4, t5, t6 or 61 2tan , tan , , tan2 2 2

θθ θ ……

Now 11 1

0

2( 3) 2( 3)( 1) ( 1) nnn

aa aS t Sb c b c a

− − −= = − = = − − − ∑ Q

22 1 2

( 3 ) ( 3 )( 1) b c b cS t tb c b c− −= = − =− −

∑

33 1 2 3

4( 5) 4( 5)( 1) a aS t t tb c b c

− + += = − =− −

∑

44 1 2 3 4

( 3 ) ( 3 )( 1) b c b cS t t t tb c b c

− + += = − = −− −

∑

55 1 2 3 4 5

2( 3) 2( 3)( 1) a aS t t t t tb c b c

− − −= = − =− −

∑

− + += = − = −− −

∑ 66 1 2 3 4 5 6

( ) ( )( 1) b c b cS t t t t t tb c b c

∴ 3 5 6 1 3 51 2 4

2 4 6

tan2 2 2 2 2 2 1

S S SS S S

θ θ θ − +θ θ θ + + + + + = − + −

⇒θ + θ + θ + θ + θ + θ − + = = ∞ − + − = Q

1 2 3 4 5 6 1 3 52 4 6tan , ( 1 0)

2 0S S S

S S S

⇒ 1 2 3 4 5 6tan tan2 2

nθ + θ + θ + θ + θ + θ π = π +

⇒ 1 2 3 4 5 6 2 (2 1)2

n nπ θ + θ + θ + θ + θ + θ = π + = + π

6.5 (III) Expansion of sinmθθθθθ, cosmθθθθθ, sinmθθθθθ.cosmθθθθθ in a series of sines or cosines of

multiple of θθθθθ

If z = cosθ + isinθ then1 cos siniz

= θ − θ

By De. Moivre‘s theorem,

1cos sin and cos sinppz p i p p i p

z= θ + θ = θ − θ

∴ 1 2coszz

+ = θ and − = θ + = θ1 12 sin ; 2cosppz i z p

z z and 1 2 sinp

pz i pz

− = θ


These results are used in the expansion of sinθ, cosθ and their product in a series of sinesor cosines of multiple of θ.

Example 42: If sin4 θθθθθ cos3 θθθθθ = A1cos θθθθθ + A3 cos3 θθθθθ + A5 cos5 θθθθθ + A7 cos 7θθθθθ, prove thatA1 + 9A3 + 25A5 + 49A7 = 0.

Solution: Let x = cisθ, and x–1 = cis(–θ) ⇒ 1 2cos

1 2 sin

xx

x ix

+ = θ − = θ

…(1)

Now,

4 3

4 3

1 1

sin cos2 2

x xx x

i

− + θ θ =

4 2 37 2 4 3

1 4 1 3 14 6 32

x x x xx x x x

= − + − + + + +

7 5 37 7 5 3

1 1 1 1 13 32

x x x xx x x x

= + − + − + + + Using Result (1)

4 37

1sin cos [2cos7 2cos5 6cos3 6cos ]2

θ θ = θ − θ − θ + θ

⇒ 4 36 6 6 6

1 1 3 3sin cos cos7 cos 5 cos3 cos2 2 2 2

θ θ = θ − θ − θ + θ …(2)

Given sin4θcos3θ = A1cosθ + A3 cos3θ + A5cos5θ + A7 cos7θ, …(3)On comparing (2) and (3), we get

7 5 3 16 6 6 61 1 3 3, , ,2 2 2 2

A A A A= = − = − = …(4)

Now, 1 3 5 7 6 6 6 63 3 1 19 25 49 9 25 492 2 2 2

A A A A + + + = + − + − +

61 (3 27 25 49) 02

= − − + =

ASSIGNMENT 5

1. Express sin 6sin

θθ as a polynomial in cosθ.

2. Show that 2 4 6sin7 7 56sin 112sin 64sin .sin

θ = − θ + θ − θθ

3. Show that 2(1 + cos 8θ) = (x4 – 4x2 + 2)2, where x = 2cosθ.


4. Show that tan3 5

2 45 105 ,1 10 5

t t tt t

− +θ =− +

where t = tan θ.

5. Show that 26sin7θ = 35sinθ – 21sin3θ + 7sin5θ – 7θ.

6. Show that 7 1cos (cos7 7 cos 5 21cos 3 35cos .64

θ = θ + θ + θ + θ [Madras, 2003 ]

7. Show that sin8θ = 2–7(cos8θ – 8cos6θ + 28cos4 θ – 56cos2 θ + 35) [Madras, 2001]

8. Show that 32 sin4θ cos2θ = cos 6θ – 2cos4θ – cos 2θ + 2.

9. Show that 5 2 1sin cos (sin 7 3sin 5 sin 3 5sin ).64

θ θ = θ − θ + θ + θ [Madras, 2003]

10. Expand cos5θsin7θ in a series of sine of multiples of θ. [Madras, 2001]11. If cos5θ = Acosθ + Bcos3θ + Ccos5θ, evaluate sin5θ in terms of A, B, C.

6.6 COMPLEX FUNCTIONS

Definition: If for each value of the complex variable z(= x + iy) in a region R, we have one ormore values of ω(= u + iv), then ω said to be a complex function of z and we writeω = u(x, y) + iv(x, y) = f(z) where u, v are real functions of x and y.

If to each value of z, there corresponds one and only one value of ω, then ω is said to be

single valued function of z otherwise a multi-valued function . For example, 1z

ω = is a single

valued function and zω = is a multi valued function of z. The former is defined at allpoints of z-plane except at z = 0 and the latter assumes two values for each value of z exceptat z = 0.

Broadly we classify them under the following type:

(I) Exponential Function of Complex Variable(II) Circular Functions of Complex Variable

(III) Hyperbolic Functions of Complex Variable(IV) Inverse Hyperbolic Functions of Complex Variable(V) Logarithmic Function of a Complex Variable

I Exponential Function of Complex Variable

1. Definition: When x is a real, we are already familiar with the exponential function2

11! 2! !

nx x x xe

n= + + + … + + ………∞

In the same way, we define the exponential function of the complex variable z = x + iy,as:

ez or 2

exp( ) 11! 2! !

nz z zzn

= + + + ……… + + ………∞


2. Properties(i) Exponential form of z = reiθ

Substituting x = 0 in equation (1), we get

2 3 4( ) ( ) ( )

11! 2! 3! 4!

iy iy iy iy iye = + + + + + ………∞

2 4 3 5

12! 4! 3! 5!y y y y

i y = − + − ……… + − + − ………

= cosy + i siny

Thus, ez = ex e iy = ex(cosy + i siny)

Also x + iy = r (cosθ + isinθ) = re iθ. Thus, z = re iθ

(ii) ez is a periodic function having imaginary period 2πi, [ez+2nπi = ez e2nπi = ez]

(iii) ez is not zero for any value of z, since ez = ex+iy = re iθ or ex eiy = reiθ .

As 0, 0, 1

0.

iyx

iyz x x

r e y e

e e e e

= > = = ∴ = = ≠

II Circular Functions of Complex Variable

1. We know that ei θ = cosθ + isinθ and e–iθ = cosθ – isinθSolving these equations, the circular functions of real angles can be written as:

cos , sin and so on.2 2

i i i ie e e ei

θ θ θ − θ+ −θ = θ =

Identically, we can write the circular functions of complex variable z by the equations:

sincos , sin , tan

2 2 cos

iz iz iz ize e e e zz z zi z

− −+ −= = =

2. Properties:

(i) Circular functions are periodic: sinz, cosz are periodic functions having real period2π while tan z have period π. Mathematically: sin(z + 2nπ) = sinz, tan(z + nπ) = tanzetc.

(ii) cosz, secz are even functions, while sinz, cosec z are odd functions.Mathematically: cos(–z) = cos z and sin(–z) = –sinz

(iii) Zeros of sin z are given by z = ±2nπ, zeros of cos z are given by

1 (2 1) , 0, 1, 2,2

z n n= ± + π = …

(iv) All the formulae for real circular functions are valid for complex circular functionse.g. sin2z + cos2z = 1, sin(z1 ± z2) = sinz1cosz2 ± cosz1sinz2

3. Euler’s theorem eiz = cosz + i sin z.


III Hyperbolic Functions of Complex Variable

1. Definitions: Let z be a real or complex number

(i)2

z ze e−+ is defined as hyperbolic cosine of z and written as coshz.

