This document provides solutions to problems from a complex analysis homework assignment. The solutions involve applying theorems related to contour integrals, Cauchy's integral formula, and derivatives to evaluate definite integrals over curves and find derivatives of functions. Key steps include using the Cauchy-Goursat theorem when functions are analytic over regions, and applying Cauchy's integral formula and its generalization to derivatives to evaluate integrals that involve singularities inside curves.

1. NCU Math, Spring 2014: Complex Analysis Homework Solution 5
Text Book: An Introduction to Complex Analysis
Problem. 16.3
Sol:
(a)
Since (eπz
π ) = eπz
, Theorem 14.1, and Theorem 14.2, we have
ˆ i
2
i
eπz
dz =
eπz
π
i
2
i
=
ei π
2
π
−
eiπ
π
=
i
π
−
−1
π
=
1 + i
π
.
(b)
Since (2 sin(z
2 )) = cos(z
2 ), Theorem 14.1, and Theorem 14.2, we have
ˆ π+2i
0
cos(
z
2
)dz = 2 sin(
z
2
)
π+2i
0
= 2 sin(
π
2
+ i) − 2 sin(0)
= 2 ·
ei( π
2 +i)
− e−i( π
2 +i)
2i
=
e−1+i π
2 − e1−i π
2
i
=
e−1
i + ei
i
= e−1
+ e.
(c)
Since (1
4 (z − 3)4
) = (z − 3)3
, Theorem 14.1, and Theorem 14.2, we have
ˆ 3
1
(z − 3)3
dz =
1
4
(z − 3)4
3
1
=
1
4
(3 − 3)4
−
1
4
(1 − 3)4
= 0 − 4
= −4.
1

2. Problem. 16.8
Sol:
(a)
Since
f(z) =
1
z2 + 2z + 2
=
1
(z + 1)2 + 1
=
1
(z + 1 + i)(z + 1 − i)
,
we can get that f(z) is analytic if z = −1 ± i. Because −1 ± i are not in closed disc |z| ≤ 1, we have
´
γ
f(z)dz = 0
by Theorem 15.2 (Cauchy-Goursat Theorem).
(b)
It is easy to see that f(z) is analytic in C by virtue of z and e−z
are analytic in C. So by Theorem 15.2
(Cauchy-Goursat Theorem), we have
´
γ
f(z)dz = 0 .
(c)
Since f(z) = 1
(2z−i)2 , we can get that f(z) is analytic if z = i
2 . Because i
2 are in closed disc |z| ≤ 1, we can
not use Theorem 15.2 (Cauchy-Goursat Theorem). But from Theorem 16.1, we have
´
γ
f(z)dz =
´
|z− i
2 |= 1
4
f(z)dz.
This implies that
ˆ
γ
f(z)dz =
ˆ
|z− i
2 |= 1
4
1
4(z − i
2 )2
dz
=
ˆ 2π
0
1
4(1
4 eiθ)2
1
4
ieiθ
dθ
=
ˆ 2π
0
ie−iθ
dθ
= −eiθ 2π
0
= 0.
Problem. 16.9
Sol:
2

3. (b)
It is easy to see that z
1−ez is analytic if z = 2kπi for all k ∈ Z. Because for all k ∈ Z, 2kπi is not in the domain
between the circle |z| = 4 and the square whose sides lie along the lines x = ±1, y = ±1, we have
´
γ
f(z)dz = 0 by
Theorem 15.2 (Cauchy-Goursat Theorem).
Problem. 16.10
Sol:
(a)
Since f(z) = ez
(z+3i)(z−3i) , we can get that f(z) is analytic if z = ±3i. Because f(z) is analytic in the domain
between the circles |z| = 1 − δ and |z| = 2 + δ for δ > 0 small, we have
´
γ
f(z)dz = 0 by Theorem 15.2 (Cauchy-
Goursat Theorem).
Problem. 16.11
Sol:
(b)
Since 1
3z2+1 = 1
3(z+ i√
3
)(z− i√
3
)
, we can get that f(z) is analytic if z = ± i√
3
. So
ˆ
γ
1
3z2 + 1
dz =
ˆ
γ
1
2
√
3i(z − i√
3
)
dz −
ˆ
γ
1
2
√
3i(z + i√
3
)
dz
= 2
ˆ
|z|=1
1
2
√
3i(z − i√
3
)
dz − 2
ˆ
|z|=1
1
2
√
3i(z + i√
3
)
dz
But we can have that
´
|z|=1
1
2
√
3i(z− i√
3
)
dz = 2πi
2
√
3i
= π√
3
and
´
|z|=1
1
2
√
3i(z+ i√
3
)
dz = 2πi
2
√
3i
= π√
3
by virtue of Example
16.1. Thus
´
γ
1
3z2+1 dz = 0.
Problem. 16.12
Sol:
It is easy to see that R+z
(R−z)z is analytic if z = R and z = 0. Then
ˆ
γr
R + z
(R − z)z
dz =
ˆ
γr
R + z
R(R − z)
dz +
ˆ
γr
R + z
Rz
dz.
3

