18 Integration in the Complex Plane

EXERCISES 18.1
Contour Integrals

2
16 736
3. 2
z dz = (3 + 2i) 3
(3 + 2i)3 = −48 +
t2 dt = i
C −2 3 3


2 
2 
2 1 21
6. |z|2 dz = 2t5 + dt − i t2 + 4 dt = 21 + ln 4 − i
C 1 t 1 t 8
9. Using y = −x + 1, 0 ≤ x ≤ 1, z = x + (−x + 1)i, dz = (1 − i) dx,


0
7 1
(x2 + iy 3 ) dz = (1 − i) [x2 + (1 − x)3 i] dx = − + i.
C 1 12 12




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

0 ≤ y ≤ 1, respectively. Now


1
sin z dz = sin x dx = 1 − cos 1
C1 0


1
sin z dz = i sin(1 + iy) dy = cos 1 − cos(1 + i).
C2 0

Thus


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





z
15. We have ˇ ze dz = zez dz + zez dz + zez dz + zez dz
C C1 C2 C3 C4
On C1 , y = 0, 0 ≤ x ≤ 1, z = x, dz = dx,


1 1

zez dz = xex dx = xex − ex  = 1.
C1 0 0

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


1
z
ze dz = i (1 + iy)e1+iy dy = iei+1 .
C2 0

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


0
zez dz = (x + i)ex+i dx = (i − 1)ei − ie1+i .
C3 1

285
18.1 Contour Integrals

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


0
zez dz = − yeiy dy = (1 − i)ei − 1.
C4 1

Thus z
ˇ ze dz = 1 + ie
i+1
+ (i − 1)ei − ie1+i + (1 − i)ei − 1 = 0.
C




18. We have ˇ (2z − 1) dz = (2z − 1) dz + (2z − 1) dz + (2z − 1) dz
C C1 C2 C3
On C1 , y = 0, 0 ≤ x ≤ 1, z = x, dz = dx,


1
(2z − 1) dz = (2x − 1) dx = 0.
C1 0

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


1
1
(2z − 1) dz = −2 y dy + i dy = −1 + i.
C2 0 0

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


0
(2z − 1) dz = (1 + i) (2x − 1 + 2ix) dx = 1 − i.
C3 1

Thus ˇ (2z − 1) dz = 0 − 1 + i + 1 − i = 0.
C

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

1
4 5
(z 2 − z + 2) dz = (1 − i) [x2 − (1 − x)2 − x + 2 + (3x − 2x2 − 1)i] dx = − i.
C 0 3 3

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

π/2
π/2
−2it −it −it
(z − z + 2) dz =
2
(−e − ie + 2)e dt = (−e−3it − ie−2it + 2e−it ) dt
C 0 0
1 1 1 1 4 5
= − ie−3πi/2 + e−πi + 2ie−πi/2 + i − − 2i = − i.
3 2 3 2 3 3
√
27. The length of the line segment from z = 0 to z = 1 + i is 2 . In addition, on this line segment

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


  √

Thus  (z + 4) dz  ≤ 6 2 .
2
C
30. With zk∗ = zk ,


n
z dz = lim zk (zk − zk−1 )
C P →0
k=1

= lim [(z12 − z1 z0 ) + (z22 − z2 z1 ) + · · · + (zn2 − zn zn−1 )]. (1)

P →0

With zk∗ = zk−1 ,


n
z dz = lim zk−1 (zk − zk−1 )
C P →0
k=1

= lim [(z0 z1 − z02 ) + (z1 z2 − z12 ) + · · · + (zn−1 zn − zn−1

2
)]. (2)
P →0

286
 18.2 Cauchy-Goursat Theorem

Adding (1) and (2) gives

1 2
2 z dz = lim (zn2 − z02 ) or z dz = (z − z02 ).
C P →0 C 2 n

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

2π
2π
−it it
ˇ f (z) dz = (e )(ie dt) = 2i dt = 4πi.
C 0 0
   
Thus circulation = Re ˇ f (z) dz = 0, and net flux = Im ˇ f (z) dz = 4π.
C C

EXERCISES 18.2
Cauchy-Goursat Theorem

z
3. f (z) = is discontinuous at z = − 32 but is analytic within and on the circle |z| = 1.
2z + 3
ez
6. f (z) = is discontinuous at z = − 52 and at z = −3 but is analytic within and on the circle
2z 2 + 11z + 15
|z| = 1.

9. By the principle of deformation of contours we can choose the more convenient circular contour C1 defined by
|z| = 1. Thus
 
1 1
ˇ z dz = ˇ z dz = 2πi
C C1

by (4) of Section 18.2.