(ii)2

z ze e−− is defined as hyperbolic sine of z and written as sinh z.

Thus, cosh and sinh2 2

z z z ze e e ez z− −+ −= =

Also define sinh 1tanh ; cothcosh tanh

z z z z

z z z zz e e e ez zz e e z e e

− −

− −− += = = =+ −

1 2 1 2sech ; cosech

cosh sinhz z z zz z

z e e z e e− −= = = =

+ −2. Properties:

(i) Periodic functions: sinhz and coshz are periodic functions having imaginary period2πi.Mathematically we can write: sinh(z + 2πi) = sinh z; cosh(z + 2πi) = coshz

(ii) Even and odd functions: coshz is an even function while sinh z is an odd function(iii) sinh0 = 0, cosh0 = 1, tanh 0 = 0(iv) Relations between hyperbolic and circular functions:

Since for all values of θ, sin and cos2 2

i i i ie e e ei

θ − θ θ − θ− +θ = θ =

Substituting θ = ix, we have

2sin . sinh ,2 2 2 2

x x x x x x x xe e e e e e e eix i i i xi i i

− − − −− − − −= = − = = = (eiθ = ei.ix = e–x)

and cos cosh2

x xe eix x− += =

Thus, in brief we have:(a) sin ix = isinhx sinh ix = i sin x sinh–1x = –isin–1(ix)(b) cos ix = coshx cosh ix = cosx cosh–1x = i cos–1ix(c) tan ix = i tanhx tanh ix = i tanx tanh–1x = –i tan–1(ix)(d) cot ix = –i cothx coth ix = –icotx coth–1x = –i cot–1(ix)

(v) Some important formulae of hyperbolic functions:1. cosh2x – sinh2x = 12. sech2x – tanh2x = 13. coth2x – cosech2x = 14. sinh(x ± y) = sinhx coshy ± coshx sinhy5. cosh(x ± y) = coshx coshy ± sinhx sinhy

6.tanh tanh

tanh( )1 tanh tanh

x yx y

x y+± =

±


7. sinh2x = 2 sinh x coshx8. cosh2x = cosh2x + sinh2x = 2cosh2x – 1 = 1 + 2sinh2x

9. 22 tanhtanh 2

1 tanhxx

x=

+10. sinh3x = 3sinhx + 4sinh3x

11. cosh3x = 4cosh3x – 3cosh x

12.3

23 tanh tanhtanh 3

1 3tanhx xx

x+=

+

13. sinh sinh 2sinh cosh2 2

x y x yx y

+ −+ =

14. sinh sinh 2 cosh sinh2 2

x y x yx y

+ −− =

15. cosh cosh 2cosh cosh2 2

x y x yx y

+ −+ =

16. cosh cosh 2sinh sinh2 2

x y x yx y

+ −− =

Proof (1): For all values of θ, we know that cos2θ + sin2θ = 1∴ Substituting θ = ix, we get cos2ix + sin2ix = 1 or cosh2x – sinh2x = 1

Also we can write 2 2

2 2cosh sinh2 2

x x x xe e e ex x− − + − − = −

2 2 2 21 2 2 12

x x x xe e e e− −= + + − − + =

Identically, we can establish formulae (2) and (3).

Proof (4):1sinh( ) sin ( ) [sin cos cos sin ]x y i x y i ix iy ix iyi

+ = + = − +

= – i [i sinh x coshy + coshx i sinh y] = sinhx coshy + coshx sinhy

Also we can write, sinh cosh cosh sinh2 2 2 2

y y y yx x x xe e e e e e e ex y x y− −− −− + + −+ = ⋅ + ⋅

( )

sinh( )2

x y x ye e x y+ − +−= = +

Identically, we can establish the formulae (5) and (6).

Proof (12):3

23 tan tantan 3

1 3 tanA AA

A−=

−

Substituting A = ix, 3

23 tan tantan 3

1 3 tanix ixix

ix−=

−

3 3

2 2

3( tanh ) ( tanh ) 3 tanh tanhtanh 3 or tanh 31 3( tanh ) 1 3 tanh

i x i x x xi x xi x x

− += =− +


Identically, we can establish the formulae (7) to (11).

Proof (16): cos cos 2sin sin2 2

C D C DC D + −− = −

Substituting and , cos cos 2sin sin2 2

x y x yC ix D iy ix iy i i

+ −= = − = −

or cosh cosh 2 sinh sinh 2sinh sinh2 2 2 2

x y x y x y x yx y i i

+ − + − − = − =

Identically, we can establish the formulae (13) to (15).

Example 43: If tan tanh ,2 2x u= prove that [KUK, 2008]

(i) tanx = sinhu and cosxcoshu = 1; (ii) log tan4 2e

xu π = +

Solution: Given tan

2tanh tan2 2 1

xu x= = or

/2 /2

/2 /2

tan2

1

u u

u u

xe ee e

−

−− =+

By componendo and dividendo,

/2 /2 /2 /2

/2 /2 /2 /2

1 tan( ) ( ) 2( ) ( ) 1 tan

2

u u u u

u u u u

xe e e e

xe e e e

− −

− −

++ + − =+ − − −

/2

/22 tan2 4 2

u

ue xe−

π = +

tan4 2

u xe π = +

Implying log tan .4 2

xu π = + This is the desired second part of the problem.

Now tanh tan2 2u x= implies tan

4 2u xe π = + … (1)

tan4 2

u xe− π = − … (2)

From (1) and (2), we have

sin sin1 1 4 2 4 2tan tan

2 2 4 2 4 2 2 cos cos4 2 4 2

u ux x

e e x xx x

− π π + − − π π = + − − = − π π + −


sin cos sin cos4 2 4 2 4 2 4 2sinh

2cos cos4 2 4 2

x x x x

ux x

π π π π + − − − + = π π + −

sin( )sin cos sin cos

cos cos2cos cos2 2

A BA B B AC D C D C D

− −= = + − +

wher ,4 2

xA π= + , ,4 2 2 4 2 2 4 2

x C D x C D xB π + π − π= − = + = −

sinsin4 2 4 2sinh tancoscos cos

2

x xxu xxx

π π + − + = = =π + …(3)

Now 2 2 2 22

1cosh 1 sinh 1 tan seccos

u u x xx

= + = + = = , (using sinhu = tanx)

or cosx cosh u = 1Hence the complete solution of the part (i).

Alternately: We have tanh tan2 2u x=

11 tan1 12tanh tan log log tan

2 2 2 2 4 21 tan2

xu x x

x−

+ π = = = + −

log tan4 2

xu π = + This is the second part of the question.Now we have to prove cosx coshu = 1

L.H.S.

2 2

2 2

1 tan 1 tanh2 2

1 tan 1 tanh2 2

x u

x u

− += ⋅

+ −

But tan tanh ,2 2x u= = (given)

So L.H.S. 2 2

2 2

1 tan 1 tan2 2 1 R.H.S

1 tan 1 tan2 2

x x

x x

− += ⋅ = =

+ −

Converse: tanh tan if log tan2 2 4 2u x xu π = = +

[PTU, 2005]


Given log tan4 2

xu π = + implies tan4 2

u xe π = +

2 2

1 tan2

1 tan2

u ux

e e x

+⋅ =

−

/2

/2

1 tan2

1 tan2

u

u

xe

xe−

+=

−

By componendo and dividendo, we have /2 /2

/2 /2tan

2

u u

u ue e xe e

−

−− =+

tanh tan2 2u x=

Example 44: If coshx = secθθθθθ, prove that log tan4 2

x π θ = − and 2 2tanh tan .

