Chapt_19

Prévia do material em texto

1919 Series and Residues
EXERCISES 19.1
Sequences and Series
1. 5i, −5, −5i, 5, 5i 2. 2− i, 1, 2 + i, 3, 2− i
3. 0, 2, 0, 2, 0 4. 1 + i, 2i, −2 + 2i, −4, −4− 4i
5. Converges. To see this write the general term as
3i + 2/n
1 + i
.
6. Converges. To see this write the general term as
(
2
5
)n 1 + n2−ni
1 + 3n5−ni
.
7. Converges. To see this write the general term as
(i + 2/n)2
i
.
8. Diverges. To see this consider the term
n
n + 1
in and take n to be an odd positive integer.
9. Diverges. To see this write the general term as
√
n
(
1 +
1√
n
in
)
.
10. Converges. The real part of the general term converges to 0 and the imaginary part of the general term converges
to π.
11. Re(zn) =
8n2 + n
4n2 + 1
→ 2 as n →∞, and Im(zn) = 6n
2 − 4n
4n2 + 1
→ 3
2
as n →∞.
12. Write zn =
(
1
4
+
1
4
i
)n
in polar form as zn =
(√
2
4
)n
cosnθ + i
(√
2
4
)n
sinnθ. Now
Re(zn) =
(√
2
4
)n
cosnθ → 0 as n →∞ and Im(zn) =
(√
2
4
)n
sinnθ → 0 as n →∞
since
√
2/4 < 1.
13. Sn =
1
1 + 2i
− 1
2 + 2i
+
1
2 + 2i
− 1
3 + 2i
+
1
3 + 2i
− 1
4 + 2i
+ · · ·+ 1
n + 2i
− 1
n + 1 + 2i
=
1
1 + 2i
− 1
n + 1 + 2i
Thus, lim
n→∞Sn =
1
1 + 2i
=
1
5
− 2
5
i.
14. By partial fractions,
i
k(k + 1)
=
i
k
− i
k + 1
and so
Sn = i− i2 +
i
2
− i
3
+
i
3
− i
4
+ · · ·+ i
n
− i
n + 1
= i− i
n + 1
.
Thus lim
n→∞Sn = i.
896
19.1 Sequences and Series
15. We identify a = 1 and z = 1− i. Since |z| = √2 > 1 the series is divergent.
16. We identify a = 4i and z = 1/3. Since |z| = 1/3 < 1 the series converges to
4i
1− 1/3 = 6i.
17. We identify a = i/2 and z = i/2. Since |z| = 1/2 < 1 the series converges to
i/2
1− i/2 = −
1
5
+
2
5
i.
18. We identify a = 1/2 and z = i. Since |z| − 1 the series is divergent.
19. We identify a = 3 and z = 2/(1 + 2i). Since |z| = 2/√5 < 1 the series converges to
3
1− 2
1 + 2i
=
9
5
− 12
5
i.
20. We identify a = −1/(1 + i) and z = i/(1 + i). Since |z| = 1/√2 < 1 the series converges to
− 1
1 + i
1− i
1 + i
= −1.
21. From lim
n→∞
∣∣∣∣∣∣∣∣
1
(1− 2i)n+2
1
(1− 2i)n+1
∣∣∣∣∣∣∣∣ =
1
|1− 2i| =
1√
5
we see that the radius of convergence is R =
√
5 . The circle of convergence is |z − 2i| = √5 .
22. From lim
n→∞
∣∣∣∣∣∣∣∣∣
1
n + 1
(
i
1 + i
)n+1
1
n
(
i
1 + i
)n
∣∣∣∣∣∣∣∣∣ = limn→∞
n
n + 1
∣∣∣∣ i1 + i
∣∣∣∣ = 1√2
we see that the radius of convergence is R =
√
2 . The circle of convergence is |z| = √2 .
23. From lim
n→∞
∣∣∣∣∣∣∣∣∣
(−1)n+1
(n + 1)2n+1
(−1)n
n2n
∣∣∣∣∣∣∣∣∣ = limn→∞
n
2(n + 1)
=
1
2
we see that the radius of convergence is R = 2. The circle of convergence is |z − 1− i| = 2.
24. From lim
n→∞
∣∣∣∣∣∣∣∣
1
(n + 1)2(3 + 4i)n+1
1
n2(3 + 4i)n
∣∣∣∣∣∣∣∣ = limn→∞
(
n
n + 1
)2 1
|3 + 4i| =
1
5
we see that the radius of convergence is R = 5. The circle of convergence is |z + 3i| = 5.
25. From lim
n→∞
n
√
|1 + 3i|n = |1 + 3i| =
√
10
we see that the radius of convergence is R = 1/
√
10 . The circle of convergence is |z − i| = 1/√10 .
897
19.1 Sequences and Series
26. From lim
n→∞
n
√∣∣∣∣ 1nn
∣∣∣∣ = limn→∞ 1n = 0
we see that the radius of convergence is ∞. The power series with center 0 converges absolutely for all z.
27. From lim
n→∞
n
√∣∣∣∣ 152n
∣∣∣∣ = limn→∞ 125 = 125
we see that the radius of convergence is R = 25. The circle of convergence is |z − 4− 3i| = 25.
28. From lim
n→∞
n
√∣∣∣∣(−1)n (1 + 2i2
)n∣∣∣∣ = limn→∞
∣∣∣∣1 + 2i2
∣∣∣∣ = √52
we see that the radius of convergence is R = 2/
√
5 . The circle of convergence is |z + 2i| = 2/√5 .
29. The circle of convergence is |z − i| = 2. Since the series of absolute values
∞∑
k=1
∣∣∣∣ (z − 1)kk2k
∣∣∣∣ = ∞∑
k=1
|z − 1|k
k2k
=
∞∑
k=1
2k
k2k
=
∞∑
k=1
1
k
is the divergent harmonic series. But z = −2+ i is on the circle of convergence and (z− i)k = (−2)k. The series
∞∑
k=1
(−2)k
k2k
=
∞∑
k=1
(−1)k
k
is convergent.
30. (a) The circle of convergence is |z| = 1. Since the series of absolute values
∞∑
k=1
∣∣∣∣zkk2
∣∣∣∣ = ∞∑
k=1
|z|k
k2
=
∞∑
k=1
1
k2
converges, the given series is absolutely convergent for every z on |z| = 1. Since absolute convergence
implies convergence, the given series converges for all z on |z| = 1.
(b) The circle of convergence is |z| = 1. On the circle, n|z|n →∞ as n →∞. This implies nzn �→ 0 as n →∞.
Thus by Theorem 19.3 the series is divergent for every z on the circle |z| = 1.
EXERCISES 19.2
Taylor Series
1.
z
1 + z
= z[1− z + z2 − z3 + · · ·] = z − z2 + z3 − z4 + · · · =
∞∑
k=1
(−1)k+1zk; R = 1
2.
1
4− 2z =
1
4
·
[
1 +
z
2
+
z2
22
+
z3
23
+ · · ·
]
=
1
4
∞∑
k=0
zk
2k
; R = 2
3. Differentiating
1
1 + 2z
= 1− 2z + 22z2 − 23z3 + · · · gives −2
(1 + 2z)2
= −2 + 2 · 22z − 3 · 23z2 + · · ·. Thus
1
(1 + 2z)
= 1− 2 · (2z) + 3 · (2z)2 − · · · =
∞∑
k=1
(−1)k−1k(2z)k−1 where R = 1
2
.
898
19.2 Taylor Series
4. Using the binomial series gives
z
(1− z) = z
[
1 + 3z +
3 · 4
2!
z +
3 · 4 · 5
3!
z3 + · · ·
]
= z + 3z2 +
3 · 4
2!
z3 +
3 · 4 · 5
3!
z4 + · · · where R = 1.
5. Replacing z in ez =
∞∑
k=0
zk
k!
by −2z gives e−2z =
∞∑
k=0
(−1)k
k!
(2z)k where R = ∞.
6. Replacing z in ez =
∞∑
k=0
zk
k!
by −z2 and multiplying the result by z gives ze−z2 =
∞∑
k=0
(−1)k
k!
z2k+1 where
R = ∞.
7. Subtracting the series for ez and e−z gives sinh z =
1
2
(ez − e−z) =
∞∑
k=0
z2k+1
(2k + 1)!
where R = ∞.
8. Adding the series for ez and e−z gives cosh z =
1
2
(ez + e−z) =
∞∑
k=0
z2k
(2k)!
where R = ∞.
9. Replacing z in cos z =
∞∑
k=0
(−1)k z
2k
(2k)!
by z/2 gives cos
z
2
=
∞∑
k=0
(−1)k
(2k)!
(z
2
)2k
where R = ∞.
10. Replacing z in sin z =
∞∑
k=0
(−1)k z
2k+1
(2k + 1)!
by 3z gives sin 3z =
∞∑
k=0
(−1)k (3z)
2k+1
(2k + 1)!
where R = ∞.
11. Replacing z in sin z =
∞∑
k=0
(−1)k z
2k+1
(2k + 1)!
by z2 gives sin z2 =
∞∑
k=0
(−1)k z
4k+2
(2k + 1)!
where R = ∞.
12. Using the identity cos z =
1
2
(1 + cos 2z) and the series cos z =
∞∑
k=0
(−1)k z
2k
(2k)!
gives
cos2 z =
1
2
+
1
2
∞∑
k=0
(−1)k (2z)
2k
(2k)!
