18 Integration in the Complex Plane

3Exercises 18.1

779

1. C

(z + 3) dz = (2+ 4i) 1

3(2t + 3) dt + i 1

(4t 1) dt = (2+ 4i)[14 + 14i] = 28+ 84i

2. C

2(2z z) dz =0

2[t 3(t2 + 2)i](1+ 2ti) dt = 02

2(6t3 + 13t) dt + i 0

(t2 + 2) dt = 50 + 20 i3

3. C

z2 dz = (3+ 2i)3 21

t2 dt = 16 (3+ 2i)3 =3

73648 + 3 i

14. C

(3z2 2z) dz = 0

(15t4 + 4t3 + 3t2 2t) dt + i 0

(6t5 + 12t3 6t2 ) dt = 2+ 0i = 2

5. Using z = eit , /2 t /2, and dz = ieit dt, 1 + z dz =

/2

(1 + eit ) dt = (2 + )i.C z

2 2 2 1

/2

21

26. C

|z|

dz = 1

2t5 +

tdt i1

t2 +

t4 dt = 21+ ln 4 8 i7. Using z = eit = cos t + i sin t, dz = ( sin t + i cos t) dt and x = cos t,

Re(z) dz =C

2

0

cos t( sin t + i cos t) dt =

2

0

sin t cos t dt + i

2

0

cos2 t dt

21 2= 0

1 sin 2t dt + 2 i

2

0

(1 + cos 2t) dt = i.

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

1

5

2

2it it

C (z + i)3 z + i +8

dz = i0

[e

5+ 8e

] dt = 10i.

9. Using y = x + 1, 0 x 1, z = x + (x + 1)i, dz = (1 i) dx,0 7 1 (x2 + iy3 ) dz = (1 i) C 1

[x2 + (1 x)3 i] dx =12

+ 12 i.10. Using z = eit , t 2, dz = ieit dt, x = cos t = (eit + eit )/2, y = sin t = (eit eit )/2i,

(x3 iy3 ) dz =

1 i

2(e3it + 3eit + 3eit + e3it )eit dt +

1 i

2

(e3it 3eit + 3eit e3it )eit dtC 8

1 2= i8

8

(2e4it + 6) dt = 3 i.4

11. C

ez dz = C1

ez dz + C2

ez dz where C1 and C2 arethelinesegments y = 0, 0 x 2 and y = x + 2,1 x 2, respectively. Now

2 ez dz = C1 01

ex dx = e2 1

1 ez dz = (1 i) C2 2

ex+(x+2)i dx = (1 i) 2

e(1i)x dx = e1i e2(1i) = e e2 .Exercises 18.1

In the second integral we have used the fact that ez has period 2i. Thus

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

12. C

sin z dz = C1

sin z dz + C2

sin z dz where C1 and C2 arethelinesegments y = 0, 0 x 1, and x = 1,0 y 1, respectively. Now

sin z dz = C1 01

1sin x dx = 1 cos 1

Thus

sin z dz = i C2 0

sin(1 + iy) dy = cos 1 cos(1 + i). 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.C

13. We have C

Im(z i) dz =C1

(y 1) dz +C2

(y 1) dzOn C1 , z = eit , 0 t /2, dz = ieit dt, y = sin t = (eit eit )/2i, = (y 1) dz =C1

1 /22 0

[eit eit 2i]eit dt =

1 /22 0

4[e2it 1+ 2ieit ] dt = 1

12 i.

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

x dx =

2 + 2 i.

Thus C

Im(z i) dz = 1 4

1 12 i + 2

1 + 2 i =

3 2 4 .

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

3/2

sin t dt + 2i

3/2

cos t dt = 2i(2) = 4i.C

15. We have

/2

zez dz =

zez dz +

/2

zez dz +

zez dz +

zez dzC C1

C2 C3 C4On C1 , y = 0, 0 x 1, z = x, dz = dx,

1 1 zez dz = C1 0

x x x

xe dx = xe e = 1. 0

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

1 zez dz = i C2 0

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

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

0 zez dz = C3 1

(x + i)ex+i dx = (i 1)ei ie1+i .Exercises 18.1

On C4 , x = 0, 0 y 1, z = iy, dz = i dy,0 zez dz = C4 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) dzOn C1 , y = x2 , 1 x 0, z = x + ix2 , dz = (1+ 2xi) dx,0 f (z) dz = C1 1

2(1+ 2xi) dx = 2 2i.On C2 , y = x2 , 0 x 1, z = x + ix2 , dz = (1+ 2xi) dx,1 f (z) dz = C2 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 dzOn C1 , y = 0, 0 x 1, z = x, dz = dx,

1 x dz = C1 0

1

1x dx = 2 .On C2 , x = 1, 0 y 1, z = 1 + iy, dz = i dy,

C2

x dz = i 0

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

x dx = 2

2 i.