12. By Theorem 18.4 and (4) of Section 18.2,
    
1 1 1
ˇ z + dz = ˇ z dz + ˇ z 2 dz = 0 + 0 = 0.
C z2 C C
  
2z + 1 1 1
15. By partial fractions, ˇ dz = ˇ dz + ˇ dz.
C z(z + 1) C z C z+1

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

 
1 1
ˇ z dz + ˇ z + 1 dz = 2πi + 0 = 2πi.
C C
  
(b) By writing ˇ = ˇ + ˇ where C1 and C2 are the circles |z| = 1/2 and |z + 1| = 1/2, respectively,
C C1 C2

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

     
1 1 1 1 1 1
ˇ dz + ˇ dz = ˇ dz + ˇ dz + ˇ dz + ˇ dz
C z C z + 1 C1 z C1 z + 1 C2 z C2 z + 1
= 2πi + 0 + 0 + 2πi = 4πi.

287
18.2 Cauchy-Goursat Theorem

2z + 1
(c) Since f (z) = is analytic within and on C it follows from Theorem 18.4 that
z(z + 1)

2z + 1
ˇ z 2 + z dz = 0.
C
  
18. (a) By writing ˇ = ˇ + ˇ where C1 and C2 are the circles |z + 2| = 1 and |z − 2i| = 1, respectively, we
C C1 C2

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

      
3 1 3 1 3 1
ˇ − dz = ˇ dz − ˇ dz + ˇ dz − ˇ dz
C z + 2 z − 2i C1 z+2 C1 z − 2i C2 z + 2 C2 z − 2i
= 3(2πi) − 0 + 0 − 2πi = 4πi.
  
8z − 3 8z − 3 8z − 3
21. We have ˇ dz = ˇ dz − ˇ dz
C z −z
2
C1 z − z
2
C2 z − z
2

where C1 and C2 are the closed portions of the curve C enclosing z = 0 and z = 1, respectively. By partial
fractions, Theorem 18.4, and (4) of Section 18.2,
  
8z − 3 1 1
ˇ z2 − z dz = 5 ˇ dz + 3 ˇ dz = 5(0) + 3(2πi) = 6πi
C1 C1 z − 1 C1 z
  
8z − 3 1 1
ˇ z2 − z dz = 5 ˇ z−1 dz + 3 ˇ z dz = 5(2πi) + 3(0) = 10πi.
C1 C2 C2


8z − 3
Thus ˇ dz = 6πi − 10πi = −4πi.
C z −z
2

  
2 2
24. Write ˇ (z + z + Re(z)) dz = ˇ (z + z) dz + ˇ Re(z) dz.
C C C

By Theorem 18.4, ˇ (z 2 + z) dz = 0. However, since Re(z) = x is not analytic,
C
   
ˇ x dz = ˇC x dz + ˇC x dz + ˇC x dz
C 1 2 3

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
0
1 1
ˇ x dz = x dx + i dy + (1 + 2i) x dx = + 2i − (1 + 2i) = i.
C 0 0 1 2 2

EXERCISES 18.3
Independence of Path

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

2−i 2−i

2z dz = 2z dz = z 2  = 48 + 24i.
C −2+7i −2+7i

288
 18.4 Cauchy’s Integral Formulas

1 1

6. (3z 2 − 4z + 5i) dz = z 3 − 2z 2 + 5iz  = −19 − 3i
−2i −2i

1−i 1−i
1 
3  7 22
9. 2
(2z + 1) dz = (2z + 1)  =− − i
−i/2 6 −i/2 6 3

1+2i 1+2i
1 z2  1 1 1
12. z2
ze dz = e  = [e−3+4i − e−2i ] = (e−3 cos 4 − cos 2) + (e−3 sin 4 + sin 2)i = 0.1918 + 0.4358i
1−i 2 1−i 2 2 2

2πi 2πi

15. cosh z dz = sinh z  = sinh 2πi − sinh πi = i sin 2π − i sin π = 0
πi πi

4+4i 
4+4i
1  √ π √ π 
18. dz = Lnz  = Ln(4 + 4i) − Ln(1 + i) = loge 4 2 + i − loge 2 + i = loge 4 = 1.3863
1+i z 1+i 4 4
21. Integration by parts gives