= 1 +
∞∑
k=1
(−1)k 2
2k−1
(2k)!
z2k where R = ∞.
13. Using (6) of Section 19.1,
1
z
=
1
1 + (z − 1) = 1− (z − 1) + (z − 1)
2 − (z − 1)3 + · · · =
∞∑
k=0
(−1)k(z − 1)k where R = 1.
14. Using (6) of Section 19.1,
1
z
=
1
1 + i + (z − 1− i) =
1
1 + i
· 1
1 +
z − 1− i
1 + i
=
1
1 + i
[
1− (z − 1− i)
1 + i
+
(z − 1− i)2
(1 + i)2
− (z − 1− i)
3
(1 + i)3
+ · · ·
]
=
1
1 + i
− (z − 1− i)
(1 + i)2
+
(z − 1− i)2
(1 + i)3
− (z − 1− i)
3
(1 + i)4
+ · · · =
∞∑
k=0
(−1)k (z − 1− i)
k
(1 + i)k+1
where R =
√
2
.
15. Using (5) of Section 19.1,
1
3− z =
1
3− 2i− (z − 2i) =
1
3− 2i ·
1
1− z − 2i
3− 2i
=
1
3− 2i
[
1 +
z − 2i
3− 2i +
(z − 2i)2
(3− 2i)2 +
(z − 2i)3
(3− 2i)3 + · · ·
]
=
1
3− 2i +
z − 2i
(3− 2i)2 +
(z − 2i)2
(3− 2i)3 +
(z − 2i)3
(3− 2i)4 + · · · =
∞∑
k=0
(z − 2i)k
(3− 2i)k+1 where R =
√
13.
899
19.2 Taylor Series
16. Using (6) of Section 19.1,
1
1 + z
=
1
1− i + z + i =
1
1− i ·
1
1 +
z + i
1− i
=
1
1− i
[
1− z + i
1− i +
(z + i)2
(1− i)2 −
(z + i)3
(1− i)3 + · · ·
]
=
1
1− i −
z + i
(1− i)2 +
(z + i)2
(1− i)3 −
(z + i)3
(1− i)4 +· · · =
∞∑
k=0
(−1)k (z + i)
k
(1− i)k+1 where R =
√
2 .
17. Using (5) of Section 19.1,
z − 1
3− z = (z − 1) ·
1
2− (z − 1) =
(z − 1)
2
· 1
1− z − 1
2
=
z − 1
2
[
1 +
z − 1
2
+
(z − 1)2
22
+
(z − 1)3
23
+ · · ·
]
=
z − 1
2
+
(z − 1)2
22
+
(z − 1)3
23
+
(z − 1)4
24
+ · · · =
∞∑
k=1
(z − 1)k
2k
where R = 2.
18. Using (5) of Section 19.1,
1 + z
1− z = −1 +
2
1− z = −1 +
2
1− i− (z − i) = −1 +
2
1− i ·
1
1− z − i
1− i
= −1 + 2
1− i
[
1 +
z − i
1− i +
(z − i)2
(1− i)2 +
(z − i)3
(1− i)3 + · · ·
]
= −1 + 2
1− i +
2(z − i)
(1− i)2 +
2(z − i)2
(1− i)3 +
2(z − i)3
(1− i)4 + · · · = −1 +
∞∑
k=0
2(z − i)k
(1− i)k+1 where R =
√
2 .
19. Using (8) of Section 19.2,
cos z =
√
2
2
−
√
2
2 · 1!
(
z − π
4
)
−
√
2
2 · 2!
(
z − π
4
)2
+
√
2
2 · 3!
(
z − π
4
)3
+ · · · where R = ∞.
20. Using the identity sin z = cos(z − π/2) and (14) of Section 19.2, sin z =
∞∑
k=0
(−1)k (z −
π
2 )
2k
(2k)!
where R = ∞.
21. Using ez = e3i · ez−3i and (12) of Section 19.2, ez = e3i
∞∑
k=0
(z − 3i)k
k!
where R = ∞.
22. Using (z − 1)e−2z = e2(z − 1)e−2(z−1) and (12) of Section 19.2,
(z − 1)e−2z = e2
∞∑
k=0
(−1)k2k
k!
(z − 1)k+1 where R = ∞.
23. Using (8) of Section 19.2, tan z = z +
1
3
z3 +
2
15
z5 + · · · .
24. Using (8) of Section 19.2, e1/(1+z) = e− ez + 3e
2
z2 − · · · .
25. Using (5) of Section 19.1,
f(z) =
1
z − 2i −
1
z − i = −
1
2i
· 1
1− z/2i +
1
i
1
1− z/i
= − 1
2i
(
1 +
z
2i
+
z
(2i)2
+
z3
(2i)3
+ · · ·
)
+
1
i
(
1 +
z
i
+
z2
i2
+
z3
i3
+ · · ·
)
= − i
2
− 3
4
z +
7i
8
z2 +
15
16
z3 − · · · .
The radius of convergence is R = 1.
900
19.2 Taylor Series
26. Using (6) and (5), respectively, of Section 19.1,
f(z) =
2
z + 1
− 1
z − 3 = 2 ·
1
1 + z
+
1
3
· 1
1− z/3 = 2(1− z + z
2 − z3 + · · ·) + 1
3
(
1 +
z
3
+
z2
32
+
z3
33
+ · · ·
)
=
7
3
− 17
9
z +
55
27
z2 − 161
81
z3 + · · · .
27. The distance from 2 + 5i to i is |2 + 5i− i| = |2 + 4i| = 2√5 .
28. The distance from πi to 0 is |πi| = π.
29. The Taylor series are
f(z) =
∞∑
k=0
(−1)k(z+1)k where R = 1; and f(z) =
∞∑
k=0
(−1)k (z − i)
k
(2 + i)k+1
where R =
√
5 .
30. The series are
f(z) =
∞∑
k=0
(−1)k (z − 3)
k
3k+1
where R = 3
and
f(z) =
∞∑
k=0
(−1)k (z − 1− i)
k
(1 + i)k+1
where R =
√
2 .
31. (a) The distance from z0 to the branch cut is one unit.
(b) The first term of the series determined by Taylor’s Theorem is
f(−1 + i) = Ln(−1 + i) = loge
√
2 +
3π
4
i =
1
2
loge 2 +
3π
4
i.
The subsequent terms of the series come from f ′(z) =
1
z
, f ′′(z) = − 1
z2
, and so on, evaluated at −1 + i.
(c) The series converges within the circle |z + 1− i| = √2 . Although the series converges
in the shaded region, it does not converge to (or represent) Ln z in this region.
32. (a) R = 1, which is the distance from the origin to z = −1.
(b) Using Taylor’s Theorem [or integrating the series for 1/(1 + z)] we obtain for R = 1,
Ln(1 + z) =
∞∑
k=1
(−1)k+1
k
zk.
(c) By replacing z in part (b) by −z we obtain for R = 1,
Ln(1− z) = −
∞∑
k=0
zk
k
.
(d) One way of obtaining the Maclaurin series for Ln
(
1 + z
1− z
)
is to use Taylor’s Theorem. Alternatively, let
us write
Ln
(
1 + z
1− z
)
= Ln(1 + z)− L(1− z)
901
19.2 Taylor Series
and subtract the series in parts (b) and (c). This gives for the common circle of convergence |z| = 1,
Ln
(
1 + z
1− z
)
= 2z +
2
3
z3 +
2
5
z5 +
2
7
z7 + · · · = 2
∞∑
k=0
1
(2k + 1)
z2k+1.
But recall that in general Ln(z1/z2) �= Lnz1−Lnz2 since Lnz1 and Lnz2 could differ by a constant multiple
of i. That is, Lnz1 − Lnz2 = Ci for some C. So
Ln
(
1 + z
1− z
)
= Ln(1 + z)− Ln(1− z)− Ci.
When z = 0 we obtain Ln 1 = Ln 1− Ln 1− Ci. Since Ln 1 = 0 we get C = 0.
33. From ez ≈ 1 + z + z
2
2
we obtain
e(1+i)/10 ≈ 1 + 1 + i
10
+
(1 + i)2
100
= 1.1 + 0.12i.
34. From sin z ≈ z − z
3
6
we obtain
sin
(
1 + i
10
)
≈ 1 + i
10
− 1
6
(
1 + i
10
)3
=
1
10
+
1
10
i− 1
6
(−2 + 2i
1000
)
=
301
3000
+
299
3000
i.
35. Using the series ez =
∞∑
k=0
zk
k!
we obtain e−t
2
=
∞∑
k=0
(−1)k t
2k
k!
. Thus
2√
π
∫ z
0
e−t
2
dt =
2√
π
∞∑
k=0
(−1)k
k!
∫ z
0
t2k dt =
2√
π
∞∑
k=0
(−1)k
k!(2k + 1)
z2k+1.
36. eiz =
∞∑
k=0
(iz)k
k!
= 1 + i
z
1!
− z
2
2!
− i z
3
3!
+
z4
4!
+ i
z5
5!
− z
6
6!
− i z
7
7!
+ · · ·
=
(
1− z
2
2!
+
z4
4!
− z
6
6!
+ · · ·
)
+ i
(
z
1!
− z
3
3!
+
z5
5!
− z
7
7!
+ · · ·
)
= cos z + i sin z
EXERCISES 19.3
Laurent Series
1. f(z) =
1
z
(
1− z
2
2!
+
z4
4!
− z
6
6!
+ · · ·
)
=
1
z
− z
2!
+
z3
4!
− z
5
6!