Thus

1 1x dz = + i

1 1i = i.

2C 2 2 2

18. We have C

(2z 1) dz =C1

(2z 1) dz +C2

(2z 1) dz +C3

(2z 1) dzOn C1 , y = 0, 0 x 1, z = x, dz = dx,1 (2z 1) dz =C1 0

(2x 1) dx = 0.On C2 , x = 1, 0 y 1, z = 1 + iy, dz = i dy,1 1 (2z 1) dz = 2C2 0

y dy + i 0

dy = 1+ i.On C3 , y = x, z = x + ix, dz = (1 + i) dx,0 (2z 1) dz = (1 + i)C3 1

(2x 1+ 2ix) dx = 1 i.Exercises 18.1

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 dzOn C1 y = 0, 0 x 1, z = x, dz = dx,

1 z2 dz = C1 0

x2 dx = 1 .3On C2 , x = 1, 0 y 1, z = 1 + iy, dz = i dy,1 2 z2 dz = C2 0

(1 + iy)2 i dy = 1+ i.

3

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

0 z2 dz = (1 + i)3 C3 1

x2 dx = 23

2 3 i.

Thus

z2 dz = 1

2 21+ i +

2i = 0.

33C 3 3

20. We have C

z2 dz = C1

z2 dz + C2

z2 dz + C3

z2 dzOn C1 , y = 0, 0 x 1, z = x, dz = dx,

1z2 dz = 0

x2 dx = 1 .3On C2 , x = 1, 0 y 1, z = 1 + iy, dz = i dy,1 2 z2 dz = C2 0

(1 iy)2 (i dy) = 1 +

3 i.

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

0 2 2 z2 dz = (1 i)2 (1 + i) C3 1

x2 dx =

33

+ 3 i.

Thus

z2 dz = 1 +1+ 2 i

2 2 2 4

33+ i = + i.

333 C

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

1 4 5 (z2 z + 2) dz = (1 i) C 0

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

3 3 i.

22. We have C

(z2 z + 2) dz = C1

(z2 z + 2) dz + C2

(z2 z + 2) dzOn C1 , y = 1, 0 x 1, z = x + i, dz = dx,

61 5 (z2 z + 2) dz = C1 0

[(x + i)2 x +2 i] dx = .On C2 , x = 1, 0 y 1, z = 1 + iy, dz = i dy,0 1 5 (z2 z + 2) dz = i C2 1

[(1 + iy)2 +1 iy] dy =

2 3 i.Exercises 18.1

Thus C

(z2 z + 2) dz =

1 5 52 3 i + 6

4= 3

53 i.

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

1 1 4 5 (z2 z + 2) dz = C 0

(5x4 + 2x3 + 7x2 3x + 1) dx + i 0

(2x5 8x3 + 3x2 1) dx =

3 3 i.24. On C , x = sin t, y = cos t, 0 t /2 or z = ieit , dz = eit dt,

/2

/2 (z2 z + 2) dz = C 0

(e2it ieit + 2)eit dt = 0

(e3it ie2it + 2eit ) dt1 3i/2

1 i

i/2 1 1 4 5

ez z

= 3 ie

5

+ 2 e + 2ie

ez

+ i 2i = i.

3 2 3 3

5 525. On C ,

|e | = e . Thus

e dz 10 =

e5 .

2 z2 + 1

|z| 1 24

C z2 +1 24 12

1 1 1

1 1 1 326. On C ,

2

= . Thus

dz

(12) = . z2 2i

|z|

|2i| 34

C z2 2i

34 2 1727. The length of the line segment from z = 0 to z = 1 + i is 2 . In addition, on this line segment

Thus

z2 dz ( + 4) 6 2.

