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1818 Integration in the Complex Plane
EXERCISES 18.1
Contour 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
(2z¯ − z) dz =
∫ 2
0
[−t− 3(t2 + 2)i](−1 + 2ti) dt =
∫ 2
0
(6t3 + 13t) dt + i
∫ 2
0
(t2 + 2) dt = 50 +
20
3
i
3.
∫
C
z2 dz = (3 + 2i)3
∫ 2
−2
t2 dt =
16
3
(3 + 2i)3 = −48 + 736
3
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
z
dz = −
∫ π/2
−π/2
(1 + eit) dt = (2 + π)i.
6.
∫
C
|z|2 dz =
∫ 2
1
(
2t5 +
2
t
)
dt− i
∫ 2
1
(
t2 +
1
t4
)
dt = 21 + ln 4− 21
8
i
7. Using z = eit = cos t + i sin t, dz = (− sin t + i cos t) dt and x = cos t,∮ˇ
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
= −1
2
∫ 2π
0
sin 2t dt +
1
2
i
∫ 2π
0
(1 + cos 2t) dt = πi.
8. Using z + i = eit, 0 ≤ t ≤ 2π, and dz = ieit dt,∮ˇ
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 = − 7
12
+
1
12
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 = 1
8
i
∫ 2π
π
(e3it + 3eit + 3e−it + e−3it)eit dt +
1
8
i
∫ 2π
π
(e3it − 3eit + 3e−it − e−3it)eit dt
=
1
8
i
∫ 2π
π
(2e4it + 6) dt =
3π
4
i.
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 = 1
2
∫ π/2
0
[eit − e−it − 2i]eit dt = 1
2
∫ π/2
0
[e2it − 1 + 2ieit] dt = 1− π
4
− 1
2
i.
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 =
1
2
+
1
2
i.
Thus
∫
C
Im(z − i) dz =
(
1− π
4
− 1
2
i
)
+
(
1
2
+
1
2
i
)
=
3
2
− 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
∮ˇ
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
∣∣∣1
0
= 1.
878
18.1 Contour Integrals
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
∮ˇ
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
∮ˇ
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 =
1
2
.
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 = −1
2
− 1
2
i.
Thus
∮ˇ
C
x dz =
1
2
+ i− 1
2
− 1
2
i =
1
2
i.
18. We have
∮ˇ
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
18.1 Contour Integrals
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
∮ˇ
C
(2z − 1) dz = 0− 1 + i + 1− i = 0.
19. We have
∮ˇ
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 =
1
3
.
On C2, x = 1, 0 ≤ y ≤ 1, z = 1 + iy, dz = i dy,∫
C2
z2 dz =
∫ 1
0
(1 + iy)2i dy = −1 + 2
3
i.
On C3, y = x, 0 ≤ x ≤ 1, z = x + ix, dz = (1 + i) dx,∫
C3
z2 dz = (1 + i)3
∫ 0
1
x2 dx =
2
3
− 2
3
i.
Thus
∮ˇ
C
z2 dz =
1
3
− 1 + 2
3
i +
2
3
− 2
3
i = 0.
20. We have
∮ˇ
C
z¯2 dz =
∫
C1
z¯2 dz +
∫
C2
z¯2 dz +
∫
C3
z¯2 dz
On C1, y = 0, 0 ≤ x ≤ 1, z = x, dz = dx, ∫
z¯2 dz =
∫ 1
0
x2 dx =
1
3
.
On C2, x = 1, 0 ≤ y ≤ 1, z = 1 + iy, dz = i dy,∫
C2
z¯2 dz = −
∫ 1
0
(1− iy)2(−i dy) = 1 + 2
3
i.
On C3, y = x, 0 ≤ x ≤ 1, z = x + ix, dz = (1 + i) dx,∫
C3
z¯2 dz = (1− i)2(1 + i)
∫ 0
1
x2 dx = −2
3
+
2
3
i.
Thus
∮ˇ
C
z¯2 dz =
1
3
+ 1 +
2
3
i− 2
3
+
2
3
i =
2
3
+
4
3
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 = 4
3
− 5
3
i.