+ · · ·
2. f(z) =
1
z5
[
z −
(
z − z
3
3!
+
z5
5!
− z
7
7!
+ · · ·
)]
=
1
3!z2
− 1
5!
+
z2
7!
− z
4
9!
+ · · ·
3. f(z) = 1− 1
1!z2
+
1
2!z4
− 1
3!z6
+ · · ·
4. f(z) =
1
z2
[
1−
(
1 +
z
1!
+
z2
2!
+
z3
3!
+ · · ·
)]
= − 1
1!z
− 1
2!
− z
3!
− z
2
4!
− · · ·
902
19.3 Laurent Series
5. f(z) =
e · ez−1
z − 1 =
e
z − 1
(
1 +
(z − 1)
1!
+
(z − 1)2
2!
+
(z − 1)3
3!
+ · · ·
)
=
e
z − 1 +
e
1!
+
e(z − 1)
2!
+
e(z − 1)2
3!
+ · · ·
6. f(z) = z
(
1− 1
2!z2
+
1
4!z4
− 1
6!z6
+ · · ·
)
= z − 1
2!z
+
1
4!z3
− 1
6!z5
+ · · ·
7. f(z) = − 1
3z
· 1
1− z
3
= − 1
3z
[
1 +
z
3
+
z2
32
+
z3
33
+ · · ·
]
= − 1
3z
− 1
32
− z
33
− z
2
34
− · · ·
8. f(z) =
1
z2
· 1
1− 3
z
=
1
z2
[
1 +
3
z
+
32
z2
+
33
z3
+ · · ·
]
=
1
z2
+
3
z3
+
32
z4
+
33
z5
+ · · ·
9. f(z) =
1
z − 3 ·
1
3 + z − 3 =
1
3(z − 3) ·
1
1 +
z − 3
3
=
1
3(z − 3)
[
1− z − 3
3
+
(z − 3)2
32
− (z − 3)
3
33
+ · · ·
]
=
1
3(z − 3) −
1
32
+
z − 3
33
− (z − 3)
2
34
+ · · ·
10. f(z) =
1
z − 3 ·
1
z − 3 + 3 =
1
(z − 3)2 ·
1
1 +
3
z − 3
=
1
(z − 3)2
[
1− 3
z − 3 +
32
(z − 3)2 −
33
(z − 3)3 + · · ·
]
=
1
(z − 3)2 −
3
(z − 3)3 +
32
(z − 3)4 −
33
(z − 3)5 + · · ·
11. f(z) =
1
3
[
1
z − 3 −
1
z
]
=
1
3
[
1
z − 4 + 1 −
1
4 + z − 4
]
=
1
3
 1
z − 4 ·
1
1 +
1
z − 4
− 1
4
· 1
1 +
z − 4
4

=
1
3
[
1
z − 4
(
1− 1
z − 4 +
1
(z − 4)2 −
1
(z − 4)3 + · · ·
)
− 1
4
(
1− z − 4
4
+
(z − 4)2
42
− (z − 4)
3
43
+ · · ·
)]
= · · · − 1
3(z − 4)2 +
1
3(z − 1) −
1
12
+
z − 4
3 · 42 −
(z − 4)2
3 · 43 + · · ·
12. f(z) =
1
3
[
1
z − 3 −
1
z
]
=
1
3
[
1
−4 + z + 1 −
1
z + 1− 1
]
=
1
3
−14 · 11− z + 1
4
− 1
z + 1
· 1
1− 1
z + 1

=
1
3
[
−1
4
(
1 +
z + 1
4
+
(z + 1)2
42
+
(z + 1)3
43
+ · · ·
)
− 1
z + 1
(
1 +
1
z + 1
+
1
(z + 1)2
+
1
(z + 1)3
+ · · ·
)]
= · · · − 1
(z + 1)2
− 1
z + 1
− 1
12
− z + 1
3 · 42 −
(z + 1)2
3 · 43 − · · ·
13. f(z) =
1
z − 2 −
1
z − 1 = −
1
2
· 1
1− z
2
− 1
z
· 1
1− 1
z
= −12
(
1 +
z
2
+
z2
22
+
z3
23
+ · · ·
)
− 1
z
(
1 +
1
z
+
1
z2
+
1
z3
+ · · ·
)
= · · · − 1
z2
− 1
z
− 1
2
− z
22
− z
2
23
− · · ·
14. f(z) =
1
z − 2 −
1
z − 1 =
1
z
· 1
1− 2
z
− 1
z
· 1
1− 1
z
=
1
z
(
1 +
2
z
+
22
z2
+
23
z3
+ · · ·
)
− 1
z
(
1 +
1
z
+
1
z2
+
1
z3
+ · · ·
)
=
1
z2
+
22 − 1
z3
+
23 − 1
z4
+
24 − 1
z5
+ · · ·
903
19.3 Laurent Series
15. f(z) =
1
z − 1 ·
−1
1− (z − 1) =
−1
z − 1 [1 + (z− 1) + (z− 1)
2 + (z− 1)3 + · · ·] = − 1
z − 1 − 1− (z− 1)− (z− 1)
2− · · ·
16. f(z) =
1
z − 2 ·
1
1 + (z − 2) =
1
z − 2 [1− (z − 2) + (z − 2)
2 − (z − 2)3 + · · ·] = 1
z − 2 − 1 + (z − 2)− (z − 2)
2 + · · ·
17. f(z) =
1/3
z + 1
+
2/3
z − 2 =
1
3(z + 1)
+
2
3
· 1−3 + (z + 1) =
1
3(z + 1)
− 2
9
· 1
1− z + 1
3
=
1
3(z + 1)
− 2
9
[
1 +
z + 1
3
+
(z + 1)2
32
+
(z + 1)3
33
+ · · ·
]
=
1
3(z + 1)
− 2
9
− 2(z + 1)
33
− 2(z + 1)
2
34
− · · ·
18. f(z) =
1
3(z + 1)
+
2
3
· 1
(z + 1)− 3 =
1
3(z + 1)
+
2
3(z + 1)
· 1
1− 3
z + 1
=
1
3(z + 1)
+
2
3(z + 1)
(
1 +
3
z + 1
+
32
(z + 1)2
+
33
(z + 1)3
+ · · ·
)
=
1
z + 1
+
2
(z + 1)2
+
2 · 3
(z + 1)3
+
2 · 32
(z + 1)4
+ · · ·
19. f(z) =
1/3
z + 1
+
2/3
z − 2 =
1
3z
· 1
1 +
1
z
− 1
3
· 1
1− z
2
=
1
3z
(
1− 1
z
+
1
z2
− 1
z3
+ · · ·
)
− 1
3
(
1 +
z
2
+
z2
22
+
z3
23
+ · · ·
)
= · · · − 1
3z2
+
1
3z
− 1
3
− z
3 · 2 −
z2
3 · 22 − · · ·
20. f(z) =
2/3
z − 2 +
1
3
1
3 + (z − 2) =
2/3
z − 2 +
1
9
· 1
1 +
z − 2
3
=
2/3
z − 2 +
1
9
(
1 +
z − 2
3
+
(z − 2)2
32
+
(z − 2)3
33
+ · · ·
)
=
2
3(z − 2) +
1
9
+
z − 2
33
+
(z − 2)2
34
+ · · ·
21. f(z) =
1
z
(1−z)−2 = 1
z
(
1 + (−2)(−z) + (−2)(−3)
z!
(−z)2 + (−2)(−3)(−4)
3!
(−z)3 + · · ·
)
=
1
z
+2+3z+4z2 + · · ·
22. f(z) =
1
z3(1− 1
z
)2
=
1
z3
(
1− 1
z
)−2
=
1
z3
(
1 + (−2)
(
−1
z
)
+
(−2)(−3)
2!
(
−1
z
)2
+
(−2)(−3)(−4)
3!
(
−1
z
)3
+ · · ·
)
=
1
z3
+
2
z4
+
3
z5
+
4
z6
+ · · ·
23. f(z) =
1
(z − 2)[1 + (z − 2)]3 =
1
z − 2 [1 + (z − 2)]
−3
=
1
z − 2
(
1 + (−3)(z − 2) + (−3)(−4)
2!
(z − 2)2 + (−3)(−4)(−5)
3!
(z − 2)3 + · · ·
)
=
1
z − 2 − 3 + 6(z − 2)− 10(z − 2)
2 + · · ·
904
19.4 Zeros and Poles
24. f(z) =
1
(z − 3)3 ·
−1
1− (z − 1) =
−1
(z − 1)3 [1 + (z − 1) + (z − 1)
2 + (z − 1)3 + · · ·]
= − 1
(z − 1)3 −
1
(z − 1)2 −
1
z − 1 − 1− (z − 1)− · · ·
25. f(z) =
3
z
+
4
z − 1 =
3
z
− 4 · 1
1− z =
3
z
− 4(1 + z + z2 + z3 + · · ·) = 3
z
− 4− 4z − 4z2 − · · ·
26. f(z) =
4
z − 1 +3·
1
1 + (z − 1) =
4
z − 1 +3(1−(z−1)+(z−1)
2−(z−1)3+· · ·) = 4
z − 1 +3−3(z−1)+3(z−1)
2−· · ·
27. f(z) = z +
2
z − 2 = 1 + (z − 1) +
2
−1 + z − 1 = 1 + (z − 1) +
2
z − 1 ·
1
1− 1
z − 1
= 1 + (z − 1) + 2
z − 1
(
1 +
1
z − 1 +
1
(z − 1)2 +
1
(z − 1)3 + · · ·
)
= · · ·+ 2
(z − 1)2 +
2
z − 1 + 1 + (z − 1)
28. f(z) = z +
2
z − 2 =
2
z − 2 + 2 + (z − 2)
EXERCISES 19.4
Zeros and Poles
1. Using e2z =
∞∑
k=0
2kzk
k!
we obtain
e2z − 1
z
=
(
1 +
2
1!
z +
22
2!
z2 +
23
3!
z3 + · · ·
)
− 1
z
=
1
z
(
2
1!
z +
22
2!
z2 +
23
3!
z3 + · · ·
)
=
2
1!
