Chapt 18 engineering mathatics solution

1818 Integration in the Complex Plane

EXERCISES 18.1Contour Integrals

1.∫

C

(z + 3) dz = (2 + 4i)[∫ 3

1

(2t + 3) dt + i

∫ 3

1

(4t − 1) dt

]= (2 + 4i)[14 + 14i] = −28 + 84i

2.∫

C

(2z̄ − z) dz =∫ 2

0

[−t − 3(t2 + 2)i](−1 + 2ti) dt =∫ 2

0

(6t3 + 13t) dt + i

∫ 2

0

(t2 + 2) dt = 50 +203

i

3.∫

C

z2 dz = (3 + 2i)3∫ 2

−2

t2 dt =163

(3 + 2i)3 = −48 +7363

i

4.∫

C

(3z2 − 2z) dz =∫ 1

0

(−15t4 + 4t3 + 3t2 − 2t) dt + i

∫ 1

0

(−6t5 + 12t3 − 6t2) dt = −2 + 0i = −2

5. Using z = eit, −π/2 ≤ t ≤ π/2, and dz = ieit dt,∫

C

1 + z

zdz = −

∫ π/2

−π/2

(1 + eit) dt = (2 + π)i.

6.∫

C

|z|2 dz =∫ 2

1

(2t5 +

2t

)dt − i

∫ 2

1

(t2 +

1t4

)dt = 21 + ln 4 − 21

8i

7. Using z = eit = cos t + i sin t, dz = (− sin t + i cos t) dt and x = cos t,∮̌C

Re(z) dz =∫ 2π

0

cos t(− sin t + i cos t) dt = −∫ 2π

0

sin t cos t dt + i

∫ 2π

0

cos2 t dt

= −12

∫ 2π

0

sin 2t dt +12

i

∫ 2π

0

(1 + cos 2t) dt = πi.

8. Using z + i = eit, 0 ≤ t ≤ 2π, and dz = ieit dt,∮̌C

[1

(z + i)3− 5

z + i+ 8

]dz = i

∫ 2π

0

[e−2it − 5 + 8eit] dt = −10πi.

9. Using y = −x + 1, 0 ≤ x ≤ 1, z = x + (−x + 1)i, dz = (1 − i) dx,∫C

(x2 + iy3) dz = (1 − i)∫ 0

1

[x2 + (1 − x)3i] dx = − 712

+112

i.

10. Using z = eit, π ≤ t ≤ 2π, dz = ieit dt, x = cos t = (eit + e−it)/2, y = sin t = (eit − e−it)/2i,∫C

(x3 − iy3) dz =18

i

∫ 2π

π

(e3it + 3eit + 3e−it + e−3it)eit dt +18

i

∫ 2π

π

(e3it − 3eit + 3e−it − e−3it)eit dt

=18

i

∫ 2π

π

(2e4it + 6) dt =3π

4i.

877

18.1 Contour Integrals

11.∫

C

ez dz =∫

C1

ez dz +∫

C2

ez dz where C1 and C2 are the line segments y = 0, 0 ≤ x ≤ 2 and y = −πx + 2π,

1 ≤ x ≤ 2, respectively. Now ∫C1

ez dz =∫ 2

0

ex dx = e2 − 1

∫C2

ez dz = (1 − πi)∫ 1

2

ex+(−πx+2π)idx = (1 − πi)∫ 1

2

e(1−πi)xdx = e1−πi − e2(1−πi) = −e − e2.

In the second integral we have used the fact that ez has period 2πi. Thus∫C

ez dz = (e2 − 1) + (−e − e2) = −1 − e.

12.∫

C

sin z dz =∫

C1

sin z dz +∫

C2

sin z dz where C1 and C2 are the line segments y = 0, 0 ≤ x ≤ 1, and x = 1,

0 ≤ y ≤ 1, respectively. Now ∫C1

sin z dz =∫ 1

0

sinx dx = 1 − cos 1

∫C2

sin z dz = i

∫ 1

0

sin(1 + iy) dy = cos 1 − cos(1 + i).

Thus∫C

sin z dz = (1−cos 1)+(cos 1−cos(1+ i)) = 1−cos(1+ i) = (1−cos 1 cosh 1)+ i sin 1 sinh 1 = 0.1663+0.9889i.

13. We have∫

C

Im(z − i) dz =∫

C1

(y − 1) dz +∫

C2

(y − 1) dz

On C1, z = eit, 0 ≤ t ≤ π/2, dz = ieit dt, y = sin t = (eit − e−it)/2i,∫C1

= (y − 1) dz =12

∫ π/2

0

[eit − e−it − 2i]eit dt =12

∫ π/2

0

[e2it − 1 + 2ieit] dt = 1 − π

4− 1

2i.

On C2, y = x + 1, −1 ≤ x ≤ 0, z = x + (x + 1)i, dz = (1 + i) dx,∫C2

(y − 1) dz = (1 + i)∫ −1

0

x dx =12

+12

i.

Thus∫

C

Im(z − i) dz =(

1 − π

4− 1

2i

)+

(12

+12

i

)=

32− π

4.

14. Using x = 6 cos t, y = 2 sin t, π/2 ≤ t ≤ 3π/2, z = 6 cos t + 2i sin t, dz = (−6 sin t + 2i cos t) dt,∫C

dz = −6∫ 3π/2

π/2

sin t dt + 2i

∫ 3π/2

π/2

cos t dt = 2i(−2) = −4i.

15. We have∮̌

C

zez dz =∫

C1

zez dz +∫

C2

zez dz +∫

C3

zez dz +∫

C4

zez dz

On C1, y = 0, 0 ≤ x ≤ 1, z = x, dz = dx,∫C1

zez dz =∫ 1

0

xex dx = xex − ex∣∣∣10= 1.

878


On C2, x = 1, 0 ≤ y ≤ 1, z = 1 + iy, dz = i dy,∫C2

zez dz = i

∫ 1

0

(1 + iy)e1+iy dy = iei+1.