2 2x θ=

[NIT Kurukshetra, 2009]

Solution: We know, cosh sec2

x xe ex−+= = θ implies e2x – 2secθex + 1 = 0

∴22 sec (4 sec 4)

2xe

θ ± θ −=

1 sinsec tan

cos+ θ= θ ± θ =

θ

2 2cos sin 2cos sin

2 2 2 21 1 cos

θ θ θ θ+ +=

− + θ

2

2 2 2

cos sin2 2

cos sin 2sin2 2 2

θ θ + =

θ θ θ + −

2

2 2



θ θ θ θ+ + = =θ θ θ θ− −

1 tan2 tan

4 21 tan2

θ+ π θ = = −θ −

Implying log tan4 2

x π θ = −

Now given coshx = secθ implies 2 2

2 2

1 tanh 1 tan2 2

1 tanh 1 tan2 2

x

x

θ+ += θ− −

… (1)


2 2

2 2

1 tanh 1 tan12 2Since cosh and seccos1 tanh 1 tan

2 2

x

x x

θ + + = θ = = θθ− −

By componendo and divedendo, (1) becomes

2 2 2 2

2 2 2 2

1 tanh 1 tanh 1 tan 1 tan2 2 2 2

1 tanh 1 tanh 1 tan 1 tan2 2 2 2

x x

x x

θ θ + − − + − − =

θ θ + + − + + −

i.e. 2 2tanh tan2 2x θ=

IV Inverse Hyperbolic Functions of Complex Variable

1. Definition: Let sinh u = z, then u is called the hyperbolic sine inverse of z and iswritten as u = sinh–1z. Similarly we define cosh–1z, tanh–1z.The inverse hyperbolic functions like other inverse functions are many valued, but weshall consider their principal values only.

2. Some Results 1 2

1 2

1

sinh log ( 1) ( )

cosh log ( 1) ( )1 1tanh log ( )2 1

z z z i

z z z iizz iiiz

−

−

−

= + + … = + − … += … −

(i) Let sinh–1z = u, then 1sinh ( )2

u uz u e e−= = −

212 or 2 1 0u u uu

z e e z ee

= − − − =

This being a quadratic in eu, we get 2

22 (4 4)

( 1)2

uz z

e z z± +

= = ± +

∴ Taking the positive sign only, 2( 1)ue z z= + + or = + + 2log ( 1)u z z

Similarly we can establish (ii).

(iii) Let tanh–1z = u, then z = tanhu, i.e. u u

u ue eze e

−

−−=+

Applying componendo and dividendo, we get 211

uu

uz e ez e−

+ = =−

or 12 log1

zuz

+ = −, whence follows the result.

Problem 45: Show that (i) 1sinh (tan ) log tan4 2

− π θ θ = + ` [KUK, 2005]

(ii) 1sec (sin ) logcot2

h− θθ = (iii) tanh–1(cos θθθθθ) = cosh–1(cosec θθθθθ)


Solution: (i) 1 2sinh (tan ) log tan (tan 1)− θ = θ + θ + ,

( )1 2Using sinh ( 1)z z x− = + + = log[tan θ ± secθ]

sin 1 1 sinlog logcos cos cos

θ + θ = ± = θ θ θ , (taking + ve sign only) … (1)

2 2sin cos 2sin cos

2 2 2 2log1 1 cos

θ θ θ θ + + = − + θ

2 2

2 2

sin cos 2sin cos2 2 2 2log

sin cos (1 cos )2 2

θ θ θ θ+ + = θ θ + − − θ

2

2 2 2

sin cos2 2log

sin cos 2sin2 2 2

θ θ + =θ θ θ + −

2

2 2

sin cos2 2log

cos sin2 2

θ θ + = θ θ −

sin cos

2 2log ,cos sin

2 2

θ θ + = θ θ−

(Taking cos2θ common)

1 tan2log

1 tan2

θ + = θ−

log tan2 2

π θ = + . Hence proved.

Alternately: Let sinh–1(tanθ) = z implying sinhz = tanθ

2tan or 2 tan 1 02

z zz ze e e e

−− = θ − θ − =

22 tan 4 tan 4

2ze θ ± θ +=


ez = tanθ ± secθimplying z = log[tanθ ± sec θ]

sin 1logcos cos

θ = ± θ θ

1 sinlog ,

cos+ θ = θ

then onwards as above (1)

(ii) 1 1 11 1sech (sin ) cosh cosh , sinsin

zz

− − − θ = = = θ θ

21 1log 1z z

= + −

( )21 1

logz

z

+ − =

21 (1 sin )

logsin

+ − θ= θ

1 coslog

sin+ θ = θ

22cos2log

2cos sin2 2

θ = θ θ

log cot .2θ= Hence proved.

(iii) As we know that 1 1 1tanh log2 1

zzz

− + = −

∴ 2

1

2

2cos1 1 cos 1 2tanh (cos ) log log2 1 cos 2 2sin

2

−

θ + θ θ = = θ− θ

12 2

cos cos2 2log log

sin sin2 2

θ θ = = θ θ

cos 2cos2 2log

sin 2cos2 2

θ θ = × θ θ

,


1 coslog

sin+ θ = θ 2 2cos (1 cos ) and 2sin cos sin

2 2 2θ θ θ = + θ = θ Q

1log cot

sin = + θ θ

( )2log cosec cosec 1= θ + θ −

= cosh–1(cosecθ), ( )2 1 log 1 coshz z z−+ − =Q

V logarithmic Function of a Complex Variable

1. Definition: If z = x + iy and w = α + i β be related that ez = w, z is called the logarithmof w to the base e and is written as z = log ew.Now w = ez = ez · 1 = ez · w2nπi, (Since 1 = cos2nπ + i sin 2nπ = ei2nπ)Taking log on both sides, Log w = Log(ez + 2nπi) = z + 2nπi, where n = 0, ±1, ±2, …

Hence the logarithm of complex number has an infinite number of values and is,therefore, a many valued function. Logw (beginning with capital letter L) is called thegeneral value, logw is the principal value of a complex number (z, say). This principalvalue is obtained by taking n = 0 in Log w.

2. We know that the logarithm of a negative quantity has no real value. But we can nowevaluate this.e.g. loge(–3) = loge3(–1) = loge3 + loge(–1) = loge3 + iπ where –1 = cis π = ei π

3. Real and imaginary part of Log(x + iy): log(x + iy) = 2inπ + log(x + iy)

= 2inπ + log[r(cosθ + i sinθ)]

where x = r cosθ, y = rsinθ so that 2 2( )r x y= + and 1tanyx

− θ = ∴ log(x + iy) = 2inπ + log(reiθ)

2 2 12 log log ( ) 2 tany

in r i x y i nx

− = π + + θ = + + π + 4. If az = N, then z is called the logarithm of N to the base a and is written as z = logaN,

where N, a and z are complex numbers.

For az = N taking logs on both sides, z logea = log N implies =log logez a Ne e implies⋅ =logez ae N. implies log ,ez a ze a⋅ = where a and z both are complex numbers.

5. log

loglog

ea

e

zz

a=

Example 46: Find the real and imaginary parts of (ααααα + iβββββ)x + iy


Solution: Substitute α = rcosθ, β = rsinθ so that 2 2 1( ) and tanr − β= α + β θ =α

Here (α + i β)x + iy = e(x + iy)Log(α + iβ), (using logezza e ⋅ α= )

= e(x + iy)[2inπ + log(α + iβ)

( ) 2 ix iy in reeθ + π+ =

= e(x + iy)[log r + i(2nπ + θ)]

= eA + iB = eA(cosB + isinB)

where A = x logr – y(2nπ + θ) and B = y logr + x(2nπ + θ)∴ Required real part = eA cosB and the imaginary part = eA sinB.

Example 47: Prove that log(1 )ii + is 2 log21 cos

8 4eπ − π

Solution:1 11 log 2 tan 1 log 1 tanlog(1 ) log(1 ) log 2L.H.S.

i ii i ii e e− − + ⋅ + ∞ + + ⋅ = = =

2 log 21 log 2

8 42 4 2ii i

e eπ ππ π − ++ ⋅ + = =

2 2log2

8 84log 2 log 2

cos sin4 4

ie e e i

ππ π− − π π = = +

∴ Required real part = eAcosB and the imaginary part= eAsinB, where π= − = π2 log 2

and8 4

A B

Example 48: If ···iii A iB

∞= + principal values only being considered, prove that

(i) tan2A B

Aπ = (ii) A2 + B2 = e–Bπππππ [NIT Kurukshetra, 2007]

Solution: Given iii A iB…∞ = + means iA + iB = A + iB (since…. iii A iB

…∞ = + )

Taking logs on both sides, (A + iB) log i = log(A + iB)

1 2 2 11 1( ) log 1 tan log( ) tan2 2

BA iB i A B iA

− − + + ∞ = + +

2 2 11( ) log( ) tan2 2

BA iB i A B iA

−π + = + + Now equating real and imaginary parts,

2 21 log( )2 2

BA B − π+ = implying A2 + B2 = e–Bπ

and 1tan2

B AA

− π= implying tan2

B AA

π=


Example 49: If (a + ib)p = mx + iy, prove that

1

2 2

2 tan

log( )

by ax a b

−

=+

Solution: Taking logs on both sides of (a + ib)p = mx + iy, we get p log(a + ib) = (x + iy)logm

implying 2 2 11 log( ) tan ( )log2

bp a b i x iy ma

− + + = + Now equating real and imaginary parts,

2 21 log( ) log2

p a b x m+ = …(i)

and 1tan logbp y ma

− = ..(ii)

Dividing the two, 1 1

2 22 2

tan 2tan

log( )log( )2

b bpy a apx a ba b

− −

= =++

Example 50: Prove that (4 1)

2nii e

π− π += and show that its values form a geometrical

progression.