+
22
2!
z +
23
3!
z2 + · · · .
From the form of the last series we see that z = 0 is a removable singularity. Define f(0) = 2.
2. Using sin 4z
∞∑
k=0
(−1)k (4z)
2k+1
(2k + 1)!
we obtain
sin 4z − 4z
z2
=
(
4
1!
z − 4
3
3!
z3 +
45
5!
z5 − 4
7
7!
z7 + · · ·
)
− 4z
z2
=
1
z2
(
−4
3
3!
z3 +
45
5!
z5 − 4
7
7!
z7 + · · ·
)
= −4
3
3!
z +
45
5!
z3 − 4
7
7!
z5 + · · · .
From the form of the last series we see that z = 0 is a removable singularity. Define f(0) = 0.
3. Since f(−2 + i) = f ′(−2 + i) = 0 and f ′′(z) = 2 for all z, z = −2 + i is a zero of order two.
4. Write f(z) = z4 − 16 = (z2 − 4)(z2 + 4) = (z − 2)(z + 2)(z − 2i)(z + 2i) to see that 2, −2, 2i, and −2i are
zeros of f . Now f ′(z) = 4z3 and f ′(2) �= 0, f ′(−2) �= 0, f ′(2i) �= 0, and f ′(−2i) �= 0. This indicates that each
zero is of order one.
5. Write f(z) = z2(z2 + 1) = z2(z − i)(z + i) to see that 0, i, and −i are zeros of f . Now f ′(z) = 4z3 + 2z and
f ′(i) �= 0 and f ′(−i) �= 0. This indicates that z = i and z = −i are zeros of order one. However f ′(0) = 0, but
f ′′(0) = 2 �= 0. Hence z = 0 is a zero of order two.
905
19.4 Zeros and Poles
6. Write f(z) = (z2 + 9)/z = (z − 3i)(z + 3i)/z to see that 3i and −3i are zeros of f . Now f ′(z) = 1− 9/z2 and
f ′(3i) = f ′(−3i) = 2 �= 0. This indicates that each zero is of order one.
7. Write f(z) = ez(ez − 1) to see that 2nπi, n = 0, ±1, ±2, . . . are zeros of f . Now f ′(z) = 2e2z − ez and
f ′(2nπi) = 2e4nπi − e2nπi = 1 �= 0. This indicates that each zero is of order one.
8. The zeros of f are the zeros of sin z, that is, nπ, n = 0, ±1, ±2, . . . . From f ′(z) = 2 sin z cos z we see f ′(nπ) = 0.
From f ′′(z) = 2(− sin2 z + cos2 z) we see f ′′(nπ) �= 0. This indicates that each zero is of order two.
9. From f(z) = z(1− cos z2) = z
(
−z
4
2!
+
z8
4!
− · · ·
)
= z5
(
− 1
2!
+
z4
4!
− · · ·
)
we see that z = 0 is a zero of order five.
10. From f(z) = z − sin z = z
3
3!
− z
5
5!
+ · · · = z3
(
1
3!
− z
2
5!
+ · · ·
)
we see that z = 0 is a zero of order three.
11. From f(z) = 1− ez−1 = −z − 1
1!
− (z − 1)
2
2!
− · · · = (z − 1)
(
1− z − 1
2!
− · · ·
)
we see that z = 1 is a zero of order one.
12. From the series ez = −
∞∑
k=0
(z − πi)k
k!
centered at πi and
f(z) = 1− πi + z + ez = 1− πi + z +
(
−1− z − πi
1!
− (z − πi)
2
2!
− (z − πi)
3
3!
− · · ·
)
= − (z − πi)
2
2!
− (z − πi)
3
3!
− · · · = (z − πi)2
(
− 1
2!
− z − πi
3!
− · · ·
)
we see that z = πi is a zero of order two.
13. From f(z) =
3z − 1
[(z − (−1 + 2i)][z − (−1− 2i)]
and Theorem 19.11 we see that −1 + 2i and −1− 2i are simple poles.
14. From f(z) =
5z2 − 6
z2
and Theorem 19.11 we see that 0 is a pole of order two.
15. From f(z) =
1 + 4i
(z + 2)(z + i)4
and Theorem 19.11 we see that −2 is a simple pole and −i is a pole of order four.
16. From f(z) =
z − 1
(z + 1)2
[
z − ( 12 +
√
3
2 i)
] [
z − ( 12 −
√
3
2 i)
]
and Theorem 19.11 we see that −1 is a pole of order two and 12 +
√
3
2 i and
1
2 −
√
3
2 i are simple poles.
17. Since sin z and cos z are analytic at nπ, n = 0, ±1, ±2, . . . , sin z has zeros of order one at nπ, and cosnπ �= 0,
it follows from Theorem 19.11 that the numbers nπ, n = 0, ±1, ±2, . . . are simple poles of f(z) = tan z.
18. From z2 sinπz = z3
(
π − π
3z2
3!
+ · · ·
)
we see z = 0 is a zero of order three. From f(z) =
cosπz
z2 sinπz
and
Theorem 19.11 we see 0 is a pole of order three. The numbers n, n± 1, ±2, . . . are simple poles.
19. From the Laurent series
f(z) =
1− cosh z
z4
=
1−
(
1 +
z2
2!
+
z4
4!
+
z6
6!
+ · · ·
)
z4
= − 1
2!z2
− 1
4!
− z
2
6!
− · · ·
we see that 0 is a pole of order two.
906
19.5 Residues and Residue Theorem
20. From the Laurent series
f(z) =
ez
z2
=
(
1 +
z
1!
+
z2
2!
+ · · ·
)
z2
=
1
z2
+
1
z
+
1
2!+ · · ·
we see that 0 is a pole of order two.
21. From 1 − ez = 1 −
(
1 +
z
1!
+
z2
2!
+ · · ·
)
= z
(
−1− z
2!
− · · ·
)
we see that z = 0 is a zero of order one. By
periodicity of ez it follows that z = 2nπi, n = 0, ±1, ±2, . . . are zeros of order one. From f(z) = 1
1− ez and
Theorem 19.11 we see that the numbers 2nπi, n = 0, ±1, ±2, . . . are simple poles.
22. z = 0 is a removable singularity of the function (sin z)/z. From f(z) =
sin z
z(z − 1) we see that only 1 is a (simple)
pole.
23. The function f(z) =
sin(1/z)
cos(1/z)
fails to be defined at z = 0 and at the solutions of cos
1
z
= 0, that is, at
1
z
= (2n+ 1)
π
2
, n = 0, ±1, ±2, . . . . Since z = 2
(2n + 1)π
, n = 0, ±1, ±2, . . . we see that in any neighborhood
of z = 0 there are points at which f is not defined and thus not analytic. Hence z = 0 is a non-isolated
singularity.
24. From the Laurent series
f(z) = z3
[
1
z
− 1
3!
(
1
z
)3
+
1
5!
(
1
z
)5
− 1
7!
(
1
z
7)
+ · · ·
]
= z2 − 1
3!
+
1
5!z2
− 1
7!z4
+ · · · , 0 < |z|,
we see that the principal part contains an infinite number of nonzero terms. Hence z = 0 is an essential
singularity.
EXERCISES 19.5
Residues and Residue Theorem
1. f(z) =
2
5(z − 1) ·
1
1 +
z − 1
5
=
2
5(z − 1)
(
1− z − 1
5
+
(z − 1)2
52
− (z − 1)
3
53
+ · · ·
)
=
2/5
z − 1 −
2
25
+
2(z − 1)
53
− 2(z − 1)
2
54
+ · · ·
Res(f(z), 1) = 2/5
2. f(z) =
1
z3
(1− z)−3 = 1
z3
(
1 + (−3)(−z) + (−3)(−4)
2!
(−z)2 + (−3)(−4)(−5)
3!
(−z)3 + · · ·
)
=
1
z3
+
3
z2
+
6
z
+ 10 + · · ·
Res(f(z), 0) = 6
3. f(z) = −3
z
− 1
z − 2 = −
3
z
+
1
2
· 1
1− z
2
= −3
z
+
1
2
(
1 +
z
2
+
z2
22
+
z3
23
+ · · ·
)
= −3
z
+
1
2
+
z
22
+
z2
23
+ · · ·
907
19.5 Residues and Residue Theorem
Res(f(z), 0) = −3
4. f(z) = (z + 3)2
(
2
z + 3
− 2
3
3!(z + 3)3
+
25
5!(z + 3)5
+ · · ·
)
= · · ·+ 2
5
5!(z + 3)3
− 2
3
3!(z + 3)
+ 2(z + 3)
Res(f(z),−3) = −4
3
5. f(z) = e−2/z
2
=
∞∑
k=0
(−2/z2)k
k!