On C3, y = 1, 0 ≤ x ≤ 1, z = x + i, dz = dx,∫C3

zez dz =∫ 0

1

(x + i)ex+idx = (i − 1)ei − ie1+i.

On C4, x = 0, 0 ≤ y ≤ 1, z = iy, dz = i dy,∫C4

zez dz = −∫ 0

1

yeiy dy = (1 − i)ei − 1.

Thus∮̌

C

zez dz = 1 + iei+1 + (i − 1)ei − ie1+i + (1 − i)ei − 1 = 0.

16. We have∫

C

f(z) dz =∫

C1

f(z) dz +∫

C2

f(z) dz

On C1, y = x2, −1 ≤ x ≤ 0, z = x + ix2, dz = (1 + 2xi) dx,∫C1

f(z) dz =∫ 0

−1

2(1 + 2xi) dx = 2 − 2i.

On C2, y = x2, 0 ≤ x ≤ 1, z = x + ix2, dz = (1 + 2xi) dx,∫C2

f(z) dz =∫ 1

0

6x(1 + 2xi) dx = 3 + 4i.

Thus∫

C

f(z) dz = 2 − 2i + 3 + 4i = 5 + 2i.

17. We have∮̌

C

x dz =∫

C1

x dz +∫

C2

x dz +∫

C3

x dz

On C1, y = 0, 0 ≤ x ≤ 1, z = x, dz = dx, ∫C1

x dz =∫ 1

0

x dx =12

.

On C2, x = 1, 0 ≤ y ≤ 1, z = 1 + iy, dz = i dy,∫C2

x dz = i

∫ 1

0

dy = i.

On C3, y = x, 0 ≤ x ≤ 1, z = x + ix, dz = (1 + i) dx,∫C3

x dz = (1 + i)∫ 0

1

x dx = −12− 1

2i.

Thus∮̌

C

x dz =12

+ i − 12− 1

2i =

12

i.

18. We have∮̌

C

(2z − 1) dz =∫

C1

(2z − 1) dz +∫

C2

(2z − 1) dz +∫

C3

(2z − 1) dz

On C1, y = 0, 0 ≤ x ≤ 1, z = x, dz = dx,∫C1

(2z − 1) dz =∫ 1

0

(2x − 1) dx = 0.

879


On C2, x = 1, 0 ≤ y ≤ 1, z = 1 + iy, dz = i dy,∫C2

(2z − 1) dz = −2∫ 1

0

y dy + i

∫ 1

0

dy = −1 + i.

On C3, y = x, z = x + ix, dz = (1 + i) dx,∫C3

(2z − 1) dz = (1 + i)∫ 0

1

(2x − 1 + 2ix) dx = 1 − i.

Thus∮̌

C

(2z − 1) dz = 0 − 1 + i + 1 − i = 0.

19. We have∮̌

C

z2 dz =∫

C1

z2 dz +∫

C2

z2 dz +∫

C3

z2 dz

On C1 y = 0, 0 ≤ x ≤ 1, z = x, dz = dx, ∫C1

z2 dz =∫ 1

0

x2 dx =13

.

On C2, x = 1, 0 ≤ y ≤ 1, z = 1 + iy, dz = i dy,∫C2

z2 dz =∫ 1

0

(1 + iy)2i dy = −1 +23

i.


z2 dz = (1 + i)3∫ 0

1

x2 dx =23− 2

3i.

Thus∮̌

C

z2 dz =13− 1 +

23

i +23− 2

3i = 0.

20. We have∮̌

C

z̄2 dz =∫

C1

z̄2 dz +∫

C2

z̄2 dz +∫

C3

z̄2 dz

On C1, y = 0, 0 ≤ x ≤ 1, z = x, dz = dx, ∫z̄2 dz =

∫ 1

0

x2 dx =13

.

On C2, x = 1, 0 ≤ y ≤ 1, z = 1 + iy, dz = i dy,∫C2

z̄2 dz = −∫ 1

0

(1 − iy)2(−i dy) = 1 +23

i.


z̄2 dz = (1 − i)2(1 + i)∫ 0

1

x2 dx = −23

+23

i.

Thus∮̌

C

z̄2 dz =13

+ 1 +23

i − 23

+23

i =23

+43

i.

21. On C, y = −x + 1, 0 ≤ x ≤ 1, z = x + (−x + 1)i, dz = (1 − i) dx,∫C

(z2 − z + 2) dz = (1 − i)∫ 1

0

[x2 − (1 − x)2 − x + 2 + (3x − 2x2 − 1)i] dx =43− 5

3i.

22. We have∫

C

(z2 − z + 2) dz =∫

C1

(z2 − z + 2) dz +∫

C2

(z2 − z + 2) dz

880


On C1, y = 1, 0 ≤ x ≤ 1, z = x + i, dz = dx,∫C1

(z2 − z + 2) dz =∫ 1

0

[(x + i)2 − x + 2 − i] dx =56

.

On C2, x = 1, 0 ≤ y ≤ 1, z = 1 + iy, dz = i dy,∫C2

(z2 − z + 2) dz = i

∫ 0

1

[(1 + iy)2 + 1 − iy] dy =12− 5

3i.

Thus∫

C

(z2 − z + 2) dz =12− 5

3i +

56

=43− 5

3i.

23. On C, y = 1 − x2, 0 ≤ x ≤ 1, z = x + i(1 − x2), dz = (1 − 2xi) dx,∫C

(z2 − z + 2) dz =∫ 1

0

(−5x4 + 2x3 + 7x2 − 3x + 1) dx + i

∫ 1

0

(2x5 − 8x3 + 3x2 − 1) dx =43− 5

3i.

24. On C, x = sin t, y = cos t, 0 ≤ t ≤ π/2 or z = ie−it, dz = e−it dt,∫C

(z2 − z + 2) dz =∫ π/2

0

(−e−2it − ie−it + 2)e−it dt =∫ π/2

0

(−e−3it − ie−2it + 2e−it) dt

= −13

ie−3πi/2 +12e−πi + 2ie−πi/2 +

13

i − 12− 2i =

43− 5

3i.