Solution: By definition, 22 log2 loglog

i

i n i en i ii ii ii e e eπ

⋅ π + ⋅ π + ⋅ = = =

2 (4 1)2 2ii n i n

e eπ π⋅ π + − π+ = =

On putting n = 0, 1, 2,…, we get the values of ii as 952 2 2, , ,e e eπ π π− −−⋅ … which form a G.P.

whose common ratio is e–2π.

Example 51: If (1 + i)(1 + 2i)…(1 + in) = (x + iy) , then show that 2.5.10 … (1 + n2) = (x2 + y2)

Solution: Given (1 + i)(1 + 2i) … (1 + in) = (x + iy),Taking log on both sides,

log(1 + i)(1 + 2i)…(1 + in) = log(x + iy)

Implies 2 2 1 2 2 1 2 2 11 1 1log(1 1 ) tan 1 log(1 2 ) tan 2 log(1 ) tan2 2 2

i i n i n− − − + + + + … + +

2 2 11 log( ) tan2

yx y i

x−= + +

On equating the real and imaginary parts,


2 2 2 2 2 2 2 21 1 1 1log(1 1 ) log(1 2 ) log(1 ) log( )2 2 2 2

n x y + + + + … + + = +

Implying 2 2 21 1log 2 log 5 log 10 log(1 ) log( )2 2

n x y ⋅ ⋅ … + = +

or 2 · 5 · 10 ·…… (1 + n2) = (x2 + y2)

Example 52: Prove that 2 22tan log .a ib abi

a ib a b − = + −

[KUK, 2006, 07]

Solution: Let a = rcosθ and b = rsinθ then tan ,ba

θ = r2 = a2 + b2 … (1)

Now, L.H.S. tan log a ibia ib

− = +

= tan i [log(a – ib) – log(a + ib)]

= tan i [log(re–iθ) – log(rei θ)]

= tan i [logr – iθ) – (logr + i θ)]

= tan i (–2iθ) = tan(–i22θ) = tan 2θ

= 22 2 2

2

22 tan 21 tan 1

baba

b a ba

θ = =− θ −−

(Using equation (1))

Example 53: If log(x + iy) = a + ib where a2 + b2 ≠≠≠≠≠ 0, show that 2 22 2

2tan log( )1

ax ya b

+ =− −

Solution: Given tanlog(x + iy) = a + ib implies log(x + iy) = tan–1(a + ib)

Now log(x + iy) = logr cisθ = tan–1(a + ib), where 2 2r x y= + and 1tanyx

−θ =

or 2 2 1 11 log( ) tan tan ( )2

yx y i a ib

x− −+ + = + …(1)

Identically, 2 2 1 11 log( ) tan tan ( )2

yx y i a ib

x− −+ − = − …(2)

Adding the two, log(x2 + y2) = tan–1(a + ib) + tan–1(a – ib)

1 ( ) ( )tan

1 ( )( )a ib a ib

a ib a ib− + + −=

− + −

12 22tan

1a

a b−=

− −

or 2 22 22tan log( )

1ax y

a b + = − −


Example 54: (i) tanhz + 2 = 0 (ii) cosz = 2

Solution: (i) tanh z + 2 = 0 ⇒ 2z z

z ze ee e

−

−− = −+

Apply Componendo and dividendo,

( ) ( ) 1 ( 2) 1( ) ( ) 1 ( 2) 3

z z z z

z z z ze e e ee e e e

− −

− −+ + − + − −= =+ − − − −

13

z

zee−

= − or 3z ie = , (taking positive sign only)

i.e. log3iz = = logr + eiθ,

For − − π = = = θ = = ∞ = π + 1 11 10, ; , tan tan

3 3 2y

x y r nx

z 12 1log 3

2i n

− = + + π

z 1 1log 32 2

i n = − + + π

ASSIGNMENT 6

1. Find the general value of (i) log(6 + 8i) (ii) log(–1) (iii) ii [MDU, 2005]2. Show that (i) log (1 + itan α) = log (sec α)+ iα, where α is an acute angle.

(ii) 3log3e

ii

−+

1 12 tan3

i n − = π −

3. Find the modulus and argument (i) (1 + i)1 – i (ii) log(1 + i)

4. If (a1 + ib1)(a2 + ib2) …… (an + ibn) = A + iB, prove that

(i) ( )( ) ( )2 2 2 2 2 2 2 21 1 2 2 n na b a b a b A B+ + …… + = +

(ii)1 21 1 1 1

1 2

tan tan tan tann

n

bb b Ba a a A

− − − −+ + … + =

5. Prove that 1log 2tan ,a ib ba ib a

−+ = − hence evaluate cos log .a ibi

a ib + −

6. If sin–1(x + iy) = log(A + iB), show that 22

2 21,

sin cosyx

u u− = where A2 + B2 = e2u.

[KUK, 2007]


7. If i(α + iβ) = α + iβ, prove that a2 + b2 = e–(4n + 1)πβ [KUK, 2005]

8. If (1 ) ,(1 )

x iy

x iyi ii

+

−+ = α + β−

prove that one of the value of 1tan log 22x y− β π= +

α9. Find all the roots of (i) sinZ = cosh4 (ii) sinhZ = i.

6.7 REAL AND IMAGINARY PART OF CIRCULAR AND HYPERBOLIC FUNCTIONS:

(1) To separate real and imaginary parts of circular function:(i) sin(x + iy); (ii) cos(x + iy); (iii) tan(x + iy);

(iv) cot(x + iy); (v) sec(x + iy); (vi) cosec(x + iy)

Proofs:

(i) sin(x + iy) = sinx cos iy + cosx sin iy = sin x coshy + icosx sinhyLikewise, cos(x + iy) = cosx coshy – isinx sinhy

(iii) Let α + iβ = tan(x + iy) then α – iβ = tan(x – iy)Adding 2α = tan(x + iy) + tan(x – iy)

i.esin( ) sin 2 sin 2

2cos( )cos( ) cos 2 cos2 cos2 cosh 2x iy x iy x xx iy x iy x iy x y

+ + −α = = =+ − + +

.

Subtracting 2iβ = tan(x + iy) + tan(x – iy)Likewise,

sin 2 sinh 2

2cos( )cos( ) cos 2 cosh 2iy i y

ix iy x iy x y

β = =+ − +

∴ sinh 2cos2 cosh 2

yx y

β =+

Likewise, sin 2 sinh 2

cot( )cosh 2 cos2

x i yx iy

y x−+ =

−(v) Let α + iβ = sec(x + iy) = sec(x + iy) then α – iβ = sec(x – iy)

Adding 2α = sec(x + iy) + sec(x – iy)

i.e. − + +α = = =

+ − + +2 cos cos 2 cos coshcos( ) cos( )

2 cos( )cos( ) cos 2 cos 2 cos 2 cosh 2x iy x yx iy x iy

x iy x iy x iy x y

subtracting, 2iβ = sec(x + iy) – sec(x – iy)

2sin sin 2 sin sinhcos( ) cos( )2cos( )cos( ) cos2 cos 2 cos2 cosh 2

x iy i x yx iy x iyi

x iy x iy x iy x y− − +β = = =

+ − + +

∴ 2sin sinh

cos2 cosh 2x y

x yβ =

+

Likewise,sin cosh cos sinh

cos ( ) 2cosh 2 cos2

x y i x yec x iy

y x−

+ =−


(2) To separate real and imaginary parts of hyperbolic functions:(i) sinh(x + iy); (ii) cosh(x + iy); (iii) tanh(x + iy)

Proof:

(i) sinh(x + iy) = (1/i) sin i (x + iy) = (1/i) sin(ix – y) = (1/i)[sin ix cosy – cos ix siny]

= (1/i)[isinhx cosy – coshx sin y]