22. We have
∫
C
(z2 − z + 2) dz =
∫
C1
(z2 − z + 2) dz +
∫
C2
(z2 − z + 2) dz
880
18.1 Contour Integrals
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 = 5
6
.
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 = 1
2
− 5
3
i.
Thus
∫
C
(z2 − z + 2) dz = 1
2
− 5
3
i +
5
6
=
4
3
− 5
3
i.
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 = 4
3
− 5
3
i.
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
= −1
3
ie−3πi/2 +
1
2
e−πi + 2ie−πi/2 +
1
3
i− 1
2
− 2i = 4
3
− 5
3
i.
25. On C,
∣∣∣∣ ezz2 + 1
∣∣∣∣ ≤ |ez||z|2 − 1 = e524 . Thus
∣∣∣∣∮ˇ
C
ez
z2 + 1
dz
∣∣∣∣ ≤ e524 · 10π = 5π12 e5.
26. On C,
∣∣∣∣ 1z2 − 2i
∣∣∣∣ ≤ 1|z|2 − |2i| = 134 . Thus
∣∣∣∣∫
C
1
z2 − 2i dz
∣∣∣∣ ≤ 134 · 12(12π) = 3π17 .
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
1
z3
dz
∣∣∣∣ ≤ 164 · 14(8π) = π32 .
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) + (z22 − z2z1) + · · ·+ (z2n − znzn−1)]. (1)
881
18.1 Contour Integrals
With z∗k = zk−1, ∫
C
z dz = lim
‖P‖→0
n∑
k=1
zk−1(zk − zk−1)
= lim
‖P‖→0
[(z0z1 − z20) + (z1z2 − z21) + · · ·+ (zn−1zn − z2n−1)]. (2)
Adding (1) and (2) gives
2
∫
C
z dz = lim
‖P‖→0
(z2n − z20) or
∫
C
z dz =
1
2
(z2n − z20).
31. (a)
∫
C
(6z + 4) dz = 6
∫
C
z dz + 4
∫
C
dz =
6
2
[(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 − z20 ] + 4[z0 − z0] = 0.
32. For f(z) = 1/z, f(z) = 1/z¯, so on z = 2eit, z¯ = 2e−it, dz = 2ieit dt, and∮ˇ
C
f(z) dz =
∫ 2π
0
1
2e−it
· 2ieit dt = 1
2
e2it
∣∣∣∣2π
0
=
1
2
[e4πi − 1] = 0.
Thus circulation = Re
(∮ˇ
C
f(z) dz
)
= 0, and net flux = Im
(∮ˇ
C
f(z) dz
)
= 0.
33. For f(z) = 2z, f(z) = 2z¯, so on z = eit, z¯ = e−it, dz = ieit dt, and∮ˇ
C
f(z) dz =
∫ 2π
0
(e−it)(ieit dt) = 2i
∫ 2π
0
dt = 4πi.
Thus circulation = Re
(∮ˇ
C
f(z) dz
)
= 0, and net flux = Im
(∮ˇ
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∮ˇ
C
f(z) dz =
∫ 2π
0
1
2eit
· 2ieit dt = i
∫ 2π
0
dt = 2πi.
Thus circulation = Re
(∮ˇ
C
f(z) dz
)
= 0, and net flux = Im
(∮ˇ
C
f(z) dz
)
= 2π.
35. For f(z) = z¯, f(z) = z so on the square we have∮ˇ
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 =
1
2∫
C2
z dz = i
∫ 1
0
(1 + iy) dy = −1
2
+ i
∫
C3
z dz =
∫ 0
1
(x + i) dx = −1
2
− i
∫
C4
z dz = −
∫ 0
1
y dy =
1
2
882
18.2 Cauchy-Goursat Theorem
and so ∮ˇ
C
f(z) dz =
1
2
+
(
−1
2
+ i
)
+
(
−1
2
− i
)
+
1
2
= 0
circulation = Re
(∮ˇ
C
f(z) dz
)
= Re(0) = 0
net flux = Im
(∮ˇ
C
f(z) dz
)
= Im(0) = 0.
EXERCISES 18.2
Cauchy-Goursat Theorem
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 the circle |z| = 1.