2 2|z + 4| |z|

|+4 |1+ i 2

+4 = 6.C

28. On C ,

1 = 1 = 1

. Thus

1 dz

1 1 (8) = .|z|

3 C z3

3 64

z 64 4 32

29. (a)

dz = lim

n zk = lim

n (zk zk

1 )C P 0 k=1

P 0 k=1= lim P 0= lim P 0

[(z1 z0 )+ (z2 z1 )+ (z3 z2 )+ + (zn1 zn2 )+ (zn zn1 )] (zn z0 ) = zn z0 (b) With zn = 2i and z0 = 2i,

k30. With z = zk ,

dz = 2i (2i) = 4i.C

nz dz = lim

zk (zk zk

1 )C P 0 k=1= lim

[(z2 z1 z0 )+ (z2 z2 z1 )+ + (z2 zn zn

1 )]. (1)

kWith z = zk1 ,

1 2 n P 0

nz dz = lim

zk

1 (zk

zk1 )C P 0 k=1= lim

[(z0 z1 z2 )+ (z1 z2 z2 )+ + (zn 1 z

z2

)]. (2)

Adding (1) and (2) gives

0 P 0

1 n

1

n12 z dz = lim

(z2 z2 ) or

z dz =

(z2 z2 ).

2n 0 n 0C P 0 CExercises 18.1

31. (a) C

(6z + 4) dz = 6 C

z dz +4 C

6dz = [(2+ 3i)22

(1 + i)2 ] + 4[(2 + 3i)

(1 + i)] =

11 + 38i(b) Sincethecontour is closed, z0 = zn and so

6 z dz +4

dz = 6[z2 z2 ]+ 4[z0 z0 ] = 0.0 0C C

32. For f (z) = 1/z, f (z) = 1/z, so on z = 2eit , z = 2eit , dz = 2ieit dt, and

it 2 1

1 2it

2

2e1 4i f (z) dz =C 0

2eit 2ie

dt = 2 e

= [ 0

1] = 0.

Thus circulation = Re C

f (z) dz = 0, net ux = Im C

f (z) dz = 0.33. For f (z) = 2z, f (z) = 2z, so on z = eit , z = eit , dz = ieit dt, and

2

2 f (z) dz =C 0

(eit )(ieit dt) = 2i0

dt = 4i.

Thus circulation = Re C

f (z) dz = 0, net ux = Im C

f (z) dz = 4.34. For f (z) = 1/(z 1), f (z) = 1/(z 1), so on z 1 = 2eit , dz = 2ieit dt, and

it 2 1

2 f (z) dz =C 0

2eit 2ie

dt = i0

dt = 2i.

Thus circulation = Re C

f (z) dz = 0, net ux = Im C

f (z) dz = 2.35. For f (z) = z, f (z) = z so on thesquarewehave

f (z) dz =C

z dz +C1

z dz +C2

z dz +C3

z dzC4wheRe 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

1

z dz = C1 0

z dz = i C2 00

1x dx = 2

1 1(1 + iy) dy = 2 + i

1 z dz = C3 1

(x + i) dx = 2 i

and so

0z dz = C4 1

1y dy = 2 1 1 1 1

2 f (z) dz = +C

2 + i +

2 i

+ = 0 2

circulation = Re C

net ux = Im

f (z) dz = Re(0) = 0

f (z) dz = Im(0) = 0.CExercises 18.2

Exercises 18.2

1. f (z) = z3 1+ 3i is a polynomial and so is an entire function.2. z2 is entire and 1 z 4

is analytic within and on thecircle |z|

= 1.3. f (z) = z 2z +3

is discontinuous at z = 3/2 but is analytic within and on thecircle |z| = 1.4. f (z) = z 3 z2 + 2z +2

is discontinuous at z =

1+ i and at z =

1

i but is analytic within and on thecircle |z| = 1.

5. f (z) = (z2

|z| = 1.

sin z 25)(z2

z

+ 9)

is discontinuous at z = 5 and at z = 3i but is analytic within and on thecircle 6. f (z) = e 2z2 + 11z + 15

is discontinuous at z = 5/2 and at z = 3 but is analytic within and on thecircle |z| = 1.

37. f (z) = tan z is discontinuous at z = 2 , 2 , ... but is analytic within and on thecircle |z| = 1.28. f (z) = z 9

is discontinuous at i,

3i, ... but is analytic within and on thecircle z

= 1.cosh z

2 2 | |9. By theprincipleof deformation of contours wecan choosethemoreconvenient circular contour C1dened by |z| = 1. Thus1 1

by (4) of Section 18.2.

dz =

C z C1

dz = 2i z

10. By theprincipleof deformation of contours wecan choosethemoreconvenient circular contour C11dened by |z (1 i)| = 16 . Thus5 1 z +1+ i dz = 5

z ( 1

i) dz = 5(2i) = 10iC

by (4) of Section 18.2.