= sinhx cosy + icoshx siny

Likewise, cosh(x + iy) = coshx cosy

(iv) If α + iβ = tanh(x + iy) = (1/i)tan(ix – y),

Then α – iβ = tanh(x – iy) = (1/i)tan(ix + y)

Adding 2α = (1/i)[tan(ix – y) + tan(ix + y)]

− + +α = = =⋅ − + + +

sin ( ) (1/ ) sin 2 sinh 22 cos( ) cos ( ) cos 2 cos 2 cos 2 cos 2

ix y ix y i ix xi ix y ix y ix y x y

Subtracting, 2iβ = (1/i)tan(ix – y) – tan(ix + y)

sin ( ) ( )2cos( )cos( )

ix y ix yi

i ix y ix y

+ − − β = −⋅ + −

sin 2 sin 2

cos2 cos2 cosh 2 cos2y y

ix y x yβ = =

+ +

Example 55: If sin(ααααα + iβββββ) = x + iy, prove that

(a)22

2 21

cosh sinhyx + =

β β (b)22

2 21

sin cosyx − =

α α

Solution: (a) x + iy = sin(α + iβ) = sinα coshβ + icosα sinhβSeparating real and imaginary parts,

x = sinα coshβ; y = cosα sinhβ

Rewrite as: sin and coscosh sinh

yxα = α =β β

Squaring and adding 22

2 21

cosh sinhyx + =

β β Hence proved the first part of the question.

(b) Again cosh and sinhsin cos

yxβ = β =α α

Squaring and subtracting, we get

22

2 21 ,

sin cosyx= −

α αbecause cosh2β � sinh2β = 1. Hence proved.


Example 56: If cosh(u + iv) = x + iy, prove that 2 22 2

2 2 2 21 and 1

cosh sinh cos siny yx x

u u v v+ = − =

[PTU, 2009]

Solution: We have x + iy = cosh(u + iv) = cos i(u + iv) = cos iu cosv + sin iu sin v = coshu cosv + isinhu sinv

∴ Separating real and imaginary parts, we get x = coshu cosv; y = sinhu sinv

i.e. cos and sincosh sinh

yx v vu u

= =

Squaring and adding, we get the first result.

Again cosh and sinhcos sin

yx v uv v

= =

Squaring and subtracting, we get the second result.

Example 57: If tan(θθθθθ + iϕϕϕϕϕ) = tanααααα + i secααααα, prove that 2 cot2

e ϕ α= ± and 12 .2

n θ = + π + α

[NIT Kurukshetra, 2005; KUK, 2009]

Solution: sin 1 sintan( ) tan seccos cos cos

ii i iα α +θ + ϕ = α + α = + =α α α

…(1)

sin 1 sintan( ) tan seccos cos cos

ii i iα α −θ − ϕ = α − α = − =α α α … (2)

We can write equations (1) and (2) in the form as θ + iϕ = tan–1(tan α + i secα) … (3)

θ – iϕ = tan–1(tan α – i secθ) … (4)Adding (3) and (4), we have

2θ = tan–1(tanα + isecα) + tan–1(tanθ – isecα)

1

2 22 tantan

1 (tan sec )− α= − α + α

− α = − α 1

22 tantan2 tan

1 1tantan

− = − α [ ]1tan cot−= − α

1tan tan2

− π = + α


1 1tan tan2 2

n n− π = π + + α = + π + α

122

n θ = + π + α Similarly subtracting (4) from (3), we have

iϕ = tan–1(tanα + isecα)– tan–1(tan α – i secα)

1

2 22 sectan

1 (tan sec )i− α= + α + α

1

22 sectan

2(1 tan )i− α= + α

22

2

sec sectan(2 )sin1 tan 1cos

i i iα αϕ = =α+ α +α

i tanh 2ϕ = i secα cos2α, because sin2α + cos2α = 1

2 2

2 2tanh 2 cose e

e e

ϕ − ϕ

ϕ − ϕ−ϕ = = α+

By componendo and dividendo

2

4 2

2

2cos1 cos 2 cot1 cos 22sin

2

e ϕ

α+ α α= = =α− α

implying 2 cot2

e ϕ α= ±

Problem 58: If x = 2cos ααααα cosh βββββ, y = 2 sin ααααα sinh βββββ, prove that sec( ) sec( )i iα + β + α − β

2 24 .x

x y=

+ [NIT Kurukshetra, 2003, 2007]

Solution: LHS = sec(α + iβ) + sec(α – iβ).

cos( ) cos( )1 1

cos( ) cos( ) cos( )cos( )i i

i i i iα + β + α − β= + =

α + β α − β α + β α − β

[2cos cos ]

(cos cos sin sin )(cos cos sin sin ) 'i

i i i iα β=

α β − α β α β + α β

CD Formula 2cos cos2 2

C D C D+ −

2 cos cosh

[(cos cosh sin sinh )(cos cosh sin sinh )]i iα β=

α β − α β α β + α β

2 24

2 2 2 2

x xy yx x x yi i

= = +− +


Example 59: Reduce tan–1(cos θθθθθ + i sin θθθθθ) to the from (a + ib). Hence show that

1tan log tan2 4 2 4 2

i n ie− θ π π π θ = + − − OR

If tan(a + ib) = cos θθθθθ + i sin θθθθθ, where letters denote the real quantities, prove that

2 4na π π= + and 2 tan ,

4 2be π θ = + where n is the integer. [PTU, 2007]

Solution: Given tan–1(cosθ + isinθ) = a + ibso that tan–1(cosθ – i sinθ) = a – ib

∴ 2a = tan–1(cosθ + isinθ) + tan–1(cosθ – i sin θ) …(1)

1 (cos sin ) (cos sin )tan

1 (cos sin )(cos sin )i i

i i− θ + θ + θ − θ= − θ + θ θ − θ

1 2costan

1 (cos sin )(cos sin )i i− θ= − θ + θ θ − θ

1

2 2 22costan

1 (cos sin )i− θ= − θ − θ

1

2 22costan

1 (cos sin )− θ= − θ + θ

= tan–1 ∞

2 or2 2 4

na n aπ π π = π + = + …(2)

Similarily, 2ib = tan–1(cosθ + isinθ) – tan–1(cosθ – i sinθ)

1 (cos sin ) (cos sin )tan

1 (cos sin )(cos sin )i i

i i− θ + θ − θ − θ= + θ + θ θ − θ

,

1

2 22 sin2 tan

1 (cos siniib − θ= + θ + θ

or tan 2ib = i sinθ implying i tanh 2b = i sin θ (³ tan ix = i tanhx) tanh2b = sinθ ...(3)

Taking inverse on both sides, 2b = tanh–1(sinθ)

1 1 sin2 log2 1 sin

b + θ = − θ

2 2

2 2

cos sin 2sin cos1 2 2 2 22 log2 cos sin 2sin cos

2 2 2 2

b

θ θ θ θ + + = θ θ θ θ+ −


2

cos sin1 2 2log4 cos sin

2 2

b

θ θ + = θ θ−

⇒

2

1 tan1 2log4 1 tan

2

b

θ + = θ−

…(4)

1 log tan2 4 2

b π θ = + ...(5)

Also from (4),

−θ − = θ+

2

1 tan1 2log4 1 tan ,2

b (on interchange of numerator and denominator in (4))

i.e. 1 log tan

2 4 2b − π θ = −

…(6)

Hence, from (2) and (6), log tan2 4 2 4 2

n ia ib π π π θ + = + − −

Also from (5), 2 log tan ,4 2

b π θ = + i.e. 2 tan4 2

be π θ = +

Example 60: Separate cos–1(cosθθθθθ + isinθθθθθ) into real and imaginary parts, where θθθθθ is a positiveacute angle.

Solution: Let cos–1(cosθ + i sin θ) = x + iy …(1)∴ cosθ + isinθ = cos(x + iy) = cosx cos iy – sinx sin iy

= cosx coshy – isinx sinh yOn comparing real and imaginary parts on both side,

}cos cos coshsin sin sinh

x yx y

θ =θ = − …(2)

Therefore, cos2θ + sin2θ = (cos x coshy)2 + (–sinx sinh y)2

⇒ 1 = cos2x(1 + sinh2y) + sin2x sinh2y

⇒ 1 = cos2x + sinh2y(cos2x + sin2x)

⇒ 1 = cos2x + sinh2y


⇒ 1 – cos2x = sinh2y

sin2x = sinh2y …(3)

Now, sinθ = –sinx sinhy ⇒ sin2θ = sin2x sinh2y

⇒ sin2θ = sin4x [using(3)]Sinceθ being given acute positive angle, sinθ is positive.