25. On C,∣∣∣∣ ez

z2 + 1

∣∣∣∣ ≤ |ez||z|2 − 1

=e5

24. Thus

∣∣∣∣∮̌C

ez

z2 + 1dz

∣∣∣∣ ≤ e5

24· 10π =

5π

12e5.

26. On C,∣∣∣∣ 1z2 − 2i

∣∣∣∣ ≤ 1|z|2 − |2i| =

134

. Thus∣∣∣∣∫

C

1z2 − 2i

dz

∣∣∣∣ ≤ 134

· 12(12π) =

3π

17.

27. The length of the line segment from z = 0 to z = 1 + i is√

2 . In addition, on this line segment

|z2 + 4| ≤ |z|2 + 4 ≤ |1 + i|2 + 4 = 6.

Thus∣∣∣∣∫

C

(z2 + 4) dz

∣∣∣∣ ≤ 6√

2 .

28. On C,∣∣∣∣ 1z3

∣∣∣∣ =1

|z|3 =164

. Thus∣∣∣∣∫

C

1z3

dz

∣∣∣∣ ≤ 164

· 14(8π) =

π

32.

29. (a)∫

C

dz = lim‖P‖→0

n∑k=1

∆zk = lim‖P‖→0

n∑k=1

(zk − zk−1)

= lim‖P‖→0

[(z1 − z0) + (z2 − z1) + (z3 − z2) + · · · + (zn−1 − zn−2) + (zn − zn−1)]

= lim‖P‖→0

(zn − z0) = zn − z0

(b) With zn = −2i and z0 = 2i,∫

C

dz = −2i − (2i) = −4i.

30. With z∗k = zk, ∫C

z dz = lim‖P‖→0

n∑k=1

zk(zk − zk−1)

= lim‖P‖→0

[(z21 − z1z0) + (z2

2 − z2z1) + · · · + (z2n − znzn−1)]. (1)

881


With z∗k = zk−1, ∫C


n∑k=1

zk−1(zk − zk−1)

= lim‖P‖→0

[(z0z1 − z20) + (z1z2 − z2

1) + · · · + (zn−1zn − z2n−1)]. (2)

Adding (1) and (2) gives

2∫

C


(z2n − z2

0) or∫

C

z dz =12(z2

n − z20).

31. (a)∫

C

(6z + 4) dz = 6∫

C

z dz + 4∫

C

dz =62[(2 + 3i)2 − (1 + i)2] + 4[(2 + 3i) − (1 + i)] = −11 + 38i

(b) Since the contour is closed, z0 = zn and so

6∫

C

z dz + 4∫

C

dz = 6[z20 − z2

0 ] + 4[z0 − z0] = 0.

32. For f(z) = 1/z, f(z) = 1/z̄, so on z = 2eit, z̄ = 2e−it, dz = 2ieit dt, and∮̌C

f(z) dz =∫ 2π

0

12e−it

· 2ieit dt =12e2it

∣∣∣∣2π

0

=12[e4πi − 1] = 0.

Thus circulation = Re(∮̌

C

f(z) dz

)= 0, and net flux = Im

(∮̌C

f(z) dz

)= 0.

33. For f(z) = 2z, f(z) = 2z̄, so on z = eit, z̄ = e−it, dz = ieit dt, and∮̌C

f(z) dz =∫ 2π

0

(e−it)(ieit dt) = 2i

∫ 2π

0

dt = 4πi.


C

f(z) dz


(∮̌C

f(z) dz

)= 4π.

34. For f(z) = 1/(z − 1), f(z) = 1/(z − 1), so on z − 1 = 2eit, dz = 2ieitdt, and∮̌C

f(z) dz =∫ 2π

0

12eit

· 2ieit dt = i

∫ 2π

0

dt = 2πi.


C

f(z) dz


(∮̌C

f(z) dz

)= 2π.

35. For f(z) = z̄, f(z) = z so on the square we have∮̌C

f(z) dz =∫

C1

z dz +∫

C2

z dz +∫

C3

z dz +∫

C4

z dz

where C1 is y = 0, 0 ≤ x ≤ 1, C2 is x = 1, 0 ≤ y ≤ 1, C3 is y = 1, 0 ≤ x ≤ 1, and C4 is x = 0, 0 ≤ y ≤ 1. Thus∫C1

z dz =∫ 1

0

x dx =12∫

C2

z dz = i

∫ 1

0

(1 + iy) dy = −12

+ i

∫C3

z dz =∫ 0

1

(x + i) dx = −12− i

∫C4

z dz = −∫ 0

1

y dy =12

882

18.2 Cauchy-Goursat Theorem

and so ∮̌C

f(z) dz =12

+(−1

2+ i

)+

(−1

2− i

)+

12

= 0

circulation = Re(∮̌

C

f(z) dz

)= Re(0) = 0

net flux = Im(∮̌

C

f(z) dz

)= Im(0) = 0.

EXERCISES 18.2Cauchy-Goursat Theorem

1. f(z) = z3 − 1 + 3i is a polynomial and so is an entire function.

2. z2 is entire and1

z − 4is analytic within and on the circle |z| = 1.

3. f(z) =z

2z + 3is discontinuous at z = −3/2 but is analytic within and on the circle |z| = 1.

4. f(z) =z − 3

z2 + 2z + 2is discontinuous at z = −1 + i and at z = −1 − i but is analytic within and on the circle

|z| = 1.

5. f(z) =sin z

(z2 − 25)(z2 + 9)is discontinuous at z = ±5 and at z = ±3i but is analytic within and on the circle

|z| = 1.

6. f(z) =ez

2z2 + 11z + 15is discontinuous at z = −5/2 and at z = −3 but is analytic within and on the circle

|z| = 1.

7. f(z) = tan z is discontinuous at z = ±π

2, ±3π

2, . . . but is analytic within and on the circle |z| = 1.

8. f(z) =z2 − 9cosh z

is discontinuous atπ

2i, ±3π

2i, . . . but is analytic within and on the circle |z| = 1.