1
ez cos z dz = ez (cos z + sin z) + C
2
and so

i i 1
1 
ez cos z dz = ez (cos z + sin z)  = [ei (cos i + sin i) − eπ (cos π + sin π)]
π 2 π 2
1
= [(cos 1 cosh 1 − sin 1 sinh 1 + eπ ) + i(cos 1 sinh 1 + sin 1 cosh 1) = 11.4928 + 0.9667i.
2
24. Integration by parts gives

z 2 ez dz = z 2 ez − 2zez + 2ez + C

and so
πi πi

z 2 ez dz = ez (z 2 − 2z + 2)  = eπi (−π 2 − 2πi + 2) − 2 = π 2 − 4 + 2πi.
0 0

EXERCISES 18.4
Cauchy’s Integral Formulas

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

ez
ˇ dz = 2πieπi = −2πi.
C z − πi
1
6. By Theorem 18.9 with f (z) = cos z,
3
 1  
cos z 1 π π
3
ˇ π dz = 2πi 3 cos 3 = 3 i.
C z−
3
z2 + 4
9. By Theorem 18.9 with f (z) = ,
z−i
z2 + 4
  
z−i 12
ˇ dz = 2πi − = −8π.
C z − 4i 3i

289
18.4 Cauchy’s Integral Formulas

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

z 2πi
ˇC (z − (−i))4 dz = 3! (0) = 0.

2z + 5
15. (a) By Theorem 18.9 with f (z) = ,
z−2
2z + 5
  
z−2 5
ˇ dz = 2πi − = −5πi.
C z 2
(b) Since the circle |z − (−1)| = 2 encloses only z = 0, the value of the integral is the same as in part (a).
2z + 5
(c) From Theorem 18.9 with f (z) = ,
z
2z + 5
  
z 9
ˇ z−2 dz = 2πi = 9πi.
C 2
(d) Since the circle |z − (−2i)| = 1 encloses neither z = 0 nor z = 2 it follows from the Cauchy-Goursat
Theorem, Theorem 18.4, that 
2z + 5
ˇ z(z − 2) dz = 0.
C

1 1 2
18. (a) By Theorem 18.10 with f (z) = , f  (z) = − , and f  (z) = ,
z−4 (z − 4)2 (z − 4)3
1
  
z−4 2πi 2 π
ˇ dz = =− i.
C z3 2! −64 32
(b) By the Cauchy-Goursat Theorem, Theorem 18.4,

1
ˇ z 3 (z − 4) dz = 0.
C

1 1
  
1 (z − 1)2 z3
21. We have ˇ dz = ˇ dz + ˇ dz
C z 3 (z − 1)2 C1 z3 C2 (z − 1)2
where C1 and C2 are the circles |z| = 1/3 and |z − 1| = 1/3, respectively. By Theorem 18.10,
1 1
 
(z − 1)2 2πi z3 2πi
ˇ z3
dz =
2!
(6) = 6πi, ˇ (z − 1)2 dz = 1! (−3) = −6πi.
C1 C2

1
Thus ˇ z 3 (z − 1)2 dz = 6πi − 6πi = 0.
C

eiz eiz
  
eiz (z + i)2 (z − i)2
24. We have ˇ dz = ˇ dz − ˇ dz
C (z 2 + 1)2 C1 (z − i)2 C2 (z − (−i))2

where C1 and C2 are the closed portions of the curve C enclosing z = i and z = −i, respectively. By
Theorem 18.10,
eiz eiz
     
(z + i)2 2πi −4e −1
−1
(z − i)2 2πi 0
ˇ dz = = πe , ˇC dz = = 0.
C1 (z − i)2 1! −8i 2 (z − (−i))2 1! 8i

290
 CHAPTER 18 REVIEW EXERCISES


eiz
Thus ˇ dz = πe−1 .
C (z 2 + 1)2

CHAPTER 18 REVIEW EXERCISES

3. True 6. π(−16 + 8i)

9. True (Use partial fractions and write the given integral as two integrals.)
12. 12π


2
2
136 88
15. |z 2 | dz = (t4 + t2 ) dt + 2i (t5 + t3 ) dt = + i
C 0 0 15 3

1−i 1−i

18. (4z − 6) dz = 2z 2 − 6z  = 12 + 20i
3i 3i

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


2π 2π
1 1 
−2 −1 2
(e−2it + e−it + eit + e2it )eit dt = −e−it + it + e2it + e3it  = 2πi.
ˇ (z + z + z + z ) dz = i 2 3 
C 0 0

cos z sin z − cos z − z sin z

24. By Theorem 18.10 with f (z) = and f  (z) = ,
z−1 (z − 1)2
cos z
  
z−1 2πi −1
ˇ dz = = −2πi.
C z2 1! 1

27. Using the principle of deformation of contours we choose C to be the more convenient circular contour |z+i| = 1
4 .
On this circle z = −i + 1 it
4e and dz = 1 it
4 ie
dt. Thus

2π  
z 1 it
ˇ z+i dz = i e − i dt = 2π.
C 0 4


 
30. We have  Ln(z + 1) dz  ≤ |max of Ln(z + 1) on C| · 2,
 
C

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,

  
 
 Ln(z + 1) dz  ≤ 1 loge 10 + π 2 = loge 10 + π .
  2 4 2
C

291