∴ 2sin sin sin 12

x π= θ = = …(4)

⇒ sinx = ± 1 ⇒ sin sin2

x π= ±

⇒ 2

x π= ± i.e. x lies between and2 2π π−

Further, 1sin sin sin sinx x −= θ ⇒ = θ , (From (4))

Also, sinθ = –sinx sinhy, [from(2)]

⇒ sin sin sinh yθ = − θ

⇒ sinh sin (say)y = − θ = α

⇒ 1 2sinh log 1y −= α = α + α +

⇒ log sin sin 1y = − θ + θ +

Example 61: If sin–1(u + iv) = ααααα + iβββββ, prove that sin2ααααα and cosh2βββββ are the roots of theequation x2 + (1 + u2 + v2)x + u2 = 0 [NIT Kurukshetra, 2008, 2009]

Solution: Given sin–1(u + iv) = α + iβTaking inverses, u + iv = sin(α + iβ) = sinα coshβ + i sinh β cosαOn comparing real and imaginary parts, u = sinα coshβ and v = sinh β cosα

u2 = sin2α cosh2β = sin2α(1 + sinh2β) = sin2α + sin2α sinh2β …(1)

v2 = cos2α sinh2β = cos2α (cosh2β – 1) = cos2α cosh2β – cos2α …(2)

and 1 + u2 + v2 = sum of the roots

= 1 + [sin2α + sin2α sinh2β] + [cos2α cosh2β – cos2α]

= 1 + [(1 – cos2α) + sin2α (cosh2β – 1)] + [cos2α cosh2β – cos2α]

= 2 – 2cos2α + sin2α cosh2β + cos2α cosh2β – sin2α = [2(1 – cos2α) – sin2α] + cosh2β (sin2α + cos2α)

= sin2α + cosh2β

Alternately: x2 – (sum of the roots)x + product of the roots = 0


implies x2 – [1 + u2 + v2]x + u2 = 0

x2 + [1+ sin2α cosh2β + cos2α sinh2β]x – sin2α cosh2β = 0

x2 + [1 + sin2α cosh2β + cos2α(cosh2β – 1)]x – sin2α cosh2β = 0

x2 + [(1 – cos2α) + cosh2β(sin2α + cos2α)]x – sin2α cosh2β = 0

x2 + [sin2α + cosh2β)x – sin2α cosh2β = 0

(x – sin2α)(x – cosh2β) = 0Implies x = sin2α, cosh2β are the roots of the above equation.

Example 62: If cos–1(x + iy) = ααααα + iβββββ, so that(i) x2sec2ααααα – y2cosec2ααααα = 1, (ii) x2sech2βββββ + y2cosech2βββββ = 1

Solution: We have, cos–1(x + iy) = α + iβ implies x + iy = cos(α + iβ) …(1)and cos–1(x – iy) = α – iβ implies x – iy = cos(α – iβ) ...(2)

Adding (1) and (2), 2x = cos(α + iβ) + (α – iβ)

α + β + α − β α + β − α + β= = α β( ) ( )2cos cos 2cos cos

2 2i i i i i

x = cos α coshβ …(3)Subtracting equation (ii) from (i), similarly we get

2 iy = cos(α + iβ) – cos(α – iβ)

( ) ( )2sin sin 2 sin sinh

2 2i i i i iα + β + α − β α − β − α − β= = − α − β

y = – sinα sinhβ …(4)From (3) and (4), we get

22

2 22 2

cosh sinh 1cos sin

yx − = β − β =α α

and 22

2 22 2

sin cos 1cosh sinh

yx + = α + α =β β

Example 63: Prove that 1sin (cosec ) 2 ( 1) ( 1) logcot2 2

n nn i− π θ θ = π + − + −

Solution: Let sin–1(cscθ) = z implies sin z = csc θ

csc2

iz ize ei

−− = θ or e2 iz – 2 icscθ e iz – 1 = 0

[ ] [ ]2 22 csc 4 (csc ) 4 2 csc 2 cot

csc cot2 2

izi i i i

e i θ ± θ − θ ± θ = = = θ ± θ

iz = log[i(csc θ ± cotθ)]


2 1 coslog logsin

i

iz eπ + θ = + θ

1 1 coslog 2 ( 1) log cot

2 sin 2 2nz n i

iπ + θ π θ = + = + − − θ

Alternately: Let sin–1(cscθ) = α + i βImplying cscθ = sin(α + iβ) = sin α cos iβ + cos α sin i β

= sinα coshβ + icosα sinh β …(1)Equating real and imaginary parts,

csc sin cosh ( )

0 cos sinh ( )iii

θ = α β … = α β …

…(2)

From (ii), Either cosα = 0 implying 2πα = …(3)

or sinh β = 0 implying β = 0

But β ≠ 0, since if β = 0 then coshβ = 1 and sin sin 1,2πα = = which is contrary to hold (i).

Substituting the value of α in (ii), csc 1 cosh2

e eβ −β+θ = ⋅ β =

Implying e 2β – 2csc θ eβ + 1 = 0, which is a quadratic in eβ and gives

e2β = csc θ + cot θ or log cot2θβ = …(4)

ASSIGNMENT 7

1. If cos(α + iβ) = r(cosθ + isin θ), prove that

(i) 2 sin( )sin( )

e β α − θ=α + θ (ii)

sin( )1 log2 sin( )

α − θβ =α + θ [Madras, 2003; KUK, 2005]

2. If cos(θ + i ϕ) = cosα + isinα, pave that(i) *sin2θ = ±sin α (ii)**cos2θ + cosh2θ = 2 [Madras,**2000, *03; KUK, 2005, 2008]

3. If tan(A + iB) = x + iy,(i) x2 + y2 + 2x cot 2A = 1 (ii) x2 + y2 – 2y coth2B + 1 = 0 (iii) xsinh2B = ysin2A

4. If tan(x + iy) = sin(u + iv), prove that sin 2 tan .sinh 2 tan

x uy v

=

5. If tanh ,4ia ib v π + = + prove that a2 + b2 =1.

6. Prove that


(i) ( )− = π + + +1 2sin ( ) 2 log 1ix n i x x (ii) 1sin (cos ) log cot .2 2

ec i− π θθ = +

7. If tan(θ + iϕ) = e iα, show that 12 2

n π θ = + and 1 log tan2 4 2

π α ϕ = +

8. Separate tan–1(x + iy) into real and imaginary parts.9. Separate sin–1(cosθ + i sin θ) into real and imaginary parts, where θ is a positive acute

angle. [PTU, 2006; KUK, 2007]

6.8 SUM OF THE SERIES

Example 64: Find the sum of the series

e ee

α αα β − β + β − ……∞

3 5cos cos3 cos5

3 5

Solution: Let 3 5

cos cos 3 cos 53 5

e eC eα α

α= β − β + β − ……∞

3 5

sin sin 3 sin 53 5

e eS eα α

α= β − β + β − ……∞

So that 3 5

(cos sin ) (cos 3 sin 3 ) (cos5 sin 5 )3 5

e eC iS e i i iα α

α+ = β + β − β + β + β + β − ……∞

3( ) 5( )

3 5

i ii e ee

α+ β α+ βα+ β= − + − ……∞

3 5

,3 5z zz= − + − ……∞ where z = eα + iβ

= tan–1z = tan–1e(α + iβ) …(1)

Similarly, C – iS = tan–1e(α – iβ) …(2)

Thus 2C = tan–1e(α + iβ) + tan–1e(α – iβ)

α β − βα+ β α − β

− −α+ β α − β α β α − β

+ += = − − 1 1 ( )tan tan

1 1 ( )( )

i ii i

i i i ie e ee e

e e e e e e

1 12

2 cos 2 costan tan1

ee e e

α− −

α −α α ⋅ β β = = − −

1 costan

2e e

−α −α

β = − −

1 costan

sinh− β = − α


Therefore, 1 1cos1 1tan tan cos cosech2 sinh 2

C − −β = − = − β α α

Alternately: From(1), C + iS = tan–1e(α + iβ) = tan–1eα(cosβ + isinβ) = tan–1(eαcos β + i eαsinβ)

is comparable to 1 12 2 2 2

1 2 2tan ( ) tanh ,2 1 1

a ba ib ia b a b

− − + = + − − + +

where a = eαcosβ and b = eαsinβ.