9. By the principle of deformation of contours we can choose the more convenient circular contour C1 defined by|z| = 1. Thus ∮̌

C

1z

dz =∮̌

C1

1z

dz = 2πi

by (4) of Section 18.2.

10. By the principle of deformation of contours we can choose the more convenient circular contour C1 defined by

|z − (−1 − i)| =116

. Thus ∮̌C

5z + 1 + i

dz = 5∮̌

C1

1z − (−1 − i)

dz = 5(2πi) = 10πi

by (4) of Section 18.2.

883


11. By Theorem 18.4 and (4) of Section 18.2,∮̌C

(z +

1z

)dz =

∮̌C

z dz +∮̌

C

1z

dz = 0 + 2πi = 2πi.

12. By Theorem 18.4 and (4) of Section 18.2,∮̌C

(z +

1z2

)dz =

∮̌C

1z

dz +∮̌

C

1z2

dz = 0 + 0 = 0.

13. Since f(z) =z

z2 − π2is analytic within and on C it follows from Theorem 18.4 that

∮̌C

z

z2 − π2dz = 0.

14. By (4) of Section 18.2,∮̌

C

10(z + i)4

dz = 0.

15. By partial fractions,∮̌

C

2z + 1z(z + 1)

dz =∮̌

C

1z

dz +∮̌

C

1z + 1

dz.

(a) By Theorem 18.4 and (4) of Section 18.2,∮̌C

1z

dz +∮̌

C

1z + 1

dz = 2πi + 0 = 2πi.

(b) By writing∮̌

C

=∮̌

C1

+∮̌

C2

where C1 and C2 are the circles |z| = 1/2 and |z + 1| = 1/2, respectively,

we have by Theorem 18.4 and (4) of Section 18.2,∮̌C

1z

dz +∮̌

C

1z + 1

dz =∮̌

C1

1z

dz +∮̌

C1

1z + 1

dz +∮̌

C2

1z

dz +∮̌

C2

1z + 1

dz

= 2πi + 0 + 0 + 2πi = 4πi.

(c) Since f(z) =2z + 1

z(z + 1)is analytic within and on C it follows from Theorem 18.4 that∮̌

C

2z + 1z2 + z

dz = 0.


C

2z

z2 + 3dz =

∮̌C

1z +

√3 i

dz +∮̌

C

1z −

√3 i

dz.

(a) By Theorem 18.4, ∮̌C

1z +

√3 i

dz +∮̌

C

1z −

√3 i

dz = 0 + 0 = 0.

(b) By Theorem 18.4 and (4) of Section 18.2,∮̌C

1z +

√3 i

dz +∮̌

C

1z −

√3 i

dz = 0 + 2πi = 2πi.

(c) By writing∮̌

C

=∮̌

C1

+∮̌

C2

where C1 and C2 are the circles |z +√

3 i| = 1/2 and |z −√

3 i| = 1/2,

respectively, we have by Theorem 18.4 and (4) of Section 18.2,∮̌C

1z +

√3 i

dz +∮̌

C

1z −

√3 i

dz =∮̌

C1

1z +

√3 i

dz +∮̌

C1

1z −

√3 i

dz +∮̌

C2

1z +

√3 i

dz +∮̌

C2

1z −

√3 i

dz

= 2πi + 0 + 0 + 2πi = 4πi.

884



C

−3z + 2z2 − 8z + 12

dz =∮̌

C

1z − 2

dz − 4∮̌

C

1z − 6

dz.

(a) By Theorem 18.4 and (4) of Section 18.2,∮̌C

1z − 2

dz − 4∮̌

C

1z − 6

dz = 0 − 4(2πi) = −8πi.

(b) By writing∮̌

C

=∮̌

C1

+∮̌

C2

where C1 and C2 are the circles |z − 2| = 1 and |z − 6| = 1, respectively,

we have by Theorem 18.4 and (4) of Section 18.2,∮̌C

1z − 2

dz − 4∮̌

C

1z − 6

dz =∮̌

C1

1z − 2

dz − 4∮̌

C1

1z − 6

dz +∮̌

C2

1z − 2

dz − 4∮̌

C2

1z − 6

dz

= 2πi − 4(0) + 0 − 4(2πi) = −6πi.

18. (a) By writing∮̌

C

=∮̌

C1

+∮̌

C2

where C1 and C2 are the circles |z + 2| = 1 and |z − 2i| = 1, respectively, we

have by Theorem 18.4 and (4) of Section 18.2,∮̌C

(3

z + 2− 1

z − 2i

)dz =

∮̌C1

3z + 2

dz −∮̌

C1

1z − 2i

dz +∮̌

C2

3z + 2

dz −∮̌

C2

1z − 2i

dz

= 3(2πi) − 0 + 0 − 2πi = 4πi.

19. By partial fractions,∮̌C

z − 1z(z − i)(z − 3i)

dz =13

∮̌C

1z

dz +(−1

2+

12

i

) ∮̌C

1z − i

dz +(

16− 1

2i

) ∮̌C

1z − 3i

dz.

By Theorem 18.4 and (4) of Section 18.2,∮̌C

z − 1z(z − i)(z − 3i)

dz = 0 +(−1

2+

12

i

)2πi + 0 = π(−1 − i).

20. By partial fractions, ∮̌C

1z3 + 2iz2

dz =14

∮̌C

1z

dz − 12

i

∮̌C

1z2

dz − 14

∮̌C

1z + 2i

dz.

By Theorem 18.4 and (4) of Section 18.2,∮̌C

1z3 + 2iz2

dz =14

2πi − 12

i(0) − 14

(0) =π

2i.

21. We have∮̌

C

8z − 3z2 − z

dz =∮̌

C1

8z − 3z2 − z

dz −∮̌

C2

8z − 3z2 − z

dz

where C1 and C2 are the closed portions of the curve C enclosing z = 0 and z = 1, respectively. By partialfractions, Theorem 18.4, and (4) of Section 18.2,∮̌

C1

8z − 3z2 − z

dz = 5∮̌

C1

1z − 1

dz + 3∮̌

C1

1z

dz = 5(0) + 3(2πi) = 6πi∮̌C1

8z − 3z2 − z

dz = 5∮̌

C2

1z − 1

dz + 3∮̌

C2

1z

dz = 5(2πi) + 3(0) = 10πi.