C1

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

1 z + dz =

z dz +

z1dz = 0+ 2i = 2i.

z C C C

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

1 1 z + dz =

dz +

1dz = 0+0 = 0.

zz 2 C C

C z2

13. Since f (z) = z

Cis analytic within and on C it follows from Theorem 18.4 that

z dz = 0.z2 2

14. By (4) of Section 18.2, C

10 (z + i)4 dz = 0.

z2 2

15. By partial fractions,

2z +1

1dz =

dz +

1 dz.

C z(z + 1)

C z

C z +1 Exercises 18.2

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

1 1

dz +C z C

z +1 dz = 2i + 0 = 2i.

(b) By writing C

+ C1 C2

where C1 and C2 arethecircles |z| = 1/2 and |z + 1| = 1/2, respectively,

wehaveby Theorem 18.4 and (4) of Section 18.2,

1 1 1

1 1 1

dz +

=C1C z C

z +1 dz =

dz +

C2z C1

z +1 dz +

dz +

z C2

z +1 dz

(c) Since f (z) =

2z +1 z(z + 1)

= 2i +0+0+ 2i = 4i.

is analytic within and on C it follows from Theorem 18.4 that

2z +1

C z2 + z dz = 0.

16. By partial fractions,

2z

dz =

1e e

dz +

1e e

dz.C

(a) By Theorem 18.4,

z2 +3

C z + 3 i

1

C z 3 i

1

C z + 3 i

dz +

C z

3 i

dz = 0+0 = 0.(b) By Theorem 18.4 and (4) of Section 18.2,

1 dz +

1 dz = 0 + 2i = 2i.

C z + 3 i

C z 3 i

(c) By writing C

=

+ C1 C2

where C1 and C2 arethecircles |z + 3 i| = 1/2 and |z 3 i| = 1/2,

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

1 dz +

1

1 dz =

1 dz +

1

2 dz +

1 dz +

1 dz

C z + 3 i

C z 3 i

C z + 3 i

C1 z 3 i

C z + 3 i

C2 z 3 i

= 2i +0+0+ 2i = 4i.

17. By partial fractions,

3z +2 dz =

1 dz

1 4 dz.

C z2 8z + 12

C z 2

C z 6

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

1

1

4

= 0 4(2i) =

8i.

=dz

C z 2

dzC z 6

(b) By writing C

+ C1 C2

where C1 and C2 arethecircles |z 2| = 1 and |z 6| = 1, respectively,

wehaveby Theorem 18.4 and (4) of Section 18.2,

4 1

dzC z 2 C

1

dz

=z 6 C1

1

dz

4z 2 C1

1

dz

+z 6 C2

1

dz

4z 2 C2

1 dz z 6

= 2i 4(0) + 0 4(2i) = 6i.Exercises 18.2

18. (a) By writing

=C

+ C1 C2

where C1 and C2 arethecircles |z + 2| = 1 and |z 2i| = 1, respectively, wehave by Theorem 18.4 and (4) of Section 18.2,

3

1

C1dz =

3

1

3

C2+

1

C z +2 z 2i

z +2 dz

dz

C1z 2i

z +2 dz

dz

C2z 2i

19. By partial fractions,

= 3(2i) 0+0 2i = 4i.

z 1

= 1 1

1 1+ +

1

+ 1 1

1 dz

C z(z i)(z 3i)

dz

3 C z

2 2 i

dz

C z i

6 2 i

dz.

C z 3i

By Theorem 18.4 and (4) of Section 18.2,

z 1

= 0 +

1 + 1 2i +0 = ( 1

i).

20. By partial fractions,

dz

C z(z i)(z 3i)

2 2 i

1 1 1 1 1

C1 1

CC z3 + 2iz2 dz = 4

dz i

z 2 C

z2 dz 4

z + 2i dz.By Theorem 18.4 and (4) of Section 18.2,

1 1 1 1

C z3 + 2iz2 dz = 4 2i 2 i(0) 4 (0) = 2 i.

21. We have

8z 3 dz =

8z 3 dz

8z 3 dz

C z2 z

C z2 z

C z2 z

12where C1 and C2 aretheclosed portions of thecurve C enclosing z = 0 and z = 1, respectively. By partial fractions, Theorem 18.4, and (4) of Section 18.2,

8z 3 dz

1C z2 z

dz 8z 3

= 5 C1

= 5

1 dz z 1

1

+3 C1

+3

1dz = 5(0) + 3(2i) = 6i z

1= 5(2i) + 3(0) = 10i.