3. f(z) =
z
2z + 3
is discontinuous at z = −3/2 but is analytic within and on the circle |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 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 + 15
is 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 − 9
cosh z
is discontinuous at
π
2
i, ±3π
2
i, . . . 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 ∮ˇ
C
1
z
dz =
∮ˇ
C1
1
z
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)| = 1
16
. Thus ∮ˇ
C
5
z + 1 + i
dz = 5
∮ˇ
C1
1
z − (−1− i) dz = 5(2πi) = 10πi
by (4) of Section 18.2.
883
18.2 Cauchy-Goursat Theorem
11. By Theorem 18.4 and (4) of Section 18.2,∮ˇ
C
(
z +
1
z
)
dz =
∮ˇ
C
z dz +
∮ˇ
C
1
z
dz = 0 + 2πi = 2πi.
12. By Theorem 18.4 and (4) of Section 18.2,∮ˇ
C
(
z +
1
z2
)
dz =
∮ˇ
C
1
z
dz +
∮ˇ
C
1
z2
dz = 0 + 0 = 0.
13. Since f(z) =
z
z2 − π2 is analytic within and on C it follows from Theorem 18.4 that
∮ˇ
C
z
z2 − π2 dz = 0.
14. By (4) of Section 18.2,
∮ˇ
C
10
(z + i)4
dz = 0.
15. By partial fractions,
∮ˇ
C
2z + 1
z(z + 1)
dz =
∮ˇ
C
1
z
dz +
∮ˇ
C
1
z + 1
dz.
(a) By Theorem 18.4 and (4) of Section 18.2,∮ˇ
C
1
z
dz +
∮ˇ
C
1
z + 1
dz = 2πi + 0 = 2πi.
(b) By writing
∮ˇ
C
=
∮ˇ
C1
+
∮ˇ
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,∮ˇ
C
1
z
dz +
∮ˇ
C
1
z + 1
dz =
∮ˇ
C1
1
z
dz +
∮ˇ
C1
1
z + 1
dz +
∮ˇ
C2
1
z
dz +
∮ˇ
C2
1
z + 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∮ˇ
C
2z + 1
z2 + z
dz = 0.
16. By partial fractions,
∮ˇ
C
2z
z2 + 3
dz =
∮ˇ
C
1
z +
√
3 i
dz +
∮ˇ
C
1
z −√3 i dz.
(a) By Theorem 18.4, ∮ˇ
C
1
z +
√
3 i
dz +
∮ˇ
C
1
z −√3 i dz = 0 + 0 = 0.
(b) By Theorem 18.4 and (4) of Section 18.2,∮ˇ
C
1
z +
√
3 i
dz +
∮ˇ
C
1
z −√3 i dz = 0 + 2πi = 2πi.
(c) By writing
∮ˇ
C
=
∮ˇ
C1
+
∮ˇ
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,∮ˇ
C
1
z +
√
3 i
dz +
∮ˇ
C
1
z −√3 i dz =
∮ˇ
C1
1
z +
√
3 i
dz +
∮ˇ
C1
1
z −√3 i dz +
∮ˇ
C2
1
z +
√
3 i
dz +
∮ˇ
C2
1
z −√3 i dz
= 2πi + 0 + 0 + 2πi = 4πi.
884
18.2 Cauchy-Goursat Theorem
17. By partial fractions,
∮ˇ
C
−3z + 2
z2 − 8z + 12 dz =
∮ˇ
C
1
z − 2 dz − 4
∮ˇ
C
1
z − 6 dz.
(a) By Theorem 18.4 and (4) of Section 18.2,∮ˇ
C
1
z − 2 dz − 4
∮ˇ
C
1
z − 6 dz = 0− 4(2πi) = −8πi.
(b) By writing
∮ˇ
C
=
∮ˇ
C1
+
∮ˇ
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,∮ˇ
C
1
z − 2 dz − 4
∮ˇ
C
1
z − 6 dz =
∮ˇ
C1
1
z − 2 dz − 4
∮ˇ
C1
1
z − 6 dz +
∮ˇ
C2
1
z − 2 dz − 4
∮ˇ
C2
1
z − 6 dz
= 2πi− 4(0) + 0− 4(2πi) = −6πi.