Thus∮̌

C

8z − 3z2 − z

dz = 6πi − 10πi = −4πi.

885


22. By choosing the more convenient contour C1 defined by |z − z0| = r where r is small enough so that the circleC1 lies entirely within C we can write∮̌

C

1(z − z0)n

dz =∮̌

C1

1(z − z0)n

dz.

Let z − z0 = reit, 0 ≤ t ≤ 2π and dz = ireit dt. Then for n = 1:∮̌C1

1z − z0

dz =∫ 2π

0

1reit

ireit dt = i

∫ 2π

0

dt = 2πi.

For n �= 1:∮̌C1

1(z − z0)n

dz =i

rn−1

∫ 2π

0

e(1−n)it dt =i

rn−1

e(1−n)it

i(1 − n)

∣∣∣∣2π

0

=1

rn−1(1 − n)[e2π(1−n)i − 1] = 0

since e2π(1−n)i = 1.

23. Write∮̌

C

(ez

z + 3− 3z̄

)dz =

∮̌C

ez

z + 3dz − 3

∮̌C

z̄ dz.

By Theorem 18.4,∮̌

C

ez

z + 3dz = 0. However, since z̄ is not analytic,

∮̌C

z̄ dz =∫ 2π

0

e−it(ieit dt) = 2πi.

Thus∮̌

C

(ez

z + 3− 3z̄

)dz = 0 − 3(2πi) = −6πi.

24. Write∮̌

C

(z2 + z + Re(z)) dz =∮̌

C

(z2 + z) dz +∮̌

C

Re(z) dz.

By Theorem 18.4,∮̌

C

(z2 + z) dz = 0. However, since Re(z) = x is not analytic,∮̌C

x dz =∮̌

C1

x dz +∮̌

C2

x dz +∮̌

C3

x dz

where C1 is y = 0, 0 ≤ x ≤ 1, C2 is x = 1, 0 ≤ y ≤ 2, and C3 is y = 2x, 0 ≤ x ≤ 1. Thus,∮̌C

x dz =∫ 1

0

x dx + i

∫ 2

0

dy + (1 + 2i)∫ 0

1

x dx =12

+ 2i − 12(1 + 2i) = i.

EXERCISES 18.3Independence of Path

1. (a) Choosing x = 0, −1 ≤ y ≤ 1 we have z = iy, dz = i dy. Thus∫C

(4z − 1) dz = i

∫ 1

−1

(4iy − 1) dy = −2i.

(b)∫

C

(4z − 1) dz =∫ i

−i

(4z − 1) dz = 2z2 − z∣∣∣i−i

= −2i

886

18.3 Independence of Path

2. (a) Choosing the line y = 13x, 0 ≤ x ≤ 3 we have z = x + 1

3xi, dz = (1 + 13 i) dx. Thus∫

C

ez dz =∫ 3

0

e(1+ 13 i)x

(1 +

13

i

)dx = e(1+ 1

3 i)x∣∣∣30= e3+i − e0 = (e3 cos 1 − 1) + ie3 sin 1.

(b)∫

C

ez dz =∫ 3+i

0

ez dz = ez∣∣∣3+i

0= e3+i − e0 = (e3 cos 1 − 1) + ie3 sin 1

3. The given integral is independent of the path. Thus∫C

2z dz =∫ 2−i

−2+7i

2z dz = z2∣∣∣2−i

−2+7i= 48 + 24i.

4. The given integral is independent of the path. Thus∫C

6z2 dz =∫ 2−i

2

6z2 dz = z3∣∣∣2−i

2= −15 − 24i.

5.∫ 3+i

0

z2 dz =13z3

∣∣∣∣3+i

0

= 6 +263

i

6.∫ 1

−2i

(3z2 − 4z + 5i) dz = z3 − 2z2 + 5iz∣∣∣1−2i

= −19 − 3i

7.∫ 1+i

1−i

z3 dz =14z4

∣∣∣∣1+i

1−i

= 0

8.∫ 2i

−3i

(z3 − z) dz =14z4 − 1

2z2

∣∣∣∣2i

−3i

=1234

9.∫ 1−i

−i/2

(2z + 1)2 dz =16(2z + 1)3

∣∣∣∣1−i

−i/2

= −76− 22

3i

10.∫ i

1

(iz + 1)3 dz =14i

(iz + 1)4∣∣∣∣i1

= −i

11.∫ i

i/2

eπz dz =1π

eπz∣∣∣ii/2

= − 1π− 1

πi

12.∫ 1+2i

1−i

zez2dz =

12ez2

∣∣∣∣1+2i

1−i

=12[e−3+4i − e−2i] =

12(e−3 cos 4 − cos 2) +

12(e−3 sin 4 + sin 2)i = 0.1918 + 0.4358i

13.∫ π+2i

π

sinz

2dz = −2 cos

z

2

∣∣∣∣π+2i

π

= −2[cos

(π

2+ i

)− cos

π

2

]= 2i sin

π

2sinh 1 = 2.3504i

14.∫ πi

1−2i

cos z dz = sin z∣∣∣πi

1−2i= sinπi − sin(1 − 2i) = i sinhπ − [sinh 1 cosh 2 − i cos 1 sinh 2]

= − sin 1 cosh 2 + i(sinhπ + cos 1 sinh 2) = −3.1658 + 13.5083i

15.∫ 2πi

πi

cosh z dz = sinh z∣∣∣2πi

πi= sinh 2πi − sinhπi = i sin 2π − i sinπ = 0

16.∫ 1+ π

2 i

i

sinh 3z dz =13

cosh 3z

∣∣∣∣1+ π2 i

i

=13

[cosh

(3 +

3π

2i

)− cosh 3i

]=

13

[cosh 3 cos

3π

2+ i sinh 3 sin

3π

2− cos 3

]= −1

3cos 3 − 1

3i sinh 3 = 0.3300 − 3.3393i

887

18.3 Independence of Path

17.∫ 4i

−4i

1z

dz = Lnz∣∣∣4i

−4i= Ln4i − Ln(−4i) = loge 4 +

π

2i −

(loge 4 − π

2i)