4. Because of Theorem 15.2 (Cauchy-Goursat Theorem) and Theorem 17.1 (Cauchy's Integral Formula), we have´
γr
R+z
R(R−z) dz = 0 and
´
γr
R+z
Rz dz = R
R · 2πi = 2πi. Thus 1
2πi
´
γr
R+z
(R−z)z dz = 1.
By the denition, we can get that
1
2πi
ˆ
γr
R + z
(R − z)z
dz =
1
2πi
ˆ 2π
0
R + reiθ
(R − reiθ)reiθ
ireiθ
dθ
=
1
2π
ˆ 2π
0
R + reiθ
R − reiθ
dθ
From Problem 3.17, we have
Re
R + reiθ
R − reiθ
=
R2
− r2
R2 − 2Rr cos θ + r2
.
This implies that
1
2π
ˆ 2π
0
R2
− r2
R2 − 2Rr cos θ + r2
dθ =
1
2π
ˆ 2π
0
Re
R + reiθ
R − reiθ
dθ
= Re
1
2π
ˆ 2π
0
R + reiθ
R − reiθ
dθ
= Re
1
2πi
ˆ
γr
R + z
(R − z)z
dz
= 1.
Problem. 18.4
Sol:
By Theorem 18.2 (Cauchy's Integral Formula for Derivatives), we have L.H.S. = 2πif (z0). From Theorem 17.1
(Cauchy's Integral Formula), we can get that R.H.S. = 2πif (z0). Thus L.H.S. = R.H.S.
Problem. 18.6
Sol:
For z is inside γ, we have
ˆ
γ
ξ3
+ 7ξ
(ξ − z)3
dξ =
2πi
2!
· 6z
= 6πiz
by Theorem 18.2 (Cauchy's Integral Formula for Derivatives) and (ξ3
+ 7ξ) = 6ξ.
4

5. It is easy to see that ξ3
+7ξ
(ξ−z)3 is analytic if ξ = z. So for z is outside γ, we can get that
´
γ
ξ3
+7ξ
(ξ−z)3 dξ = 0 by Theorem
15.2 (Cauchy-Goursat Theorem).
Thus
g(z) =
6πiz if z is inside γ,
0 if z is outside γ.
Problem. 18.7
Sol:
It is easy to see that f(z) is analytic in C. By Theorem 18.2 (Cauchy's Integral Formula for Derivatives) and
f (z) = ez
−e−z
/2, we have
ˆ
γ
f(z)
z4
dz =
2πi
3!
· (
1 − 1
2
)
= 0.
Problem. 18.9
Sol:
(b)
It is easy to see that cos z is analytic in C. By Theorem 18.2 (Cauchy's Integral Formula for Derivatives) and
(cos z) = sin z, we have
ˆ
γ
f(z)
z4
dz =
2πi
3!
· sin 0
= 0.
(c)
It is easy to see that z is analytic in C. By Theorem 18.2 (Cauchy's Integral Formula for Derivatives) and
(z) = 1, we have
ˆ
γ
z
(2z + 1)2
dz =
ˆ
γ
z
4(z + 1
2 )2
dz
=
1
4
·
2πi
1!
· 1
=
πi
2
.
5