18. (a) By writing
∮ˇ
C
=
∮ˇ
C1
+
∮ˇ
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,∮ˇ
C
(
3
z + 2
− 1
z − 2i
)
dz =
∮ˇ
C1
3
z + 2
dz −
∮ˇ
C1
1
z − 2i dz +
∮ˇ
C2
3
z + 2
dz −
∮ˇ
C2
1
z − 2i dz
= 3(2πi)− 0 + 0− 2πi = 4πi.
19. By partial fractions,∮ˇ
C
z − 1
z(z − i)(z − 3i) dz =
1
3
∮ˇ
C
1
z
dz +
(
−1
2
+
1
2
i
) ∮ˇ
C
1
z − i dz +
(
1
6
− 1
2
i
) ∮ˇ
C
1
z − 3i dz.
By Theorem 18.4 and (4) of Section 18.2,∮ˇ
C
z − 1
z(z − i)(z − 3i) dz = 0 +
(
−1
2
+
1
2
i
)
2πi + 0 = π(−1− i).
20. By partial fractions, ∮ˇ
C
1
z3 + 2iz2
dz =
1
4
∮ˇ
C
1
z
dz − 1
2
i
∮ˇ
C
1
z2
dz − 1
4
∮ˇ
C
1
z + 2i
dz.
By Theorem 18.4 and (4) of Section 18.2,∮ˇ
C
1
z3 + 2iz2
dz =
1
4
2πi− 1
2
i(0)− 1
4
(0) =
π
2
i.
21. We have
∮ˇ
C
8z − 3
z2 − z dz =
∮ˇ
C1
8z − 3
z2 − z dz −
∮ˇ
C2
8z − 3
z2 − z dz
where C1 and C2 are the closedportions of the curve C enclosing z = 0 and z = 1, respectively. By partial
fractions, Theorem 18.4, and (4) of Section 18.2,∮ˇ
C1
8z − 3
z2 − z dz = 5
∮ˇ
C1
1
z − 1 dz + 3
∮ˇ
C1
1
z
dz = 5(0) + 3(2πi) = 6πi∮ˇ
C1
8z − 3
z2 − z dz = 5
∮ˇ
C2
1
z − 1 dz + 3
∮ˇ
C2
1
z
dz = 5(2πi) + 3(0) = 10πi.
Thus
∮ˇ
C
8z − 3
z2 − z dz = 6πi− 10πi = −4πi.
885
18.2 Cauchy-Goursat Theorem
22. By choosing the more convenient contour C1 defined by |z − z0| = r where r is small enough so that the circle
C1 lies entirely within C we can write∮ˇ
C
1
(z − z0)n dz =
∮ˇ
C1
1
(z − z0)n dz.
Let z − z0 = reit, 0 ≤ t ≤ 2π and dz = ireit dt. Then for n = 1:∮ˇ
C1
1
z − z0 dz =
∫ 2π
0
1
reit
ireit dt = i
∫ 2π
0
dt = 2πi.
For n �= 1:∮ˇ
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) [e
2π(1−n)i − 1] = 0
since e2π(1−n)i = 1.
23. Write
∮ˇ
C
(
ez
z + 3
− 3z¯
)
dz =
∮ˇ
C
ez
z + 3
dz − 3
∮ˇ
C
z¯ dz.
By Theorem 18.4,
∮ˇ
C
ez
z + 3
dz = 0. However, since z¯ is not analytic,
∮ˇ
C
z¯ dz =
∫ 2π
0
e−it(ieit dt) = 2πi.
Thus
∮ˇ
C
(
ez
z + 3
− 3z¯
)
dz = 0− 3(2πi) = −6πi.
24. Write
∮ˇ
C
(z2 + z + Re(z)) dz =
∮ˇ
C
(z2 + z) dz +
∮ˇ
C
Re(z) dz.
By Theorem 18.4,
∮ˇ
C
(z2 + z) dz = 0. However, since Re(z) = x is not analytic,∮ˇ
C
x dz =
∮ˇ
C1
x dz +
∮ˇ
C2
x dz +
∮ˇ
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,∮ˇ
C
x dz =
∫ 1
0
x dx + i
∫ 2
0
dy + (1 + 2i)
∫ 0
1
x dx =
1
2
+ 2i− 1
2
(1 + 2i) = i.