= · · · − 2
3
3!z6
+
22
2!z4
− 2
1!z2
+ 1; Res(f(z), 0) = 0
6. f(z) =
e−2
(z − 2)2 e
−(z−2) =
e−2
(z − 2)2
(
1− z − 2
1!
+
(z − 2)2
2!
− (z − 2)
3
3!
+ · · ·
)
=
e−2
(z − 2)2 −
e−2
z − 2 +
e−2
2
− e
−2(z − 2)
3!
+ · · ·
Res(f(z), 2) = −e−2
7. Res(f(z), 4i) = lim
z→4i
(z − 4i) · z
(z − 4i)(z + 4i) = limz→4i
z
z + 4i
=
1
2
Res(f(z),−4i) = lim
z→−4i
(z + 4i) · z
(z − 4i)(z + 4i) = limz→−4i
z
z − 4i =
1
2
8. Res(f(z), 1/2) = lim
z→1/2
(z − 1/2) 4z + 8
2(z − 1/2) = limz→1/2(2z + 4) = 5
9. Res(f(z), 1) = lim
z→1
(z − 1) 1
z2(z + 2)(z − 1) = limz→1
1
z2(z + 2)
=
1
3
Res(f(z),−2) = lim
z→−2
(z + 2)
1
z2(z + 2)(z − 1) = limz→−2
1
z2(z − 1) = −
1
12
Res(F (z), 0) =
1
1!
lim
z→0
d
dz
[
z2 · 1
z2(z + 2)(z − 1)
]
= lim
z→0
−2z − 1
(z + 2)2(z − 1)2 = −
1
4
10. Res(f(z), 1 + i) =
1
1!
lim
z→1+i
d
dz
[
(z − 1− i)2 · 1
(z − 1− i)2(z − 1 + 1)2
]
= lim
z→1+i
−2
(z − 1 + i)3 = −
1
4
i
Res(f(z), 1− i) = 1
1!
lim
z→1−i
d
dz
[
(z − 1 + i)2 · 1
(z − 1− i)2(z − 1 + i)2
]
= lim
z→1−i
−2
(z − 1− i)3 =
1
4
i
11. Res(f(z),−1) = lim
z→−1
(z + 1) · 5z
2 − 4z + 3
(z + 1)(z + 2)(z + 3)
= lim
z→−1
5z2 − 4z + 3
(z + 2)(z + 3)
= 6
Res(f(z),−2) = lim
z→−2
(z + 2) · 5z
2 − 4z + 3
(z + 1)(z + 2)(z + 3)
= lim
z→−2
5z2 − 4z + 3
(z + 1)(z + 3)
= −31
Res(f(z),−3) = lim
z→−3
(z + 3) · 5z
2 − 4z + 3
(z + 1)(z + 2)(z + 3)
= lim
z→−3
5z2 − 4z + 3
(z + 1)(z + 2)
= 30
12. Res(f(z),−3) = lim
z→−3
(z + 3) · 2z − 1
(z − 1)4(z + 3) = limz→−3
2z − 1
(z − 1)4 = −
7
256
Res(f(z), 1) =
1
3!
lim
z→1
d3
dz3
[
(z − 1)4 · 2z − 1
(z − 1)4(z + 3)
]
=
1
6
lim
z→1
−42
(z + 3)4
= − 7
256
13. Res(f(z), 0) =
1
1!
lim
z→0
d
dz
[
z2 · cos z
z2(z − π)3
]
= lim
z→0
−(z − π) sin z − 3 cos z
(z − π)4 = −
3
π4
Res(f(z), π) =
1
2!
lim
z→π
d2
dz2
[
(z − π)3 · cos z
z2(z − π)3
]
=
1
2
lim
z→π
−z2 cos z + 4z sin z + 6 cos z
z4
=
π2 − 6
2π4
908
19.5 Residues and Residue Theorem
14. Using
d
dz
(ez − 1) = ez and the result in (4) in the text,
Res(f(z), 2nπi) =
ez
ez
∣∣∣∣
z=2nπi
= 1.
15. Using
d
dz
cos z = − sin z and the result in (4) in the text,
Res
(
f(z), (2n + 1)
π
2
)
=
1
− sin z
∣∣∣∣
z=(2n+1) π2
=
1
− sin(2n + 1)π2
= (−1)n+1.
16. z = 0 is a pole of order two. Thus by (2) in the text and L’Hoˆpital’s rule,
Res(f(z), 0) =
1
1!
lim
z→0
d
dz
[
z2 · 1
z sin z
]
= lim
z→0
sin z − z cos z
sin2 z
= lim
z→0
cos z + z sin z − cos z
2 sin z cos z
= lim
z→0
z
2 cos z
= 0.
For the simple poles at z = nπ, n = ±1, ±2, . . . we have from (4) in the text,
Res(f(z), nπ) =
1
z cos z + sin z
∣∣∣∣
z=nπ
=
(−1)n
nπ
.
17. (a)
∮ˇ
C
1
(z − 1)(z + 2)2 dz = 0 by Theorem 18.4.
(b)
∮ˇ
C
1
(z − 1)(z + 2)2 dz = 2πiRes(f(z), 1) =
2π
9
i
(c)
∮ˇ
C
1
(z − 1)(z + 2)2 dz = 2πi [Res(f(z), 1) + Res(f(z),−2)] = 2πi
[
1
9
+
(
−1
9
)]
= 0
18. (a)
∮ˇ
C
z + 1
z2(z − 2i) dz = 2πiRes(f(z), 0) = π
(
−1 + 1
2
i
)
(b)
∮ˇ
C
z + 1
z2(z − 2i) dz = 2πiRes(f(z), 2i) = π
(
1− 1
2
i
)
(c)
∮ˇ
C
z + 1
z2(z − 2i) dz = 2πi[Res(f(z), 0) + Res(f(z), 2i)] = 2πi
[
1
4
+
1
2
i +
(
−1
4
− 1
2
i
)]
= 0
19. (a) From the Laurent series z3e−1/z
2
= · · · − 1
3!z3
+
1
2!z
− z + z3 we see Res(f(z), 0) = 1/2. Hence∮ˇ
C
z3e−1/z
2
dz = 2πiRes(f(z), 0) = πi.
(b)
∮ˇ
C
z3e−1/z
2
dz = 2πiRes(f(z), 0) = πi
(c)
∮ˇ
C
z3e1/z
2
dz = 0 by Theorem 18.4.
20. (a)
∮ˇ
C
1
z sin z
dz = 0 by Theorem 18.4.
(b) z = 0 is a pole of order two (see Problem 16). Thus∮ˇ
C
1
z sin z
dz = 2πiRes(f(z), 0) = 2πi(0) = 0.
(c)
∮ˇ
C
1
z sin z
dz = 2πi[Res(f(z),−π) + Res(f(z), 0) + Res(f(z), π)] = 2πi
[
1
π
+ 0 +
(
− 1
π
)]
= 0
909
19.5 Residues and Residue Theorem
21.
∮ˇ
C
1
z2 + 4z + 13
dz = 2πiRes(f(z),−2 + 3i) = π
3
22.
∮ˇ
C
1z3(z − 1)4 dz = 2πiRes(f(z), 1) = −20πi
23.
∮ˇ
C
z
z4 − 1 dz = 2πi[Res(f(z),−1) + Res(f(z), 1) + Res(f(z),−i) + Res(f(z), i)] = 2πi
[
1
4
+
1
4
− 1
4
− 1
4
]
= 0
24.
∮ˇ
C
z
(z + 1)(z2 + 1)
dz = 2πi[Res(f(z), i) + Res(f(z),−i)] = 2πi
[
1
4
− 1
4
i +
1
4
+
1
4
i
]
= πi
25.
∮ˇ
C
zez
z2 − 1 dz = 2πi[Res(f(z), 1) + Res(f(z),−1)] = 2πi
[
e
2
+
e−1
2
]
= 2πi cosh 1
26.
∮ˇ
C
ez
z3 + 2z2
dz = 2πi[Res(f(z), 0) + Res(f(z),−2)] = 2πi
[
1
2
+
e−2
4
]
= πi
(
1 +
1
2
e−2
)
27.
∮ˇ
C
tan z
z
dz = 2πiRes
(
f(z),
π
2
)
) = −4i. Note: z = 0 is not a pole. See Example 1, Section 19.4.
28.
∮ˇ
C
cotπz
z2
dz = 2πiRes(f(z), 0) = 2πi
(
−π
3
)
= −2π
2
3
i
Note: z = 0 is a pole of order three. Use L’Hoˆpital’s rule (or Mathematica) to show that
Res(f(z), 0) =
1
2
lim
z→0
d2
dz2
z cotπz =
1
2
lim
z→0
[−2π csc2 πz + 2π2z cotπz csc2 πz] = 1
2
(
−2π
3
)
= −π
3
.
29.
∮ˇ
C
cotπz dz = 2πi[Res(f(z), 1) + Res(f(z), 2) + Res(f(z), 3)] = 2πi
[
1
π
+
1
π
+
1
π
]
= 6i
30.
∮ˇ
C
2z − 1
z2(z3 + 1)
dz = 2πi
[
Res(f(z), 0) + Res(f(z),−1) + Res
(
f(z),
1
2
+
√
3
2
i
)]
= 2πi[
2 + (−1) +
(
−1
2
− 1
6
√
3 i
)]
= π
(√
3
3
+ i
)
31.
∮ˇ
C
eiz + sin z
(z − π)4 dz = 2πiRes(f(z), π) = π
(
−1
3
+
1
3
i
)
32.