C1 z2 z

dzC2 z 1

dzC2 z

Thus C

8z 3 dz z2 z

= 6i

10i =

4i.

22. By choosing themoreconvenient contour C1 dened by |z z0 | = r where r is small enough so that the circleC1 lies entirely within C wecan write

=dz 1

1C (z z0 )n C

1 dz.(z z0 )n

Let z z0 = reit , 0 t 2 and dz = ireit dt. Then for n = 1:

1 2 2

C1 1

dz =

ireit dt = i

dt = 2i.

For n = 1:

z z0

2

0 reit

0

(1n)it 2 1 i dz =

e(1n)it dt = i

e

1 = [e2(1n)i

1] = 0

C1 (z z0 )n

rn1 0

rn1 i(1 n) 0

rn1 (1 n) since e2(1n)i = 1.Exercises 18.2

z ez

23. Write Cz

e z +3 3z

dz = C

z +3

dz 3 C

z dz.

By Theorem 18.4, C

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

z dz =C

2

0

eit (ieit dt) = 2i.

z

Thus C

e z +3 3z

dz = 0 3(2i) = 6i.

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,

x dz = C 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,

1 2

1 0 1 1 x dz =C

x dx + i0

dy + (1+ 2i)0

x dx = 2 + 2i 2 (1+ 2i) = i.

Exercises 18.3

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

1 (4z 1) dz = iC 1

i i

(4iy 1) dy = 2i.(b) C

(4z 1) dz =i

(4z 1) dz = 2z2 z

i

= 2i2. (a) Choosing theline y = 1 x, 0 x 3 we have z = x + 1 xi, dz = (1 + 1 i) dx. Thus

ez dz =

33e(1+

11+ i

3

1 3dx = e = e

3

e = (e

cos 1

1) + ie

sin 1.

13 i)x C 0 3

(1+ 3 i)x 0

3+i 0 3 3

(b) C

ez dz = 0

3+i

3+iz z 3+i

e dz = e = e 0

e0 = (e3 cos 1

1) + ie3 sin 13. The given integral is independent of the path. Thus2i

2i 2z dz = C 2+7i

2z dz = z2 2+7i

= 48 + 24i.

4. The given integral is independent of the path. Thus2i

2i 6z2 dz = C 2

6z2 dz = z3 2

= 15 24i.

35.

3+i

z2 dz = 1 z3

3+i

26

3= 6 + i 0 0Exercises 18.3

6.

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

= 19 3i

12i

1+i

7.

z3 dz = 1 z4

1+i= 0

2i1i

2i8.3i

4

(z3 z) dz =

1i

1

26z44

z1 2i2 = 3i

1234

9.

1i

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

1i

7 22

6 3= ii/2

i

i/2

i

10.

(iz + 1)3 dz = 1 (iz + 1)4 = i

1

11.

4i 1

i ez dz = 1 ez i = 1 1 i

i/2

i/2

12.

1+2i

22 1 2

2zez dz = ez

1+2i

1

2= [e3+4i

2e2i ] = 1 (e3 cos 4

1

2 cos 2) + (e3 sin 4 + sin 2)i = 0.1918 + 0.4358i1i

1i

+2i

+2i

13.

sin z dz = 2 cos z

= 2

cos

+ i cos

= 2i sin sinh 1 = 2.3504i

2i

2 2 214. 12i

i

cos z dz = sin z 12i

= sin i

sin(1

2i) = i sinh

[sinh 1 cosh 2

i cos 1 sinh 2]

2i

= sin 1 cosh 2 + i(sinh + cos 1 sinh 2) = 3.1658 + 13.5083i15. i

2i

cosh z dz = sinh z = sinh 2i i

sinh i = i sin 2

i sin = 0

16. i

21+ i

sinh 3z dz

z1= cosh 33

1+ i

2 =

1 cosh3

3+ 32

ii cosh 3 i1 3 3 1 1= cosh 3 cos3 2

+ i sinh 3 sin 2

cos 3

= 3 cos 3

3 i sinh 3 = 0.3300 3.3393i

4i17.

1 4idz = Lnz

= Ln4i

Ln(

4i) = log

4+ i

log 4

i = i4i z4+4i

4i

e 2

e 2

18.