EXERCISES 18.3
Independence 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 + 13xi, dz = (1 + 13 i) dx. Thus∫
C
ez dz =
∫ 3
0
e(1+
1
3 i)x
(
1 +
1
3
i
)
dx = e(1+
1
3 i)x
∣∣∣3
0
= 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 =
1
3
z3
∣∣∣∣3+i
0
= 6 +
26
3
i
6.
∫ 1
−2i
(3z2 − 4z + 5i) dz = z3 − 2z2 + 5iz
∣∣∣1
−2i
= −19− 3i
7.
∫ 1+i
1−i
z3 dz =
1
4
z4
∣∣∣∣1+i
1−i
= 0
8.
∫ 2i
−3i
(z3 − z) dz = 1
4
z4 − 1
2
z2
∣∣∣∣2i
−3i
=
123
4
9.
∫ 1−i
−i/2
(2z + 1)2 dz =
1
6
(2z + 1)3
∣∣∣∣1−i
−i/2
= −7
6
− 22
3
i
10.
∫ i
1
(iz + 1)3 dz =
1
4i
(iz + 1)4
∣∣∣∣i
1
= −i
11.
∫ i
i/2
eπz dz =
1
π
eπz
∣∣∣i
i/2
= − 1
π
− 1
π
i
12.
∫ 1+2i
1−i
zez
2
dz =
1
2
ez
2
∣∣∣∣1+2i
1−i
=
1
2
[e−3+4i − e−2i] = 1
2
(e−3 cos 4− cos 2) + 1
2
(e−3 sin 4 + sin 2)i = 0.1918 + 0.4358i
13.
∫ π+2i
π
sin
z
2
dz = −2 cos z
2
∣∣∣∣π+2i
π
= −2
[
cos
(π
2
+ i
)
− cos π
2
]
= 2i sin
π
2
sinh 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+π2 i
i
sinh 3z dz =
1
3
cosh 3z
∣∣∣∣1+π2 i
i
=
1
3
[
cosh
(
3 +
3π
2
i
)
− cosh 3i
]
=
1
3
[
cosh 3 cos
3π
2
+ i sinh 3 sin
3π
2
− cos 3
]
= −1
3
cos 3− 1
3
i sinh 3 = 0.3300− 3.3393i
887
18.3 Independence of Path
17.
∫ 4i
−4i
1
z
dz = Lnz
∣∣∣4i
−4i
= Ln4i− Ln(−4i) = loge 4 +
π
2
i−
(
loge 4−
π
2
i
)
= πi
18.
∫ 4+4i
1+i
1
z
dz = Lnz
∣∣∣4+4i
1+i
= Ln(4 + 4i)− Ln(1 + i) = loge 4
√
2 +
π
4
i−
(
loge
√
2 +
π
4
i
)
= loge 4 = 1.3863
19.
∫ 4i
−4i
1
z2
dz = −1
z
∣∣∣∣4i
−4i
= −
[
1
4i
−
(
1
−4i
)]
=
1
2
i
20.
∫ 1+√3 i
1−i
(
1
z
+
1
z2
)
dz = Lnz − 1
z
∣∣∣∣1+
√
3 i
1−i
= loge 2 +
π
3
i− 1
1 +
√
3 i
−
(
loge
√
2− π
4
i− 1
1− i
)
= loge
√
2 +
1
4
+ i
(
7π
12
+
√
3
4
+
1
2
)
= 0.5966 + 2.7656i
21. Integration by parts gives ∫
ez cos z dz =
1
2
ez(cos z + sin z) + C
and so∫ i
π
ez cos z dz =
1
2
ez(cos z + sin z)
∣∣∣i
π
=
1
2
[ei(cos i + sin i)− eπ(cosπ + sinπ)]
=
1
2
[(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
∣∣∣i
0
= −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 18.4
Cauchy’s Integral Formulas
1. By Theorem 18.9, with f(z) = 4, ∮ˇ
C
4
z − 3i dz = 2πi · 4 = 8πi.