= πi

18.∫ 4+4i

1+i

1z

dz = Lnz∣∣∣4+4i

1+i= Ln(4 + 4i) − Ln(1 + i) = loge 4

√2 +

π

4i −

(loge

√2 +

π

4i)

= loge 4 = 1.3863

19.∫ 4i

−4i

1z2

dz = −1z

∣∣∣∣4i

−4i

= −[

14i

−(

1−4i

)]=

12

i

20.∫ 1+

√3 i

1−i

(1z

+1z2

)dz = Lnz − 1

z

∣∣∣∣1+√

3 i

1−i

= loge 2 +π

3i − 1

1 +√

3 i−

(loge

√2 − π

4i − 1

1 − i

)

= loge

√2 +

14

+ i

(7π

12+

√3

4+

12

)= 0.5966 + 2.7656i

21. Integration by parts gives ∫ez cos z dz =

12

ez(cos z + sin z) + C

and so∫ i

π

ez cos z dz =12

ez(cos z + sin z)∣∣∣iπ=

12[ei(cos i + sin i) − eπ(cos π + sinπ)]

=12[(cos 1 cosh 1 − sin 1 sinh 1 + eπ) + i(cos 1 sinh 1 + sin 1 cosh 1) = 11.4928 + 0.9667i.

22. Integration by parts gives ∫z sin z dz = −z cos z + sin z + C

and so ∫ i

0

z sin z dz = −z cos z + sin z∣∣∣i0= −i cos i + sin i = −i cosh 1 + i sinh 1 = −0.3679i.

23. Integration by parts gives ∫zez dz = zez − ez + C

and so∫ 1+i

i

zez dz = ez(z−1)∣∣∣1+i

i= ie1+i+ei(1−i) = (cos 1+sin 1−e sin 1)+i(sin 1−cos 1+e cos 1) = −0.9056+1.7699i.

24. Integration by parts gives ∫z2ez dz = z2ez − 2zez + 2ez + C

and so ∫ πi

0

z2ez dz = ez(z2 − 2z + 2)∣∣∣πi

0= eπi(−π2 − 2πi + 2) − 2 = π2 − 4 + 2πi.

888

18.4 Cauchy’s Integral Formulas

EXERCISES 18.4Cauchy’s Integral Formulas

1. By Theorem 18.9, with f(z) = 4, ∮̌C

4z − 3i

dz = 2πi · 4 = 8πi.

2. By Theorem 18.10 with f(z) = z2 and f ′(z) = 2z,∮̌C

z2

(z − 3i)2dz =

2πi

1!2(3i) = −12π.

3. By Theorem 18.9 with f(z) = ez, ∮̌C

ez

z − πidz = 2πieπi = −2πi.

4. By Theorem 18.9 with f(z) = 1 + 2ez,∮̌C

1 + 2ez

zdz = 2πi(1 + 2e0) = 6πi.

5. By Theorem 18.9 with f(z) = z2 − 3z + 4i,∮̌C

z2 − 3z + 4i

z − (−2i)dz = 2πi(−4 + 6i + 4i) = −π(20 + 8i).

6. By Theorem 18.9 with f(z) =13

cos z,

∮̌C

13

cos z

z − π

3

dz = 2πi

(13

cosπ

3

)=

π

3i.

7. (a) By Theorem 18.9 with f(z) =z2

z + 2i,

∮̌C

z2

z + 2i

z − 2idz = 2πi

(− 4

4i

)= −2π.

(b) By Theorem 18.9 with f(z) =z2

z − 2i,

∮̌C

z2

z − 2i

z − (−2i)dz = 2πi

( −4−4i

)= 2π.

8. (a) By Theorem 18.9 with f(z) =z2 + 3z + 2i

z + 4,

∮̌C

z2 + 3z + 2i

z + 4z − 1

dz = 2πi

(4 + 2i

5

)= π

(−4

5+

85

i

).

889


(b) By Theorem 18.9 with f(z) =z2 + 3z + 2i

z − 1,

∮̌C

z2 + 3z + 2i

z − 1

z − (−4)dz = 2πi

(4 + 2i

−5

)= π

(45− 8

5i

).

9. By Theorem 18.9 with f(z) =z2 + 4z − i

,

∮̌C

z2 + 4z − i

z − 4idz = 2πi

(−12

3i

)= −8π.

10. By Theorem 18.9 with f(z) =sin z

z + πi,

∮̌C

sin z

z + πi

z − πidz = 2πi

(sinπi

2πi

)= i sinhπ.

11. By Theorem 18.10 with f(z) = ez2, f ′(z) = 2zez2

, and f ′′(z) = 4z2ez2+ 2ez2

,∮̌C

ez2

(z − i)3dz =

2πi

2![−4e−1 + 2e−1] = −2πe−1i.

12. By Theorem 18.10 with f(z) = z, f ′(z) = 1, f ′′(z) = 0, and f ′′′(z) = 0,∮̌C

z

(z − (−i))4dz =

2πi

3!(0) = 0.

13. By Theorem 18.10 with f(z) = cos 2z, f ′(z) = −2 sin 2z, f ′′(z) = −4 cos 2z, f ′′′(z) = 8 sin 2z, f (4)(z) = 16 cos 2z,∮̌C

cos 2z

z5dz =

2πi

4!(16 cos 0) =

4π

3i.

14. By Theorem 18.10 with f(z) = e−z sin z, f ′(z) = e−z cos z − e−z sin z, and f ′′(z) = −2e−z cos z,∮̌C

e−z sin z

z3dz =

2πi

2!(−2e0 cos 0) = −2πi.