∮ˇ
C
cos z
(z − 1)2(z2 + 9) dz = 2πiRes(f(z), 1) = 2πi(−0.02 cos 1− 0.1 sin 1) = −0.5966i
EXERCISES 19.6
Evaluation of Real Integrals
1.
∫ 2π
0
dθ
1 + 12 sin θ
=
∮ˇ
C
4
z2 + 4iz − 1 dz = (4)2πiRes(f(z), (
√
3− 2)i) = 4π√
3
2.
∫ 2π
0
dθ
10− 6 cos θ =
1
2
·
(−2
i
) ∮ˇ
C
dz
(3z − 1)(z − 3) = (i)2πiRes
(
f(z),
1
3
)
=
π
4
910
19.6 Evaluation of Real Integrals
3.
∫ 2π
0
cos θ
3 + sin θ
dθ =
∮ˇ
C
z2 + 1
z(z2 + 6iz − 1) dz = 2πi[Res(f(z), 0) + Res(f(z),−3 + 2
√
2 i)] = 0
4.
∫ 2π
0
1
1 + 3 cos2 θ
dθ =
4
i
∮ˇ
C
z
3z4 + 10z2 + 3
dz =
(
4
i
)
2πi
[
Res
(
f(z), (
√
3
3
i
)
+ Res
(
f(z),−
√
3
3
i
)]
= π
5.
∫ π
0
dθ
2− cos θ =
1
2
∫ 2π
0
dθ
2− cos θ = −
1
i
∮ˇ
C
dz
z2 − 4z + 1 =
(
−1
i
)
2πiRes(f(z), 2−
√
3 ) =
π√
3
6.
∫ π
0
dθ
1 + sin2 θ
=
1
2
∫ 2π
0
dθ
1 + sin2 θ
= −2
i
∮ˇ
C
z
z4 − 6z2 + 1 dz
=
(
−2
i
)
2πi[Res(f(z),
√
3− 2
√
2 ) + Res(f(z),−
√
3− 2
√
2 )] =
π√
2
7.
∫ 2π
0
sin2 θ
5 + 4 cos θ
dθ = − 1
4i
∮ˇ
C
(z2 − 1)2
z2(2z2 + 5z + 2)
dz =
(
− 1
4i
)
2πi
[
Res(f(z), 0) + Res
(
f(z),−1
2
)]
=
π
4
8.
∫ 2π
0
cos2 θ
3− sin θ dθ =
1
2i
∮ˇ
C
z4 + 2z2 + 1
z2(iz2 + 6z − i) dz =
(
1
2i
)
2πi[Res(f(z), 0) + Res(f(z), 3− 2
√
2 )i)] = π[6− 4
√
2 ]
9. We use cos 2θ = (z2 + z−2)/2.∫ 2π
0
cos 2θ
5− 4 cos θ dθ =
i
2
∮ˇ
C
z4 + 1
z2(2z2 − 5z + 2) dz =
(
i
2
)
2πi
[
Res(f(z), 0) + Res
(
f(z),
1
2
)]
=
π
6
10.
∫ 2π
0
1
cos θ + 2 sin θ + 3
dθ =
2
i
∮ˇ
C
1
(1− 2i)z2 + 6z + 1 + 2i =
(
2
i
)
2πiRes
(
f(z),−1
5
− 2
5
i
)
= π
11.
∫ ∞
−∞
1
x2 − 2x + 2 dx = 2πiRes(f(z), 1 + i) = π
12.
∫ ∞
−∞
1
x2 − 2x + 25 dx = 2πiRes(f(z), 1 + 2
√
6 i) =
π
2
√
6
13.
∫ ∞
−∞
1
(x2 + 4)2
dx = 2πiRes(f(z), 2i) =
π
16
14.
∫ ∞
−∞
x2
(x2 + 1)2
dx = 2πiRes(f(z), i) =
π
2
15.
∫ ∞
−∞
1
(x2 + 1)3
dx = 2πiRes(f(z), i) =
3π
8
16.
∫ ∞
−∞
x
(x2 + 4)3
dx = 0 (The integrand is an odd function)
17.
∫ ∞
−∞
2x2 − 1
x4 + 5x2 + 4
dx = 2πi[Res(f(z), i) + Res(f(z), 2i)] =
π
2
18.
∫ ∞
−∞
dx
(x2 + 1)2(x2 + 9)
= 2πi[Res(f(z), i) + Res(f(z), 3i)] =
5π
96
19.
∫ ∞
0
x2 + 1
x4 + 1
dx =
1
2
∫ ∞
−∞
x2 + 1
x4 + 1
dx = πiRes(f(z), i) =
π√
2
20.
∫ ∞
0
1
x6 + 1
dx =
1
2
∫ ∞
−∞
1
x6 + 1
dx = πi
[
Res
(
f(z),
√
3
2
+
1
2
i
)
+ Res(f(z), i) + Res
(
f(z),
−√3
2
+
1
2
i
)]
=
π
3
911
19.6 Evaluation of Real Integrals
21.
∫ ∞
−∞
eix
x2 + 1
dx = 2πiRes(f(z), i) = πe−1. Therefore,
∫ ∞
−∞
cosx
x2 + 1
dx = Re
(∫ ∞
−∞
eix
x2 + 1
dx
)
= πe−1.
22.
∫ ∞
−∞
e2ix
x2 + 1
dx = 2πiRes(f(z), i) = πe−2. Therefore,
∫ ∞
−∞
cos 2x
x2 + 1
dx = Re
(∫ ∞
−∞
e2ix
x2 + 1
dx
)
= πe−2.
23.
∫ ∞
−∞
xeix
x2 + 1
dx = 2πiRes(f(z), i) = πe−1i. Therefore,
∫ ∞
−∞
x sinx
x2 + 1
dx = Im
(∫ ∞
−∞
xeix
x2 + 1
dx
)
= πe−1.
24.
∫ ∞
−∞
eix
(x2 + 4)2
dx = 2πiRes(f(z), 2i) =
3e−2
16
π;
∫ ∞
−∞
cosx
(x2 + 4)2
dx = Re
(∫ ∞
−∞
eix
(x2 + 4)2
dx
)
=
3e−2
16
π.
Therefore, ∫ ∞
0
cosx
(x2 + 4)2
dx =
1
2
(
3e−2
16
π
)
=
3e−2
32
π.
25.
∫ ∞
−∞
e3ix
(x2 + 1)2
dx = 2πiRes(f(z), i) = 2πe−3;
∫ ∞
−∞
cos 3x
(x2 + 1)2
dx = Re
(∫ ∞
−∞
e3ix
(x2 + 1)2
dx
)
= 2πe−3.
Therefore, ∫ ∞
0
cos 3x
(x2 + 1)2
dx =
1
2
(2πe−3) = πe−3.
26.
∫ ∞
−∞
eix
x2 + 4x + 5
dx = 2πiRes(f(z),−2 + i) = πe−1−2i. Therefore
∫ ∞
−∞
sinx
x2 + 4x + 5
dx = Im
(∫ ∞
−∞
eix
x2 + 4x + 5
dx
)
= −πe−1 sin 2
27.
∫ ∞
−∞
e2ix
x4 + 1
dx = 2πi
[
Res
(
f(z),
1√
2
+
1√
2
i
)
+ Res
(
f(z),− 1√
2
+
1√
2
i
)]
= 2πi
[(
−
√
2
8
−
√
2
8
i
)
e(−
√
2+
√
2 i) +
(√
2
8
−
√
2
8
i
)
e(−
√
2−√2 i)
]
= πe−
√
2
[√
2
2
cos
√
2 +
√
2
2
sin
√
2
]
∫ ∞
−∞
cos 2x
x4 + 1
dx = Re
(∫ ∞
−∞
e2ix
x4 + 1
dx
)
= πe−
√
2
[√
2
2
cos
√
2 +
√
2
2
sin
√
2
]
Therefore ∫ ∞
0
cos 2x
x4 + 1
dx = πe−
√
2
√
2
4
(cos
√
2 + sin
√
2 ).
28.
∫ ∞
−∞
xeix
x4 + 1
dx = 2πi
[
Res
(
f(z),
1√
2
+
1√
2
i
)
+ Res
(
f(z),− 1√
2
+
1√
2
i
)]
= 2πi
[
− i
4
e(−1/
√
2+i/
√
2 ) +
i
4
e(−1/
√
2−i/√2)
]
=
(
πe−1/
√
2 sin
1√
2
)
i∫ ∞
−∞
x sinx
x4 + 1
dx = Im
(∫ ∞
−∞
xeix
x4 + 1
dx
)
= πe−1/
√
2 sin
1√
2
Therefore ∫ ∞
0
x sinx
x4 + 1
dx =
π
2
e−1/
√
2 sin
1√
2
.
912
19.6 Evaluation of Real Integrals
29.
∫ ∞
−∞
eix
(x2 + 1)(x2 + 9)
dx = 2πi[Res(f(z), i) + Res(f(z), 3i)] = 2πi
[
− i
16
e−1 +
i
48
e−3
]
=
1
8
e−1 − 1
24
e−3
30.
∫ ∞
−∞
xeix
(x2 + 1)(x2 + 4)
dx = 2πi[Res(f(z), i) + Res(f(z), 2i)] = 2πi
[
1
6
e−1 − 1
6
e−2
]
=
π
3
(e−1 − e−2)i;∫ ∞
−∞
x sinx
(x2 + 1)(x2 + 4)
dx = Im
(∫ ∞
−∞
xeix
(x2 + 1)(x2 + 4)
dx
)
=
π
3
(e−1 − e−2).