2. By Theorem 18.10 with f(z) = z2 and f ′(z) = 2z,∮ˇ
C
z2
(z − 3i)2 dz =
2πi
1!
2(3i) = −12π.
3. By Theorem 18.9 with f(z) = ez, ∮ˇ
C
ez
z − πi dz = 2πie
πi = −2πi.
4. By Theorem 18.9 with f(z) = 1 + 2ez,∮ˇ
C
1 + 2ez
z
dz = 2πi(1 + 2e0) = 6πi.
5. By Theorem 18.9 with f(z) = z2 − 3z + 4i,∮ˇ
C
z2 − 3z + 4i
z − (−2i) dz = 2πi(−4 + 6i + 4i) = −π(20 + 8i).
6. By Theorem 18.9 with f(z) =
1
3
cos z,
∮ˇ
C
1
3
cos z
z − π
3
dz = 2πi
(
1
3
cos
π
3
)
=
π
3
i.
7. (a) By Theorem 18.9 with f(z) =
z2
z + 2i
,
∮ˇ
C
z2
z + 2i
z − 2i dz = 2πi
(
− 4
4i
)
= −2π.
(b) By Theorem 18.9 with f(z) =
z2
z − 2i ,
∮ˇ
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
,
∮ˇ
C
z2 + 3z + 2i
z + 4
z − 1 dz = 2πi
(
4 + 2i
5
)
= π
(
−4
5
+
8
5
i
)
.
889
18.4 Cauchy’s Integral Formulas
(b) By Theorem 18.9 with f(z) =
z2 + 3z + 2i
z − 1 ,
∮ˇ
C
z2 + 3z + 2i
z − 1
z − (−4) dz = 2πi
(
4 + 2i
−5
)
= π
(
4
5
− 8
5
i
)
.
9. By Theorem 18.9 with f(z) =
z2 + 4
z − i ,
∮ˇ
C
z2 + 4
z − i
z − 4i dz = 2πi
(
−12
3i
)
= −8π.
10. By Theorem 18.9 with f(z) =
sin z
z + πi
,
∮ˇ
C
sin z
z + πi
z − πi dz = 2πi
(
sinπi
2πi
)
= i sinhπ.
11. By Theorem 18.10 with f(z) = ez
2
, f ′(z) = 2zez
2
, and f ′′(z) = 4z2ez
2
+ 2ez
2
,∮ˇ
C
ez
2
(z − i)3 dz =
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,∮ˇ
C
z
(z − (−i))4 dz =
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,∮ˇ
C
cos 2z
z5
dz =
2πi
4!
(16 cos 0) =
4π
3
i.
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,∮ˇ
C
e−z sin z
z3
dz =
2πi
2!
(−2e0 cos 0) = −2πi.
15. (a) By Theorem 18.9 withf(z) =
2z + 5
z − 2 ,
∮ˇ
C
2z + 5
z − 2
z
dz = 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
,
∮ˇ
C
2z + 5
z
z − 2 dz = 2πi
(
9
2
)
= 9πi.
(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 ∮ˇ
C
2z + 5
z(z − 2) dz = 0.
890
18.4 Cauchy’s Integral Formulas
16. By partial fractions, ∮ˇ
C
z
(z − 1)(z − 2) dz = 2
∮ˇ
C
dz
z − 2 −
∮ˇ
C
dz
z − 1 .
(a) By the Cauchy-Goursat Theorem, Theorem 18.4,∮ˇ
C
z
(z − 1)(z − 2) dz = 0.
(b) As in part (a), the integral is 0.
(c) By Theorem 18.4,
∮ˇ
C
dz
z − 2 = 0 whereas by Theorem 18.9,
∮ˇ
C
dz
z − 1 = 2πi. Thus∮ˇ
C
z
(z − 1)(z − 2) dz = −2πi.
(d) By Theorem 18.9,
∮ˇ
C
dz
z − 1 = 2πi and
∮ˇ
C
dz
z − 2 = 2πi. Thus∮ˇ
C
z
(z − 1)(z − 2) dz = 2(2πi)− 2πi = 3πi.
17. (a) By Theorem 18.10 with f(z) =
z + 2
z − 1− i and f
′(z) =
−3− i
(z − 1− i)2 ,
∮ˇ
C
z + 2
z − 1− i
z2
dz =
2πi
1!