1 4+4i

dz = Lnz = Ln(4 + 4i)

Ln(1 + i) = log

42+ i

log

2+ i = log

4 = 1.3863

1+i

4i19.

z

1dz =

1+i

1 4i 1=

e

1 = 1 i

4 e 4 e

z24i z

4i

4i

4i 2

20.

1+3 i

1 + 1 dz = Lnz

1 1+3 i

= loge

2+ i

1e e

e log 2 i

1

21i z z

z 1i

3 1+ 3 i

4 1 i 1

7

3 1 = loge

21. Integration by parts gives

2+ + i4

+ +12 4 2

= 0.5966 + 2.7656i ez cos z dz = 1 ez (cos z + sin z)+ C2Exercises 18.3

and so i

i ez cos z dz = 1 ez (cos z + sin z)

1= [ei (cos i + sin i)

e (cos + sin )] 2 2 1

= [(cos 1 cosh 1 sin 1 sinh 1 + e )+ i(cos 1 sinh 1 + sin 1 cosh 1) = 11.4928 + 0.9667i.2

22. Integration by parts gives

and so i

z sin z dz = z cos z + sin z + C

i

0z sin z dz = z cos z + sin z = i cos i + sin i = i cosh 1 + i sinh 1 = 0.3679i.0

23. Integration by parts gives

and so

zez dz = zez ez + C 1+i

i

zez dz = ez (z 1) 1+i i

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

24. Integration by parts gives

and so i

z2 ez dz = z2 ez 2zez + 2ez + C

i

z2 ez dz = ez (z2 2z + 2) 0 0

= ei (2 2i + 2) 2 = 2 4+ 2i.

Exercises 18.4

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

4 dz

C z 3i

= 2i

4 = 8i.

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

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

dz z2C (z 3i)2

2i

=1!

2(3i) = 12.

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

ezdz

C z i

= 2iei = 2i.

1+2ez 0dz = 2i(1+ 2e ) = 6i.

C z

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

z2 3z +4idz

C z (2i)

= 2i(4+ 6i + 4i) = (20+ 8i).Exercises 18.4

16. By Theorem 18.9 with f (z) = 3 cos z,

1cos z3

1 C z 3

dz = 2i

cos3 3

= i.3

27. (a) By Theorem 18.9 with f (z) = z ,z + 2iz2 z + 2iC z 2i

dz = 2i

4 4i

= 2.

2(b) By Theorem 18.9 with f (z) = z ,z 2iz2 z 2i C z (2i)

2

dz = 2i

4 4i

= 2.8. (a) By Theorem 18.9 with f (z) = z +3z +2i ,z +4 z2 + 3z + 2i z +4

C z 1

2

dz = 2i

4+ 2i = 5

4 5 +

8 5 i .(b) By Theorem 18.9 with f (z) = z +3z +2i ,z 1z2 + 3z + 2i z 1

C z (4)

2

dz = 2i

4+2i 5

= 45

8 5 i .9. By Theorem 18.9 with f (z) = z +4 ,z i

C

sin z

z2 +4 z i z 4i

dz = 2i

12 3i

= 8.10. By Theorem 18.9 with f (z) =

,z + i

sin zz + i

dz = 2i

sin i

= i sinh .C z i

2i

2 2 2 211. By Theorem 18.10 with f (z) = ez , f (z) = 2zez , and f (z) = 4z2 ez

+ 2ez ,

2zdze = 2i

[4e1 + 2e1 ] = 2e1 i.

C (z i)3 2!

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

dz z

C (z (i))4

2i= 3! (0) = 0.Exercises 18.4

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,

cos 2z

z 5 dz =C

2i (16 cos 0) =4!

43 i.14. By Theorem 18.10 with f (z) = ez sin z, f (z) = ez cos z ez sin z, and f (z) = 2ez cos z,ez sin z

2i 0

z 3 dz =C

( 2e

2!

cos 0) = 2i.

15. (a) By Theorem 18.9 with f (z) =

2z +5 ,z 22z +5 z 2 dz = 2iC z

5 2 = 5i.

(b) Sincethecircle |z (1)| = 2 encloses only z = 0, thevalueof theintegral is thesameas in part (a).2z +5 (c) From Theorem 18.9 with f (z) =

,z2z +5 zC z 2

dz = 2i

9 2

= 9i.

(d) Sincethecircle |z (2i)| = 1 encloses neither z = 0 nor z = 2 it follows from theCauchy-GoursatTheorem, Theorem 18.4, that

16. By partial fractions,

2z +5 dz

C z(z 2)

= 0. z

dz = 2

dz dz

C (z 1)(z 2)

C z

2 C

.z 1(a) By the Cauchy-Goursat Theorem, Theorem 18.4,

(b) As in part (a), theintegral is 0.

z dz

C (z 1)(z 2)

= 0.