Therefore, ∫ ∞
0
x sinx
(x2 + 1)(x2 + 4)
dx =
1
2
[π
3
(e−1 − e−2)
]
=
π
6
(e−1 − e−2).
31. Consider the contour integral
∮ˇ
C
eiz
z
dz. The function f(z) =
1
z
has a simple pole at z = 0. If we use the
contour C shown in Figure 19.14, it follows from the Cauchy-Goursat Theorem that∮ˇ
C
=
∫
CR
+
∫ −r
−R
+
∫
−Cr
+
∫ R
r
= 0.
Taking limits as R →∞ and as r → 0 and using Theorem 19.17 we then find
P.V.
∫ ∞
−∞
eix
x
dx− πiRes(f(z)eiz, 0) = 0 or P.V.
∫ ∞
−∞
eix
x
dx = πi.
Equating the imaginary parts of
∫ ∞
−∞
cosx + i sinx
x
dx = 0 + πi gives∫ ∞
−∞
sinx
x
dx = π.
32. Consider the contour integral
∮ˇ
C
eiz
z(z2 + 1)
dz. The function f(z) =
1
z(z2 + 1)
has simple poles at z = 0 and
at z = i. If we use the contour C shown in Figure 19.14, it follows from Theorem 19.14 that∮ˇ
C
=
∫
CR
+
∫ −r
−R
+
∫
−Cr
+
∫ R
r
= 2πiRes(f(z), i).
Taking limits as R →∞ and as r → 0 and using Theorem 19.17 we then find
P.V.
∫ ∞
−∞
eix
x(x2 + 1)
dx− πiRes(f(z)eiz, 0) = 2πiRes(f(z)eiz, i)
or
P.V.
∫ ∞
−∞
eix
x(x2 + 1)
dx = πi + 2πi
(
−e
−1
2
)
.
Equating the imaginary parts of
∫ ∞
−∞
cosx + i sinx
x(x2 + 1)
dx = 0 + π(1− e−1)i gives∫ ∞
−∞
sinx
x(x2 + 1)
dx = π(1− e−1).
33.
∫ π
0
dθ
(a + cos θ)2
=
1
2
∫ 2π
0
dθ
(a + cos θ)2
=
2
i
∮ˇ
C
z
(z2 + 2az + 1)2
dz (C is |z| = 1) = 2
i
∮ˇ
C
z
(z − r1)2(z − r2)2 dz
where r1 = −a +
√
a2 − 1 , r2 = −a−
√
a2 − 1 . Now∮ˇ
C
z
(z − r1)2(z − r2)2 dz = 2πiRes(f(z), r1) = 2πi
a
4(
√
a2 − 1 )3 =
aπ
2(
√
a2 − 1 )3 i.
913
19.6 Evaluation of Real Integrals
Thus, ∫ π
0
dθ
(a + cos θ)2
=
2
i
· aπ
2(
√
a2 − 1 )3 i =
aπ
(
√
a2 − 1 )3 .
When a = 2 we obtain∫ π
0
dθ
(2 + cos θ)2
=
2π
(
√
3 )3
and so
∫ 2π
0
dθ
(2 + cos θ)2
=
4π
3
√
3
.
34.
∫ 2π
0
sin2 θ
a + b cos θ
dθ =
i
2b
∮ˇ
C
z2 − 1
z2(z − r1)(z − r2)dz (C is |z| = 1) where r1 = (−a +
√
a2 − b2 )/b,
r2 = (−a−
√
a2 − b2 )/b. Now∮ˇ
C
z2 − 1
z2(z − r1)(z − r2) dz = 2πi[Res(f(z), 0) + Res(f(z), r1)] = 2πi
[
−2a
b
+
2
√
a2 − b2
b
]
.
Thus, ∫ 2π
0
sin2 θ
a + b cos θ
dθ =
2π
b2
(a−
√
a2 − b2 ), a > b > 0.
When a = 5, b = 4 we obtain ∫ 2π
0
sin2 θ
5 + 4 cos θ
dθ =
2π
16
(5−
√
9 ) =
π
4
.
35. Consider the contour integral
∮ˇ
C
eaz
1 + ez
dz. The function f(z) =
eaz
1 + ez
has simple poles at z = πi, 3πi,
5πi, . . . in the upper plane. Using the contour in Figure 19.15 we have from Theorem 19.14∮ˇ
C
=
∫ r
−r
+
∫
C2
+
∫
C3
+
∫
C4
= 2πiRes(f(z), πi) = −2πieaπi.
On C2, z = r + iy, 0 ≤ y ≤ π, dz = i dy, ∣∣∣∣∫
C2
eaz
1 + ez
dz
∣∣∣∣ ≤ earer − 1(2π).
Because 0 < a < 1, this last expression goes to 0 as r →∞. On C3, z = x + 2πi, −r ≤ x ≤ r, dz = dx,∫
C3
eaz
1 + ez
dz =
∫ −r
r
ea(x+2πi)
1 + ex+2πi
dx = −e2aπi
∫ r
−r
eax
1 + ex
dx.
On C4, z = −r + iy, 0 ≤ y ≤ 2π, dz = i dy,∣∣∣∣∫
C4
eaz
1 + ez
dz
∣∣∣∣ ≤ e−ar1− e−r (2π).
Because 0 < a, this last expression goes to 0 as r →∞. Hence∫ r
−r
eax
1 + ex
dx− e2aπi
∫ r
−r
eax
1 + ex
dx = −2πieaπi
gives, as r →∞,
(1− e2aπi)
∫ ∞
−∞
eax
1 + ex
dx = −2πieaπi.
That is ∫ ∞
−∞
eax
1 + ex
dx = − 2πie
aπi
1− e2aπi =
π
eaπi − e−aπi
2i
=
π
sin ax
.
914
CHAPTER 19 REVIEW EXERCISES
36. Using the Fourier sine transform with respect to y the partial differential equation becomes
d2U
dx2
− α2U = 0
and so
U(x, α) = c1 coshαx + c2 sinhαx.
The boundary condition u(0, y) becomes U(0, α) = 0 and so c1 = 0. Thus U(x, α) = c2 sinhαx. Now to evaluate
U(π, α) =
∫ ∞
0
2y
y4 + 4
sinαy dy =
∫ ∞
−∞
y
y4 + 4
sinαy dy
we use the contour integral
∫
C
zeiαz
z4 + 4
dz and∫ ∞
−∞
xeiαx
x4 + 4
dx = 2πi[Res(f(z), 1 + i) + Res(f(z),−1 + i)] = 2πi
[
−1
8
ie(−1+i)α +
1
8
ie(−1−i)α
]
=
π
2
(e−α sinα)i
∫ ∞
−∞
x sinαx
x4 + 4
dx = Im
(∫ ∞
−∞
xeiαx
x4 + 4
dx
)
=
π
2
e−α sinα.
Finally, U(π, α) =
π
2
e−α sinα = c2 sinhαπ gives c2 =
π
2
e−α sinα
sinhαπ
. Hence U(x, α) =
π
2
e−α sinα
sinhαπ
sinhαx and
u(x, y) =
∫ ∞
0
e−α sinα
sinhαπ
sinhαx sinαy dα.
CHAPTER 19 REVIEW EXERCISES
1. True 2. False 3. False 4. True
5. True 6. True 7. True 8. five
9. 1/π 10. three; −1/6 11. |z − i| = √5 12. False
13.
ez(1+i) + ez(1−i)
2
=
1
2
(
1 + z(1 + i) +
z2
2!
(1 + i)2 + · · ·
)
+
1
2
(
1 + z(1− i) + z
2
2!
(1− i)2 + · · ·
)
= 1 + z
[
(1 + i) + (1− i)
2
]
+
z2
2!
[
(1 + i)2 + (1− i)2
2
]
+ · · · = 1 +
∞∑
k=1
(
√
2 )k cos kπ4
k!
zk
Here we have used (1 + i)n = (
√
2 )nenπi/4 and (1− i)n = (√2 )ne−nπi/4 so that
(1 + i)n + (1− i)n
2
= (
√
2 )n
[
enπi/4 + e−nπi/4
2
]
= (
√
2 )n cos
nπ
4
.
14. sin
π
z
= 0 implies z =
1
n
, n = ±1, ±2, . . . . All singularities are isolated except the singularity z = 0.
15. f(z) =
1
z4
[
1−
(
1 +
iz
1!
+
i2z2
2!
+
i3z3
3!
+
i4z4
4!
+ · · ·
)]
= − i
z3
+
1
2!z2
+
i
3!z
− 1
4!
− iz
5!
+ · · ·
915
CHAPTER 19 REVIEW EXERCISES
16. ez/(z−2) = e · e2/(z−2) = e
(
1 +
2
z − 2 +
22
2!(z − 2)2 +
23
3!(z − 2)3 + · · ·
)
= e
∞∑
k=0
2k
k!
(z − 2)−k
17. (z − i)2 sin 1
z − i = (z − i)
2
[
1
z − i −
1
3!(z − i)3 +
1
5!(z − i)5 − · · ·
]
= · · ·+ 1
5!(z − i)3 −
1
3!(z − i) + (z − i)
18.
1− cos z2
z5
=
1
z5
[
1−
(
1− z
4
2!
+
z8
4!
− z
12
6!
+
z16
8!
− · · ·
)]
=
1
2!z
− z
3
4!
+
z7
6!
− z
11
8!