( −3− i
(−1− i)2
)
= −π(3 + i).
(b) By Theorem 18.9 with f(z) =
z + 2
z2
,
∮ˇ
C
z + 2
z2
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 ,
∮ˇ
C
1
z − 4
z3
dz =
2πi
2!
(
2
−64
)
= − π
32
i.
(b) By the Cauchy-Goursat Theorem, Theorem 18.4,∮ˇ
C
1
z3(z − 4) dz = 0.
19. By writing
∮ˇ
C
(
e2iz
z4
− z
4
(z − i)3
)
dz =
∮ˇ
C
e2iz
z4
dz −
∮ˇ
C
z4
(z − i)3 dz
we can apply Theorem 18.10 to each integral:∮ˇ
C
e2iz
z4
dz =
2πi
3!
(−8i) = 8π
3
,
∮ˇ
C
z4
(z − i)3 dz =
2πi
2!
(−12) = −12πi.
Thus
∮ˇ
C
(
e2iz
z4
− z
4
(z − i)3
)
dz = π
(
8
3
+ 12i
)
.
20. By writing
∮ˇ
C
(
cosh z
(z − π)3 −
sin2 z
(2z − π)3
)
dz =
∮ˇ
C
cosh z
(z − π)3 dz −
∮ˇ
C
1
8 sin
2 z
(z − π2 )3
dz
891
18.4 Cauchy’s Integral Formulas
we apply Theorem 18.4 to the first integral and Theorem 18.10 to the second:∮ˇ
C
cosh z
(z − π)3 dz = 0,
∮ˇ
C
1
8 sin
2 z
(z − π2 )3
dz =
2πi
2!
(
−1
4
sin2
π
2
)
= −π
4
i.
Thus
∮ˇ
C
(
cosh z
(z − π)3 −
sin2 z
(2z − π)3
)
dz =
π
4
, i.
21. We have
∮ˇ
C
1
z3(z − 1)2 dz =
∮ˇ
C1
1
(z − 1)2
z3
dz +
∮ˇ
C2
1
z3
(z − 1)2 dz
where C1 and C2 are the circles |z| = 1/3 and |z − 1| = 1/3, respectively. By Theorem 18.10,
∮ˇ
C1
1
(z − 1)2
z3
dz =
2πi
2!
(6) = 6πi,
∮ˇ
C2
1
z3
(z − 1)2 dz =
2πi
1!
(−3) = −6πi.
Thus
∮ˇ
C
1
z3(z − 1)2 dz = 6πi− 6πi = 0.
22. We have
∮ˇ
C
1
z2(z2 + 1)
dz =
∮ˇ
C1
1
z2(z + i)
z − i dz +
∮ˇ
C2
1
z2 + 1
z2
dz
where C1 and C2 are the circles |z − i| = 1/3 and |z| = 1/8, respectively. By Theorems 18.9 and 18.10,
∮ˇ
C1
1
z2(z + i)
z − i dz = 2πi
(
1
−2i
)
= −π,
∮ˇ
C2
1
z2 + 1
z2
dz =
2πi
1!
(0) = 0.
Thus
∮ˇ
C
1
z2(z2 + 1)
dz = −π.
23. We have
∮ˇ
C
3z + 1
z(z − 2)2 dz =
∮ˇ
C1
3z + 1
z
(z − 2)2 dz −
∮ˇ
C2
3z + 1
(z − 2)2
z
dz
where C1 and C2 are the closed portions of the curve C enclosing z = 2 and z = 0, respectively. By
Theorems 18.10 and 18.9,
∮ˇ
C1
3z + 1
z
(z − 2)2 dz =
2πi
1!
(
−1
4
)
= −π
2
i,
∮ˇ
C2
3z + 1
(z − 2)2
z
dz = 2πi
(
1
4
)
=
π
2
i.
Thus
∮ˇ
C
3z + 1
z(z − 2)2 dz = −
π
2
i− π
2
i = −πi.