15. (a) By Theorem 18.9 with f(z) =2z + 5z − 2

,

∮̌C

2z + 5z − 2

zdz = 2πi

(−5

2

)= −5πi.

(b) Since the circle |z − (−1)| = 2 encloses only z = 0, the value of the integral is the same as in part (a).

(c) From Theorem 18.9 with f(z) =2z + 5

z,

∮̌C

2z + 5z

z − 2dz = 2πi

(92

)= 9πi.

(d) Since the circle |z − (−2i)| = 1 encloses neither z = 0 nor z = 2 it follows from the Cauchy-GoursatTheorem, Theorem 18.4, that ∮̌

C

2z + 5z(z − 2)

dz = 0.

890


16. By partial fractions, ∮̌C

z

(z − 1)(z − 2)dz = 2

∮̌C

dz

z − 2−

∮̌C

dz

z − 1.

(a) By the Cauchy-Goursat Theorem, Theorem 18.4,∮̌C

z

(z − 1)(z − 2)dz = 0.

(b) As in part (a), the integral is 0.

(c) By Theorem 18.4,∮̌

C

dz

z − 2= 0 whereas by Theorem 18.9,

∮̌C

dz

z − 1= 2πi. Thus∮̌

C

z

(z − 1)(z − 2)dz = −2πi.

(d) By Theorem 18.9,∮̌

C

dz

z − 1= 2πi and

∮̌C

dz

z − 2= 2πi. Thus∮̌

C

z

(z − 1)(z − 2)dz = 2(2πi) − 2πi = 3πi.

17. (a) By Theorem 18.10 with f(z) =z + 2

z − 1 − iand f ′(z) =

−3 − i

(z − 1 − i)2,

∮̌C

z + 2z − 1 − i

z2dz =

2πi

1!

( −3 − i

(−1 − i)2

)= −π(3 + i).

(b) By Theorem 18.9 with f(z) =z + 2z2

,

∮̌C

z + 2z2

z − (1 + i)dz = 2πi

(3 + i

(1 + i)2

)= π(3 + i).

18. (a) By Theorem 18.10 with f(z) =1

z − 4, f ′(z) = − 1

(z − 4)2, and f ′′(z) =

2(z − 4)3

,

∮̌C

1z − 4z3

dz =2πi

2!

(2

−64

)= − π

32i.

(b) By the Cauchy-Goursat Theorem, Theorem 18.4,∮̌C

1z3(z − 4)

dz = 0.

19. By writing∮̌

C

(e2iz

z4− z4

(z − i)3

)dz =

∮̌C

e2iz

z4dz −

∮̌C

z4

(z − i)3dz

we can apply Theorem 18.10 to each integral:∮̌C

e2iz

z4dz =

2πi

3!(−8i) =

8π

3,

∮̌C

z4

(z − i)3dz =

2πi

2!(−12) = −12πi.

Thus∮̌

C

(e2iz

z4− z4

(z − i)3

)dz = π

(83

+ 12i

).

20. By writing∮̌

C

(cosh z

(z − π)3− sin2 z

(2z − π)3

)dz =

∮̌C

cosh z

(z − π)3dz −

∮̌C

18 sin2 z

(z − π2 )3

dz

891


we apply Theorem 18.4 to the first integral and Theorem 18.10 to the second:∮̌C

cosh z

(z − π)3dz = 0,

∮̌C

18 sin2 z

(z − π2 )3

dz =2πi

2!

(−1

4sin2 π

2

)= −π

4i.

Thus∮̌

C

(cosh z

(z − π)3− sin2 z

(2z − π)3

)dz =

π

4, i.

21. We have∮̌

C

1z3(z − 1)2

dz =∮̌

C1

1(z − 1)2

z3dz +

∮̌C2

1z3

(z − 1)2dz

where C1 and C2 are the circles |z| = 1/3 and |z − 1| = 1/3, respectively. By Theorem 18.10,

∮̌C1

1(z − 1)2

z3dz =

2πi

2!(6) = 6πi,

∮̌C2

1z3

(z − 1)2dz =

2πi

1!(−3) = −6πi.

Thus∮̌

C

1z3(z − 1)2

dz = 6πi − 6πi = 0.

22. We have∮̌

C

1z2(z2 + 1)

dz =∮̌

C1

1z2(z + i)

z − idz +

∮̌C2

1z2 + 1

z2dz

where C1 and C2 are the circles |z − i| = 1/3 and |z| = 1/8, respectively. By Theorems 18.9 and 18.10,

∮̌C1

1z2(z + i)

z − idz = 2πi

(1

−2i

)= −π,

∮̌C2

1z2 + 1

z2dz =

2πi

1!(0) = 0.

Thus∮̌

C

1z2(z2 + 1)

dz = −π.

23. We have∮̌

C

3z + 1z(z − 2)2

dz =∮̌

C1

3z + 1z

(z − 2)2dz −

∮̌C2

3z + 1(z − 2)2

zdz

where C1 and C2 are the closed portions of the curve C enclosing z = 2 and z = 0, respectively. ByTheorems 18.10 and 18.9,

∮̌C1

3z + 1z

(z − 2)2dz =

2πi

1!

(−1

4

)= −π

2i,

∮̌C2

3z + 1(z − 2)2

zdz = 2πi

(14

)=

π

2i.

Thus∮̌

C

3z + 1z(z − 2)2

dz = −π

2i − π

2i = −πi.

24. We have∮̌

C

eiz

(z2 + 1)2dz =

∮̌C1

eiz

(z + i)2

(z − i)2dz −

∮̌C2

eiz

(z − i)2

(z − (−i))2dz

892

CHAPTER 18 REVIEW EXERCISES

where C1 and C2 are the closed portions of the curve C enclosing z = i and z = −i, respectively. ByTheorem 18.10,

∮̌C1

eiz

(z + i)2

(z − i)2dz =

2πi

1!

(−4e−1

−8i

)= πe−1,

∮̌C2

eiz

(z − i)2

(z − (−i))2dz =

2πi

1!

(08i

)= 0.

Thus∮̌

C

eiz

(z2 + 1)2dz = πe−1.