+ · · ·
19. (a) f(z) =
1
z − 3 −
1
z − 1 =
1
1− z −
1
3
· 1
1− z
3
= (1 + z + z2 + z4 + · · ·)− 1
3
(
1 +
z
3
+
z2
32
+
z3
33
+ · · ·
)
=
2
3
+
8
9
z +
26
27
z2
(b) f(z) = −1
z
· 1
1− 1
z
− 1
3
· 1
1− z
3
= −1
z
(
1 +
1
z
+
1
z2
+
1
z3
+ · · ·
)
− 1
3
(
1 +
z
3
+
z2
32
+
z3
33
+ · · ·
)
= · · · − 1
z3
− 1
z2
− 1
z
− 1
3
− z
32
− z
2
33
− · · ·
(c) f(z) = −1
z
· 1
1− 1
z
+
1
z
· 1
1− 3
z
= −1
z
(
1 +
1
z
+
1
z2
+
1
z3
+ · · ·
)
+
1
z
(
1 +
3
z
+
32
z2
+
33
z3
+ · · ·
)
=
2
z2
+
8
z3
+
26
z4
+ · · ·
(d) f(z) = − 1
z − 1 −
1
2
· 1
1− z − 1
2
= − 1
z − 1 −
1
2
(
1 +
z − 1
2
+
(z − 1)2
22
+
(z − 1)3
3!
+ · · ·
)
= − 1
z − 1 −
1
2
− z − 1
22
− (z − 1)
2
23
− · · ·
20. (a) f(z) =
1
25
(
1− z
5
)−2
=
1
25
[
1 + (−2)
(
−z
5
)
+
(−2)(−3)
2!
(
−z
5
)2
+
(−2)(−3)(−4)
3!
(
−z
5
)3
+ · · ·
]
=
1
25
+ 2
z
53
+ 3
z2
54
+ 4
z3
55
+ · · ·
(b) (z − 5)−2 = 1
z2
(
1− 5
z
)−2
=
1
z2
[
1 + (−2)
(
−5
z
)
+
(−2)(−3)
2!
(
−5
z
)2
+
(−2)(−3)(−4)
3!
(
−5
z
)3
+ · · ·
]
=
1
z2
+ 2
5
z3
+ 3
52
z4
+ 4
53
z5
+ · · ·
(c)
1
(z − 5)2 is the Laurent series.
21.
∮ˇ
C
2z + 5
z(z + 2)(z − 1)4 dz = 2πi[Res(f(z), 0) + Res(f(z),−2)] =
404
81
πi
22.
∮ˇ
C
z2
(z − 1)3(z2 + 4) dx = 2πiRes(f(z), 1) =
8π
125
i
23.
∮ˇ
C
1
2 sin z − 1 dz = 2πiRes
(
f(z),
π
6
)
=
2π√
3
i
24.
∮ˇ
C
z + 1
sinh z
dz = 2πi[Res(f(z), 0) + Res(f(z), πi)] = 2πi[1 + (−πi− 1)] = 2π2
916
CHAPTER 19 REVIEW EXERCISES
25.
∮ˇ
C
e2z
z4 + 2z3 + 2z2
dz = 2πi[Res(f(z), 0) + Res(f(z),−1 + i) + Res(f(z),−1− i)]
= 2πi
[
1
2
+
e−2
4
(cos 2 + i sin 2) +
e−2
4
(cos 2− i sin 2)
]
= π(1 + e−2 cos 2)i
26.
∮ˇ
C
1
z4 − 2z2 + 4 dz = 2πi
[
Res
(
f(z),
√
6
2
+
√
2
2
i
)
+ Res
(
f(z),−
√
6
2
+
√
2
2
i
)]
=
π
2
√
2
27.
∮ˇ
C
1
z(ez − 1) dz = 2πiRes(f(z), 0) = −πi. Note: z = 0 is a pole of order two, and so
Res(f(z), 0) = lim
z→0
d
dz
z2 · 1
z2
(
1 +
z
2!
+
z2
3!
+ · · ·
) = lim
z→0
−
(
1
2!
+
2z
3!
+ · · ·
)
(
1 +
z
2!
+
z2
3!
+ · · ·
)2 = −12 .
28.
∮ˇ
C
z
(z − 1)(z + 1)10 dz = 2πi[Res(f(z), 1) + Res(f(z),−1)] = 2πi
[
1
210
+
(
− 1
210
)]
= 0
29. Using two integrals,∮ˇ
C
ze3/z dz +
∮ˇ
C
sin z
z2(z − π)3 dz = 2πiRes(f(z), 0) + 2πi[Res(f(z), 0) + Res(f(z), π)]
= 2πi · 9
2
+ 2πi
[
− 1
π3
+
2
π3
]
=
(
9π +
2
π2
)
i.
Note: In the first integral z = 0 is an essential singularity and the residue is obtained from the Laurent series
ze3/z = · · ·+ 3
3
3!z2
+
32
2!z
+ 3 + z.
30.
∮ˇ
C
cscπz dz = 2πi[Res(f(z), 0) + Res(f(z), 1) + Res(f(z), 2)] = 2πi
[
1
π
+
(
− 1
π
)
+
1
π
]
= 2i
31.
∫ ∞
−∞
x2
(x2 + 2x + 2)(x2 + 1)2
dx = 2πi[Res(f(z),−1 + i) + Res(f(z), i)] = 2πi
[
3
25
− 4
25
i− 3
25
+
9
100
i
]
=
7π
50
32.
∫ ∞
−∞
x + ai
x2 + a2
eix dx =
∫ ∞
−∞
x cosx− a sinx
x2 + a2
dx + i
∫ ∞
−∞
a cosx + x sinx
x2 + a2
dx = 0 + 2πiRes(f(z), ai) = 2πie−a
Thus, ∫ ∞
−∞
a cosx + x sinx
x2 + a2
dx = 2πe−a.
33.
∫ 2π
0
cos2 θ
2 + sin θ
dθ =
1
2
∮ˇ
C
z4 + 2z2 + 1
z2(z2 + 4iz − 1) dz (C is |z| = 1) = πi[Res(f(z), 0) + Res(f(z), (−2 +
√
3 )i)]
= πi[−4i + 2
√
3 i] = (4− 2
√
3 )π
[Note: The answer in the text is correct but not simplified.]
34.
∫ 2π
0
cos 3θ
5− 4 cos θ dθ = −1
2i
∮ˇ
C
z6 + 1
z3(z − 2)(2z − 1) dz (C is |z| = 1) = −π
[
Res(f(z), 0) + Res
(
f(z),
1
2
)]
= −π
[
21
8
+
(
−65
24
)]
=
π
12
917
CHAPTER 19 REVIEW EXERCISES
35. The integrand of
∮ˇ
C
1− eiz
z2
dz has a simple pole at z = 0. Using a contour as in Figure 19.14 of Section 19.6
we have ∮ˇ
C
=
∫
CR
+
∫ −r
−R
+
∫
−CR
+
∫ R
r
= 0.
By taking limits as R → 0 and as r → 0, and using Theorem 19.17 we find
P.V.
∫ ∞
−∞
1− eix
x2
dx− πiRes(f(z), 0) = 0.
Thus,
P.V.
∫ ∞
−∞
1− cosx− i sinx
x2
dx = π.
Equating real parts gives
P.V.
∫ ∞
−∞
1− cosx
x2
dx = π.
Finally, ∫ ∞
0
1− cosx
x2
dx =
1
2
∫ ∞
−∞
1− cosx
x2
dx =
π
2
.
36. We have
Ce−a
2z2eibz dz =
∫ r
−r
+
∫
C1
+
∫
C2
+
∫
C3
= 0
by the Cauchy-Goursat Theorem. Therefore,∫ r
−r
= −
∫
C1
−
∫
C2
−
∫
C3
.
Let C1 and C3 denote the vertical sides of the rectangle. By the ML-inequality,
∫
C1
→ 0 and
∫
C3
→ 0 as
r →∞. On C2, z = x + b2a2 i, −r ≤ x ≤ r, dz = dx,∫ ∞
−∞
e−ax
2
eibx dx = −
∫ −∞
∞
e−a
2(x+ b2a2 i)
2
eib(x+
b
2a2
i) dx =
∫ ∞
−∞
e−a
2x2e−b
2/4a2 dx∫ ∞
−∞
e−ax
2
(cos bx + i sin bx) dx = e−b
2/4a2
∫ ∞
−∞
e−a
2x2 dx.
Using the given value of
∫ ∞
−∞
e−a
2x2 dx and equating real and imaginary parts gives
∫ ∞
−∞
e−ax
2
cos bx dx =
√
π
a
e−b
2/4a2 and so
∫ ∞
0
e−ax
2
cos bx dx =
√
π
2a
e−b
2/4a2 .
37. ak =
1
2πi
∮ˇ
C
e(u/2)(z−z
−1)
zk+1
dz =
1
2πi
∫ 2π
0
e(u/2)(e
it−e−it)
(eit)k+1
ieit dt =
1
2π
∫ 2π
0
e(u/2)(2i sin t)e−kit dt
=
1
2π
∫ 2π
0
e−i(kt−u sin t)dt =
1
2π
∫ 2π
0
[cos(kt− u sin t)− i sin(kt− u sin t)] dt = 1
2π
∫ 2π
0
cos(kt− u sin t) dt
since
∫ 2π
0
sin(kt− u sin t) dt = 0. (To obtain this last result, expand the integrand and let t = 2π − x.)
918