24. We have
∮ˇ
C
eiz
(z2 + 1)2
dz =
∮ˇ
C1
eiz
(z + i)2
(z − i)2 dz −
∮ˇ
C2
eiz
(z − i)2
(z − (−i))2 dz
892
CHAPTER 18 REVIEW EXERCISES
where C1 and C2 are the closed portions of the curve C enclosing z = i and z = −i, respectively. By
Theorem 18.10,
∮ˇ
C1
eiz
(z + i)2
(z − i)2 dz =
2πi
1!
(−4e−1
−8i
)
= πe−1,
∮ˇ
C2
eiz
(z − i)2
(z − (−i))2 dz =
2πi
1!
(
0
8i
)
= 0.
Thus
∮ˇ
C
eiz
(z2 + 1)2
dz = πe−1.
CHAPTER 18 REVIEW EXERCISES
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
∮ˇ
C
(x + iy) dz =
∫ 3
−4
z dz =
z2
2
∣∣∣∣3
−4
= −7
2
.
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
2
y2
) ∣∣∣∣2
0
= 2 + 8i.
On C2, y = 2, −4 ≤ x ≤ 3, z = x + 2i, dz = dx,∫
C2
(x− 2i) dx =
∫ 3
−4
(x− 2i) dx = 1
2
x2 − 2ix
∣∣∣∣3
−4
= −7
2
− 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
2
y2
) ∣∣∣∣0
2
= −2− 6i.
Thus
∫
C
(x− iy) dz = 2 + 8i− 7
2
− 14i− 2− 6i = −7
2
− 12i.
893
CHAPTER 18 REVIEW EXERCISES
15.
∫
C
|z2| dz =
∫ 2
0
(t4 + t2) dt + 2i
∫ 2
0
(t5 + t3) dt =
136
15
+
88
3
i
16.
∫
C
eπz dz =
1
π
∫ 1+i
i
eπz(π dz) =
1
π
eπz
∣∣∣∣1+i
i
=
1
π
(1− eπ)
17. By the Cauchy-Goursat Theorem, Theorem 18.4,
∮ˇ
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∮ˇ
C
(z−2 + z−1 + z + z2) dz = i
∫ 2π
0
(e−2it + e−it + eit + e2it)eit dt = −e−it + it + 1
2
e2it +
1
3
e3it
∣∣∣∣2π
0
= 2πi.
22. By partial fractions and Theorem 18.9,∮ˇ
C
3z + 4
z2 − 1 dz =
7
2
∮ˇ
C
1
z − 1 dz −
1
2
∮ˇ
C
1
z − (−1) dz =
7
2
(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,∮ˇ
C
e−2z
z4
dz =
2πi
3!
(−8) = −8π
3
i.
24. By Theorem 18.10 with f(z) =
cos z
z − 1 and f
′(z) =
sin z − cos z − z sin z
(z − 1)2 ,
∮ˇ
C
cos z
z − 1
z2
dz =
2πi
1!
(−1
1
)
= −2πi.
25. By Theorem 18.9 with f(z) =
1
2(z + 3)
,
∮ˇ
C
1
2(z + 3)
(z − (−1/2)) dz = 2πi
(
1
5
)
=
2π
5
i.
26. Since the function 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∮ˇ
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 = 14 ieit dt. Thus∮ˇ
C
z
z + i
dz = i
∫ 2π
0
(
1
4
eit − i
)
dt = 2π.
894
CHAPTER 18 REVIEW EXERCISES
28. (a) By Theorem 18.9 with f(z) =
eiπz
2(z − 2) ,
∮ˇ
C
eiπz
2(z − 2)
z − 1/2 dz = 2πi
(
eiπ/2
−3
)
=
2π
3
.
(b) By Theorem 18.9 with f(z) =
eiπz
2z − 1 ,
∮ˇ
C
eiπz
2z − 1
z − 2 dz = 2πi
(
e2πi
3
)
=
2π
3
i.
(c) By the Cauchy-Goursat Theorem, Theorem 18.4,∮ˇ
C
eiπz
2z2 − 5z + 2 dz = 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,∮ˇC
f ′(z)
f(z)
dz =
∮ˇ
C
g′(z)
g(z)
dz + n
∮ˇ
C
1
z
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 + π4
)
2 = loge 10 +
π
2
.
895