Example 65: Sum the series 2 31 1sin sin2 sin 32 3

x x xθ − ⋅ θ + ⋅ θ − … … …∞

Solution: Let 2 31 1sin sin 2 sin 32 3

S x x x= θ − θ + ⋅ θ − …… …∞

2 31 1cos cos 2 cos32 3

C x x x= θ − θ + ⋅ θ − ………∞

Therefore,

2 3

(cos sin ) (cos 2 sin 2 ) (cos 3 sin 3 )2 3x xC iS x i i i+ = θ + θ − θ + θ + θ + θ − … …∞

2 3

2 3 ,2 3

i i ix xxe e eθ θ θ= − + − … … …∞ (a logarithmic series)

2 3 4

log(1 ), where2 3 4

iz z zz z z xe θ= − + − + … … ∞ = + =

= log[1 + x(cosθ + isinθ) = logr cisθ

2 2 1 sinlog (1 cos ) ( sin ) tan1 cos

xx x ix

− θ= + θ + θ + + θ

Equating imaginary parts, 1 sintan1 cos

xSx

− θ=+ θ

[except, when xcosθ = –1]

Example 66: Find the sum of the series cos ααααα + cos(ααααα + βββββ) + cos(ααααα + 2 βββββ)+ ………to n terms

Solution: Given C = cosα + cos(α + β) + cos(α + 2β) + … to n terms

S = sin α + sin(α + β) + sin(α + 2β) + … to n termsThus,

C + iS = eiα + ei(α + β) + ei(α + 2β) + … to n terms

= eiα(1 + eiβ + e2iβ + … to n terms)


1(1 ) ,

1

ini

ieee

βα

β −= −

(1 )sum of G.P. with terms1

na rnr

− = −Q

βα − β

β − β− −=

− −(1 )(1 )

(1 )(1 )

ini i

i ie e e

e e

α β − β − β

β − β

− − + =− − +

( 1)1

(1 1)

i in i i n

i i

e e e e

e e

( ) ( ) ( 1)

2 2cos

i i i n i ne e e eC iSα α −β α+ β α+ − β − − ++ = − β

On taking real parts only,

α − α − β + α + − β − α + β= β2

cos cos( ) cos( 1 ) cos( )

2.2sin2

n nC 21 cos 2sin

2θ − θ = Q

( ) α + − β + α + β α + β − α + − β α + α − β α − β − α + = β2

1 ( ) ( ) ( 1 )( ) ( )2sin sin 2sin sin2 2 2 2

4sin2

n n n n

− β + αα − β β β − + = β2

2 1 222sin sin 2sin sin2 2 2 2

4 sin2

n

− β + α α − β − = β

2 1 2 2sin sin2 2

2 sin2

n

(Using CD formula)

2 1 2 2 1 22 22cos sin2 2 2 2

2sin2

n n− β + α − + αα − β α − β + − = β

( 1)cos sin2 2

sin2

n n

C

− β α + β = β


Example 67: Find the sum of the series 2 31 1sin sin2 sin32 3

x x xθ − θ + θ − ………∞

Solution: Let = θ − θ + θ − ………∞2 31 1sin sin 2 sin 32 3

S x x x

= θ − θ + θ − ………∞2 31 1cos cos2 cos32 3

S x x x

∴ 2 31 1(cos sin ) (cos2 sin 2 ) (cos3 sin 3 )2 3

C iS x i x i x i+ = θ + θ − θ + θ + θ + θ − ………∞ …(1)

2 2 3 31 12 3

i i ixe x e x eθ θ θ= − + − …………∞

2 3

2 3z zz= − + − ……∞ (where z = xeiθ)

= log(1 + z) = log(1 + xei θ)

= log[(1 + xcosθ) + ixsin θ] = log(rcisα) = log r + iα …(2)

where 2 2 1 sin(1 cos ) ( sin ) and tan1 cos

xr x xx

− θ = + θ + θ α = + θ

Whence 1 sintan , except 1 and (2 1)1 cos

x x nx

− θ α = = θ = + π + θ

Example 68: Sum the series1 + xcosααααα + x2 cos2ααααα + x3cos3ααααα + ………… + xn–1cos(n – 1) ααααα

where x is less than unity. Also find the sum to infinity. [NIT Kurukshetra, 2007, 2008]

Solution: Let C = 1 + xcosα + x2cos2α + x3cos3α + … + xn–1cos (n – 1)α … (1) S = xsinα + x2sin2α + x3sin3α + … + xn–1 sin (n – 1) α …(2)

∴ C + iS = 1 + x(cosα + isinα) + x2(cos2α + isin2α) + … + xn–1 (cosn – 1α + i sinn – 1α)

= 1 + xeiα + x2e2iα + … + xn–1ei(n–1)α

2 11 ,nz z z −= + + + ……… + where z = xeiα

1(1 ) 1

1 1

n n ni

i

z x ez xe

α

α

− −= =− −

(1 ) 1

1 1

n in i

i i

x e xexe xe

α − α

α − α

− −= ×− −

1 ( 1 )

21

1 ( )

n in i n i n

i i i ix e xe x e

x e e x e e

α − α + − α

α − α α − α− − +=

− + +


∴+− α + α − α − α + − α + − α

+ =− α +

1

2

1 (cos sin ) (cos sin ) [cos( 1) sin( 1) ]1 2 cos

n nx n i n x i x n i nC iS

x x…(3)

On equating real and imaginary parts on both sides

∴1

2

1 cos cos cos( 1)1 2 cos

n nx x n x nC

x x

+− α − α + − α=

− α +…(4)

Since it is given that x is numerically less than one.⇒ xn, xn + 1 → 0 as n → ∞

∴ The sum to infinity, 2

1 (cos sin )1 2 cos

x iC iS

x x− α − α

+ =− α +

, (From (3))

and so that its real part is 2

1 cos1 2 cos

xCx x− α=

− α +

Example 69: Find the sum of the series sin αααααcos ααααα + sin2 ααααα cos2 ααααα + sin3ααααα cos3ααααα + …… + ∞∞∞∞∞

Solution: Let C = sinα cosα + sin2α cos2α + sin3α cos3α + …… + ∞ S = sinα sinα + sin2α sin2α + sin3α sin3α + …… + ∞

∴ C + iS = sinα eiα + sin2α e2iα + sin3α e3iα + …… + ∞ = z + z2 + z3 + … + ∞, where z = sin α eiα

sin .

1 1 sin .

i

iz e

z e

α

αα= =

− − α

sin . 1 sin .

1 sin . 1 sin .

i i

i ie e

e e

α − α

α − αα − α= ×

− α − α

α

α − α α − α

α − α=

− α + + α2

sin ( sin )1 sin .( ) sin

i

i i i i

ee e e e

2

sin (cos sin sin )1 2 sin cos sin

iα α + α − α=− α α + α

α α − α + α

+ =− α + α

2

2

sin (cos sin ) sin1 sin 2 sin

iC iS

∴ 2

sin (cos sin )1 sin 2 sin

Cα α − α=

− α + α

Example 70: Find the sum of the series 3 5

sin sin3 sin53 5c cc α + α + α + ………∞

Solution: Let 3 5

sin sin 3 sin 53 5c cS c= α + α + α + ………∞


3 5cos cos 3 cos 5

3 5c cC c= α + α + α + ………∞

∴ 3 5

(cos sin ) (cos 3 sin 3 ) (cos5 sin 5 )3 5c cC iS c i i i+ = α + α + α + α + α + α + ………∞

α α α= + + + ………∞3 5

3 5

3 5i i ic cce e e

= tanh–1(ceiα) …(1)Likewise,

C – iS = tanh–1(ce–i α) …(2)Subtracting (2) from (1),

1 12 tanh ( ) tanh ( )i iiS ce ce− α − − α= − , − − − − − = − 1 1 1using tanh tanh tanh

1A BA B

AB

α − α

− − −α − α

− α α = = = − − −1 1 1

2 22 sin 2 sintanh tanh tanh

1 1 1

i i

i ice ce ic ci

ce ce c c

12

2 sin2 tan1ciS i

c− α = −

or 12

1 2 sintan2 1

cSc

− α = − …(3)

Note: For finding C, take sum (1) and (2) and 1 1 1use tan tanh tanh1A BA B

AB− − − + + = +

Alternately: 1 1 1tanh ( ) log2 1

ii

iceC iS cece

α− α

α++ = =−

and proceed further for simplification and

separation of real and imaginary parts.