(c) By Theorem 18.4,

dz = 0 whereas by Theorem 18.9,

dz = 2i. Thus

C z 2

C z 1 z dz

C (z 1)(z 2)

= 2i.

(d) By Theorem 18.9,

dz = 2i and

dz = 2i. Thus

C z 1

C z 2 z dz

C (z 1)(z 2)

= 2(2i) 2i = 3i.

17. (a) By Theorem 18.10 with f (z) = z +2 and f (z) = 3 i ,z 1 i

z +2

z 1 i dz =

(z 1 i)2

2i 3 i = (3 + i).C z2

1! (1 i)2Exercises 18.4

(b) By Theorem 18.9 with f (z) = z +2 ,z2z +2 z2

dz = 2i

3+ i

= (3 + i).C z (1 + i)

1

(1 + i)2

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

z 4

, f (z) =

1

(z 4)2

, and f (z) =

, (z 4)3 z 4

dz =

2i 2

= i.C z3

2! 64 32

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

dz 1

C z3 (z 4)

= 0.

2iz 4

2iz 419. By writing

e z dz =

e dz

z dz

zC 4 (z i)3

C z4

C (z i)3we can apply Theorem 18.10 to each integral:

e2iz

2i 8

z4 2i

z 4 dz =C

( 8i) = ,

3! 3

2iz 4

C (z

i)3 dz =

8

2! (12) = 12i.

Thus

e z dz = + 12i .

zC 4 (z i)3 3

22 1 2

20. By writing

cosh z

sin z dz =

cosh z

dzdz

8 sin z

C (z )3 (2z )3

C (z )3

C (z )3

we apply Theorem 18.4 to the rst integral and Theorem 18.10 to thesecond:

21 2 cosh z

dz= 0,

8 sin z

2i

dz=

1 sin2 =

C (z )3

C (z )3

2! 4

2 4 i.

Thus

cosh z

sin2e z dz =

C (z )3 (2z )3

1 2

4 , i.

1

21. We have

1

1dz =

(z 1) dz +

z3 dz

C z3 (z 1)2

C z3

C2 (z 1)2

where C1 and C2 arethecircles |z| = 1/3 and |z 1| = 1/3, respectively. By Theorem 18.10, 1 (z 1)2

z3C1

dz =

2i2!

(6) = 6i,

1 z3C2 (z 1)2

dz =

2i1!

(3) = 6i.

Thus C

1

dz = 6i z3 (z 1)2

1 2

6i = 0.

1222. We have

1

dz =

z (z + i)

2dz +

z +1 dz

C z2 (z2 + 1)

C1 z i

C z2Exercises 18.4

where C1 and C2 arethecircles |z i| = 1/3 and |z| = 1/8, respectively. By Theorems 18.9 and 18.10,

1 z2 (z + i)C1 z i

dz = 2i

12i

= ,

1 z2 +1

2C z2

dz =

2i1!

(0) = 0.

Thus C

1 dz =z2 (z2 + 1)

3z +1

.

3z +1 2

23. We have C

3z +1 z(z 2)2

dz = C1

z dz

2 (z 2) C2

(z 2) dz z

where C1 and C2 aretheclosed portions of thecurve C enclosing z = 2 and z = 0, respectively. ByTheorems 18.10 and 18.9,

3z +1 z

dz =

2i 1

= i,

3z +1 (z 2)2

dz = 2i

1 = i.C1 (z 2)2

1! 4 2

C2 z 4 2

Thus

3z +1 =

= i.

dzC z(z 2)2

iz

2 i 2 i

eiz2

eiz2

24. We have

e

dz =

(z + i)

2dz

(z i) dz

C (z2 + 1)2

C1 (z i)2

C (z (i))2

where C1 and C2 aretheclosed portions of thecurve C enclosing z = i and z = i, respectively. ByTheorem 18.10,

eiz (z + i)2

dz =

2i

4e1

= e1 ,

eiz (z i)2

dz =

2i 0

= 0.C1 (z i)2

1! 8i

C2 (z (i))2

1! 8i

Thus C

ei zdz = e1 .(z2 + 1)2

Chapter 18 Review Exercises

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

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

9. True(Usepartial fractions and writethegiven integral as two integrals.)

10. True

11. integer not equal to 1; 1

12. 12Chapter 18 Review Exercises

13. Since f (z) = z is entire, C

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

3 z2 3 7 (x + iy) dz =C

z dz =4 2

4

= 2 .