6. (e)
By Example 16.1, we have
ˆ
γ
1
z(z + 1)
dz =
ˆ
γ
1
z
dz −
ˆ
γ
1
(z + 1)
dz
= 2πi − 2πi
= 0.
(f)
It is easy to see that ez
sin z is analytic in C. So
´
γ
ez
sin zdz = 0 by Theorem 15.2 (Cauchy-Goursat Theorem).
Since
1
(z2 + 3)2
=
z − 2
√
3i
12
√
3i(z −
√
3i)2
−
z + 2
√
3i
12
√
3i(z +
√
3i)2
and z−2
√
3i
12
√
3i(z−
√
3i)2
is analytic if z =
√
3i, we have
ˆ
γ
1
(z2 + 3)2
dz =
ˆ
γ
z − 2
√
3i
12
√
3i(z −
√
3i)2
dz −
ˆ
γ
z + 2
√
3i
12
√
3i(z +
√
3i)2
dz
= −
ˆ
γ
z + 2
√
3i
12
√
3i(z +
√
3i)2
dz
by virtue of Theorem 15.2 (Cauchy-Goursat Theorem). From Theorem 18.2 (Cauchy's Integral Formula for Deriva-
tives) and z+2
√
3i
12
√
3i(z+
√
3i)2
is analytic if z = −
√
3i, we can get that
ˆ
γ
1
(z2 + 3)2
dz =
2πi
1!
−
1
12
√
3i
= −
π
6
√
3
.
Thus
´
γ
ez
sin z + 1
(z2+3)2 dz = − π
6
√
3
.
Problem. 18.11
Sol:
Let γr be the curve of |z| = r where 0 r 1. So we have
ˆ
γr
f(z)
z2
dz ≤ 2πr ·
1
r2(1 − r)
=
2π
r(1 − r)
6

7. by Theorem 13.1 (ML-Inequality) and f(z)
z2 ≤
1
1−|z|
|z2| = 1
r2(1−r) for all |z| = r 1. This implies that
|f (0)| =
1!
2πi
ˆ
γr
f(z)
z2
dz
=
1!
2πi
·
ˆ
γr
f(z)
z2
dz
≤
1
2π
·
2π
r(1 − r)
=
1
r(1 − r)
from Theorem 18.2 (Cauchy's Integral Formula for Derivatives). Since 0 r 1 arbitrary and 1
r(1−r) ≤ 1
4 for all
0 r 1 by easy computation, we can get that |f (0)| ≤ 1
4 .
Problem. 18.13
Sol:
Fix z ∈ C. We dene γz,r by |ξ − z| = r. So we have
ˆ
γz,r
f (ξ)
(z − ξ)2
dξ ≤ 2πr ·
r + |z|
r2
= 2π(1 +
|z|
r
)
by Theorem 13.1 (ML-Inequality) and |f (ξ)| ≤ |ξ| ≤ |ξ − z| + |z| = r + |z| for all |ξ − z| = r. From Theorem 18.2
(Cauchy's Integral Formula for Derivatives), we can get that
|f (z)| =
1!
2πi
ˆ
γz,r
f (ξ)
(z − ξ)2
dξ
=
1!
2πi
·
ˆ
γz,r
f (ξ)
(z − ξ)2
dξ
≤
1
2π
· 2π(1 +
|z|
r
)
= 1 +
|z|
r
.
Since r 0 arbitrary,
|f (z)| = lim
r→∞
|f (z)|
≤ lim sup
r→∞
1 +
|z|
r
= 1.
7

8. So we have |f (z)| ≤ 1 for all z ∈ C. This implies that f (z) = c for some constant c ∈ C by Theorem 18.6
(Liouville's Theorem). Because |f (z)| ≤ 1 for all z ∈ C, we can get that |c| ≤ 1.
From Theorem 14.1, we have f (z) − f (0) =
´ z
0
f (ξ)dξ = bz. Similarly, we can get that f(z) = f(0) + f (0)z +
c
2 z2
. Because |f (z)| ≤ |z|, we have |f (0)| ≤ |0| = 0. So f (0) = 0. We dene a = f(0) ∈ C and b = c
2 ∈ C. Then
f(z) = a + bz2
where a, b ∈ C and |b| ≤ 1.
Problem. 18.15
Sol:
We dene h(z) = f(z)
g(z) . Since g(z) = 0 and f, g are entire functions, we can get that h(z) is an entire function.
Because |h(z)| = f(z)
g(z) ≤ 1 and h(z) is an entire function, we have h(z) ≡ c for some constant c ∈ C by Theorem
18.6 (Liouville's Theorem). Thus f(z) = cg(z) for some constant c ∈ C.
8