Example 71: Sum the series 2sincos sin cos2 cos3

1.2θθ + θ θ + θ + ………∞

Solution: Let 2sincos sin cos2 cos 3

1.2C θ= θ + θ θ + θ + ………∞

2sinsin sin sin 2 sin 3

1.2S θ= θ + θ θ + θ + ………∞

∴ θ θ θθ+ = + θ + + ………∞2

2 3sinsin1.2

i i iC iS e e e

2 2sin sin

11 1.2

i ii e e

eθ θ

θ θ θ= + + + ………∞

θ = + + + + ………∞

2 31

1 2! 3!i z z ze


= e iθ(e z) [where z = sinθ eiθ = sinθ (cosθ + isinθ)]

= eiθ[e sin θcosθ + i sin 2θ]

= (e i θe i sin 2θ) × e sinθ cos θ

2sin cos ( sin )ie eθ θ θ+ θ =

= e sinθ cos θ[cos(θ + sin2θ) + isin(θ + sin 2θ)]

∴ sin cos 2 sin cos 2cos( sin ) and sin( sin )C e S eθ θ θ θ= θ + θ = θ + θ

Example 72: Find the sum of the series

sin( 2 ) sin( 4 )sin

2! 4!α + β α + βα − + − ………∞

and cos( 2 ) cos( 4 )cos

2! 4!α + β α + βα − + − ………∞

Solution: Let sin( 2 ) sin( 4 )sin2! 4!

S α + β α + β= α − + − ………∞

cos( 2 ) cos( 4 )cos2! 4!

C α + β α + β= α − + − ………∞

∴ ( 2 ) ( 4 )

2! 4!

i ii e eC iS e

α+ β α+ βα+ = − + − ………∞

2 4

12! 4!

i ii e ee

β βα = − + − ………∞

2 4

1 ,2! 4!

i z ze α = − + − ………∞ where z = eiβ

[ ]cos cos( )i i ie z e eα α β = =

= eiα[cos(cosβ + isinβ)]

= ei α[cos (cosβ)cos(isin β) – sin(cosβ)sin(i sinβ)]

= eiα[cos (cosβ)cosh(sinβ) – i sin(cosβ)sinh (sin β)] C + iS = (cosα + isinα)[cosAcoshB – isinAsinhB]

Example 73: Find the sum to of the infinite series

1 1.3 1.3.51 cos cos2 cos3 ( )2 2.4 2.4.6

− θ + θ − θ + ……… −π < θ < π

Solution: Let 1 1.3 1.3.51 cos cos2 cos32 2.4 2.4.6

C = − θ + θ − θ + ………∞


and 1 1.3 1.3.50 sin sin 2 sin 32 2.4 2.4.6

S = − θ + θ − θ + ………∞

∴ 2 31 1.3 1.3.512 2.4 2.4.6

i i iC iS e e eθ θ θ+ = − + − − ………

2 3

1 1 1 1 11 1 21 2 2 2 2 212 1.2 1.2.3

i i ie e eθ θ θ

− − − − − − − − = + − + + + ………

1/2 1/2(1 ) (1 cos sin )ie iθ − −= + = + θ + θ

1222cos .2sin cos

2 2 2i

−θ θ θ = +

1/2 1/2

2 cos cos sin2 2 2

i− −θ θ θ = +

1/2

2cos cos sin2 4 4

i−θ θ θ = −

Equating the real parts, we have 1/2

2 cos cos .2 4

C−θ θ =

ASSIGNMENT 8

Sum the following series:

1.1 1cos cos2 cos32 3

θ − θ + θ − ………∞ [SVTU, 2006]

2.2 3

cos cos 2 cos 32 3x xx θ − θ + θ − ………∞

3.( 1) ( 1)( 2)sin sin 2 sin 31.2 1.2.3

n n n n nn + + +α + α + α + ………∞ [NIT Kurukshetra, 2009]

4. 2 2 3 41 1 1sin sin 2 sin sin 3 sin sin 4 sin2 3 4

θ − θ θ + θ θ − θ θ + ………∞ [SVTU, 2005; 2006]

5.2 3

cos cos( ) cos( 2 ) cos( 3 )2! 3!x xxα + α + β + α + β + α + β + ………∞

α β + = = Hint : , where i z iC iS e e z xe

6.2 3

sin sin( ) sin( 2 ) sin( 3 )2! 3!x xxα + α + β + α + β + α + β + ………∞ [PTU, 2008]

α β + = = Hint : ,i z iC iS e e z xe


7. cosα sin α + cos2α sin 2α + ……… ∞[Hint: G.P. with common ratio: cos α ei α]

8.1 1.3 1.3.51 cos2 cos 4 cos62 2.4 2.4.6

+ α − α + α − ………∞

9.1 1 21 cos cos2 cos33 9 27

x x x+ + + + ………∞

10. sin sin( 2 ) sin( 1)nα + α + β + ……… + α + − β [PTU, 2009]

[Hint: G.P. with common ratio: eiβ]

ANSWERS

Assignment 1

1. (i) − − + 1 (2 3) (2 3)2

i (ii)825

i−

2. 2sin cos sin2 2 2

iα π − α π − α + 3. 2 2 2 1

yx y x

−+ − +

4. x = ±1.5 y = ±2 5. A circle, centre (– 1, 1), radius 2 units

6. (i) 2i (ii)1 12 2

i−

Assignment 2

1. Other vertices are 2i, 3 , 3 , 2 , 3i i i i− + − − − −

2. (i) P(z) represents a circle with centre A(a) and radius ‘k’.(ii) P(z) is a straight line through A(a) making an angle α with OX.

3. (i) Annular region between the circles of radii 2 and 4 with centre (–3, 0) includingboundary of inner circle.(ii) Region of complex plane about the line y = 2.

(iii) Infinite region bounded by the line and3 2π πθ = θ = .

(iv) Region between the lines x = ±2 above the real axis.4. Ellipse with foci at z = ±1 and major axis = 3.6. (i) Right bisector of the line joining z = 3 and z = –1.

(ii) Circle through the points z = 3 and z = –1.7. Locus of P(z) is a circle (unless k = 1, when locus is the right bisector of z – a and z – b)


Assignment 3

3.π

+4( 1)

mn n

Assignment 4

2. (i) ±182 [0.98 (0.195],i

122 [ 0.195 (0.98)]i− ±

(ii) 15 4 32 cos ,

10n +

where n = 0, 1, 2, 3, 4

4. cos sin ,5 5

iπ π ± 3 3cos sin5 5

iπ π ±

5.2 2cos sin , 1,2, ,67 7n ni nπ π + = …

6.( 1 ) (1 ),

2 2i i− + −

√ √ 7. x3 – x2 – 2x + 1 = 0

Assignment 5

1. 32cos5θ – 24cos3θ + 6cosθ10. – (2)11(sin12θ – 2sin10θ – 4sin8θ + 10sin6θ + 5sin4θ – 20sin2θ)11. sin5θ = Asinθ – Bsin3θ + Csin5θ

Assignment 6

1. (i) 1 4log 10 tan 23e i n− + ± π (ii) log1 + i (2n + 1)π (iii) ( ) π π− + −2 1

2 2n

e e

3. (i) π π+ + π −

24 12 , 2 log 2

4

ne n (ii)

2

8 , log 24 ee

π− π

5.2 2

2 2a ba b

−+ 9. (i) ( 1) 4

2nZ n iπ = π + − − (ii) 2

2Z i n π = π +

Assignment 7

1.1

2 21 2tan2 1

xx y

−α =− − 9. 1cos sinx −= θ

12 2

21 tanh2 1

yx y

−β =+ +

= θ + + θ 2log sin 1 siny


Assignment 8

1. log 2cos2θ

2. 21 log(1 2 cos )2

x x+ θ + 3.π − α α

( )sin 2 sin

2 2/nn

4.2

1 sintan1 cos sin

S − θ=+ θ θ 5. C = excosβ[cos(α + xsinβ)]

6. S = e xcosβ[sin(α + xsinβ)] 7. S = cot α 8. = α + α[cos (1 cos )]C

9. 9 3cos10 6cos

xx

− −

10. { }sin ( 1) sin csc2 2 2

nS nβ β β= α + −

COMPLEX NUMBERS AND FUNCTIONS - NIT Kurukshetra

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