14. We have C

(x iy) dz =C1

(x iy) dz +C2

(x iy) dz +C3

(x iy) dzOn C1 , x = 4, 0 y 2, z = 4 + iy, dz = i dy,

2 2 (4 iy)i dy = i

i

2(4 iy) dy = i 4y y = 2+ 8i.

2 C1 0 0On C2 , y = 2, 4 x 3, z = x + 2i, dz = dx,

3

1 3 7 (x 2i) dx =C2 4

(x 2i) dx =

x2 2ix =

2 4 2

14i.On C3 , x = 3, 0 y 2, z = 3 + iy, dz = i dy,

0 0 (3 iy)i dy = i

i

2(3 iy) dy = i 3y y = 2 6i.

2 C3 2 2

Thus C

7 (x iy) dz = 2+ 8i 2

7 14i 2 6i = 2

12i.

15. C

2|z2 | dz = 0

2(t4 + t2 ) dt + 2i 01+i

(t5 + t3 ) dt =

1+i

13615

88+ 3 i

16.

ez dz = 1

ez ( dz) = 1 ez

1= (1

e )

iC i

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

ez dz = 0.

18.

1i

(4z 6) dz = 2z2 6z 1i

= 12+ 20i3i

19.

sin z dz =

1+4i

3i

sin z dz = cos z

1+4i

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

20. C

12i

(4z3 + 3z2 + 2z + 1) dz =0

1

(4z + 3z + 2z + 1) dz = z + z + z + z = 122i3 2 4 3 2

6i 0

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

(z2 + z1 + z + z2 ) dz = iC

2

0

(e2it + eit + eit + e2it )eit dt = eit + it +

1 e2it +2

e31 23it 0

= 2i.

22. By partial-fractions and Theorem 18.9,

dz3z +4

= 7 1

1 1

7

= (2i)

1(2i) = 6i.

C z2 1

2 C

dz

z 1 2 C

dzz (1) 2 2

23. By Theorem 18.10 with f (z) = e2z , f (z) = 2e2z , f (z) = 4e2z , and f (z) = 8e2z ,e2z

2i 8

z 4 dz =C

( 8) = i.

3! 3Chapter 18 Review Exercises

24. By Theorem 18.10 with f (z) =

cos z z 1

and f (z) =

cos z

sin z cos z z sin z , (z 1)2 z 1

C z2

1

dz =

2i 1!

1 1 = 2i.25. By Theorem 18.9 with f (z) = 2(z + 3) ,

12(z + 3)

dz = 2i

1 = 2 i.C (z (1/2)) 5 5

26. Sincethefunction f (z) = z/ sin z is analytic within and on the given simple closed contour C , it follows from the Cauchy-Goursat Theorem, Theorem 18.4, that z csc z dz = 0.C

427. Using theprincipleof deformation of contours wechoose C to be the more convenient circular contour |z +i| = 1 .On this circle z = i + 1 eit and dz = 1 ieit dt. Thus4 4

z

it0 2 1

C z + i dz = i

4 e i

dt = 2.

iz28. (a) By Theorem 18.9 with f (z) = e ,2(z 2)

eiz 2(z 2)C z 1/2

dz = 2i

ei/2 2= .3 3

iz(b) By Theorem 18.9 with f (z) = e ,2z 1

eiz 2z 1C z 2

dz = 2i

e2i =3

23 i.

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

dz eiz

C 2z2 5z +2

= 0.

29. For f (z) = zn g(z) we have f (z) = zn g (z)+ nzn1 g(z) and son n1f (z) = z g (z)+nz g(z) = g (z) + nf (z)

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

zn g(z)

g(z) z .

f (z)

Cg (z) 1

CC f (z) dz =

g(z) dz + n

dz = 0 + n(2i) = 2ni. z30. We have

Ln(z + 1) dz |max of Ln(z + 1) on C | 2,

C Chapter 18 Review Exercises

where 2 is the length of the line segment. Now

|Ln(z + 1)| | loge (z + 1)| + |Arg(z + 1)|.But max Arg(z + 1) = /4 when z = i and max|z + 1| = 10 when z = 2 + i. Thus,

1

Ln(z + 1) dz 2

loge 10 +

4 2 = loge 10 + 2 .C