1. True 2. False 3. True 4. True

5. 0 6. π(−16 + 8i) 7. π(6π − i) 8. a constant function

9. True (Use partial fractions and write the given integral as two integrals.)

10. True

11. integer not equal to −1; −1

12. 12π

13. Since f(z) = z is entire,∫

C

(x + iy) dz is independent of the path C. Thus

∮̌C

(x + iy) dz =∫ 3

−4

z dz =z2

2

∣∣∣∣3−4

= −72

.

14. We have∫

C

(x − iy) dz =∫

C1

(x − iy) dz +∫

C2

(x − iy) dz +∫

C3

(x − iy) dz

On C1, x = 4, 0 ≤ y ≤ 2, z = 4 + iy, dz = i dy,∫C1

(4 − iy)i dy = i

∫ 2

0

(4 − iy) dy = i

(4y − i

2y2

) ∣∣∣∣20

= 2 + 8i.

On C2, y = 2, −4 ≤ x ≤ 3, z = x + 2i, dz = dx,∫C2

(x − 2i) dx =∫ 3

−4

(x − 2i) dx =12x2 − 2ix

∣∣∣∣3−4

= −72− 14i.

On C3, x = 3, 0 ≤ y ≤ 2, z = 3 + iy, dz = i dy,∫C3

(3 − iy)i dy = i

∫ 0

2

(3 − iy) dy = i

(3y − i

2y2

) ∣∣∣∣02

= −2 − 6i.

Thus∫

C

(x − iy) dz = 2 + 8i − 72− 14i − 2 − 6i = −7

2− 12i.

893


15.∫

C

|z2| dz =∫ 2

0

(t4 + t2) dt + 2i

∫ 2

0

(t5 + t3) dt =13615

+883

i

16.∫

C

eπz dz =1π

∫ 1+i

i

eπz(π dz) =1π

eπz

∣∣∣∣1+i

i

=1π

(1 − eπ)

17. By the Cauchy-Goursat Theorem, Theorem 18.4,∮̌

C

eπz dz = 0.

18.∫ 1−i

3i

(4z − 6) dz = 2z2 − 6z∣∣∣1−i

3i= 12 + 20i

19.∫

C

sin z dz =∫ 1+4i

1

sin z dz = − cos z∣∣∣1+4i

1= cos 1 − cos(1 + 4i) = −14.2144 + 22.9637i

20.∫

C

(4z3 + 3z2 + 2z + 1) dz =∫ 2i

0

(4z3 + 3z2 + 2z + 1) dz = z4 + z3 + z2 + z∣∣∣2i

0= 12 − 6i

21. On |z| = 1, let z = eit, dz = ieit dt, so that∮̌C

(z−2 + z−1 + z + z2) dz = i

∫ 2π

0

(e−2it + e−it + eit + e2it)eit dt = −e−it + it +12e2it +

13e3it

∣∣∣∣2π

0

= 2πi.

22. By partial fractions and Theorem 18.9,∮̌C

3z + 4z2 − 1

dz =72

∮̌C

1z − 1

dz − 12

∮̌C

1z − (−1)

dz =72(2πi) − 1

2(2πi) = 6πi.

23. By Theorem 18.10 with f(z) = e−2z, f ′(z) = −2e−2z, f ′′(z) = 4e−2z, and f ′′′(z) = −8e−2z,∮̌C

e−2z

z4dz =

2πi

3!(−8) = −8π

3i.

24. By Theorem 18.10 with f(z) =cos z

z − 1and f ′(z) =

sin z − cos z − z sin z

(z − 1)2,

∮̌C

cos z

z − 1z2

dz =2πi

1!

(−11

)= −2πi.

25. By Theorem 18.9 with f(z) =1

2(z + 3),

∮̌C

12(z + 3)

(z − (−1/2))dz = 2πi

(15

)=

2π

5i.

26. Since the function f(z) = z/ sin z is analytic within and on the given simple closed contour C, it follows fromthe Cauchy-Goursat Theorem, Theorem 18.4, that∮̌

C

z csc z dz = 0.

27. Using the principle of deformation of contours we choose C to be the more convenient circular contour |z+i| = 14 .

On this circle z = −i + 14eit and dz = 1

4 ieit dt. Thus∮̌C

z

z + idz = i

∫ 2π

0

(14eit − i

)dt = 2π.

894


28. (a) By Theorem 18.9 with f(z) =eiπz

2(z − 2),

∮̌C

eiπz

2(z − 2)

z − 1/2dz = 2πi

(eiπ/2

−3

)=

2π

3.

(b) By Theorem 18.9 with f(z) =eiπz

2z − 1,

∮̌C

eiπz

2z − 1z − 2

dz = 2πi

(e2πi

3

)=

2π

3i.

(c) By the Cauchy-Goursat Theorem, Theorem 18.4,∮̌C

eiπz

2z2 − 5z + 2dz = 0.

29. For f(z) = zng(z) we have f ′(z) = zng′(z) + nzn−1g(z) and so

f ′(z)f(z)

=zng′(z) + nzn−1g(z)

zng(z)=

g′(z)g(z)

+n

z.

Thus by Theorem 18.4 and (4) of Section 18.2,∮̌C

f ′(z)f(z)

dz =∮̌

C

g′(z)g(z)

dz + n

∮̌C

1z

dz = 0 + n(2πi) = 2nπi.

30. We have∣∣∣∣∫

C

Ln(z + 1) dz

∣∣∣∣ ≤ |max of Ln(z + 1) on C| · 2,

where 2 is the length of the line segment. Now

|Ln(z + 1)| ≤ | loge(z + 1)| + |Arg(z + 1)|.

But max Arg(z + 1) = π/4 when z = i and max|z + 1| =√

10 when z = 2 + i. Thus,∣∣∣∣∫C

Ln(z + 1) dz

∣∣∣∣ ≤ (12

loge 10 +π

4

)2 = loge 10 +

π

2.

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Chapt 18 engineering mathatics solution

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Transcript of Chapt 18 engineering mathatics solution