This document contains solutions to homework problems from a complex analysis course. It solves problems involving logarithms, exponentials, trigonometric and hyperbolic functions of complex numbers. Key steps and solutions are shown for problems involving contour integrals of complex functions along curves in the complex plane. The length of one such curve is computed to be 8a. Several contour integrals are evaluated, with one found to be equal to 1 - i. Bounds on contour integrals are also determined using theorems.

1. NCU Math, Spring 2014: Complex Analysis Homework Solution 4
Text Book: An Introduction to Complex Analysis
Problem. 9.1
Sol:
(a)
Since log(−2) = ln 2 + iπ + 2kπi for all k ∈ Z, we have z = ln 2 + iπ + 2kπi for all k ∈ Z are the solutions of
ez
= −2.
(b)
Since log(1 +
√
3i) = ln 2 + iπ
3 + 2kπi for all k ∈ Z, we have z = ln 2 + iπ
3 + 2kπi for all k ∈ Z are the solutions
of ez
= 1 +
√
3i.
(c)
Since log 1 = 2kπi for all k ∈ Z, we have 2z − 1 = 2kπi for all k ∈ Z would solveexp(2z − 1) = 1. Thus
z = 1
2 + kπi for all k ∈ Z are the solutions of exp(2z − 1) = 1.
(d)
From (9.5), we can get that the solutions of sin z = 2 are
z = sin−1
(2)
= −i log(2i + (1 − 22
)
1/2
)
=
−i log(2i +
√
3i)
−i log(2i −
√
3i)
=
−i(ln(2 +
√
3) + iπ
2 + 2kπi)
−i(ln(2 −
√
3) + iπ
2 + 2kπi)
=
−i ln(2 +
√
3) + π
2 + 2kπ
−i ln(2 −
√
3) + π
2 + 2kπ
for all k ∈ Z.
Problem. 9.4
Sol:
(a)
1

2. We have
Log(−ei) = ln e + iArg(−ei)
= 1 − i
π
2
.
(b)
We have
Log(1 − i) = ln
√
2 + iArg(1 − i)
=
1
2
ln 2 − i
π
4
.
(c)
We have
log(−1 +
√
3i) = ln 2 + iArg(−1 +
√
3i)
= ln 2 + i
2π
3
.
Problem. 9.5
Sol:
(a)
From the denition, we have
Log(1 + i)2
= ln |1 + i|2
+ iArg((1 + i)2
)
= ln 2 + i
π
2
,
and
Log(1 + i) = ln |1 + i| + iArg((1 + i))
= ln
√
2 + i
π
4
.
Thus
Log(1 + i)2
= ln 2 + i
π
2
= 2(ln
√
2 + i
π
4
)
= 2Log(1 + i).
(b)
2

3. From the denition, we have
Log(−1 + i)2
= ln | − 1 + i|2
+ iArg((−1 + i)2
)
= ln 2 − i
π
2
,
and
Log(−1 + i) = ln | − 1 + i| + iArg((−1 + i))
= ln
√
2 + i
3π
4
.
Thus
Log(−1 + i)2
= ln 2 − i
π
2
= 2(ln
√
2 − i
π
4
)
= 2Log(−1 + i).
Problem. 9.7
Sol:
(a)
From the denition, we have
(1 + i)i
= eiLog(1+i)
= ei(ln
√
2+i π
4 )
= e− π
4 +i ln 2
2 .
(b)
From the denition, we have
(−1)π
= eπLog(−1)
= eπ(ln 1+iπ)
= eiπ2
.
(c)
From the denition, we have
(1 − i)4i
= e4iLog(1−i)
= e4i(ln
√
2−i π
4 )
= eπ+i2 ln 2
.
3

4. (d)
From the denition, we have
(−1 + i
√
3)
3/2
= e
3
2 Log(−1+
√
3i)
= e
3
2 (ln 2+i 2π
3 )
= e
3 ln 2
2 +iπ
= −2
√
2.
Problem. 9.9
Sol:
(a)
From the denition, we have
sin−1
√
5 = −i log(i
√
5 + (1 −
√
5
2
)
1/2
)
= −i log(i
√
5 + (−4)
1/2
)
=
−i log(i
√
5 + i2)
−i log(i
√
5 − i2)
=
−i(ln(
√
5 + 2) + π
2 i + 2kπi)
−i(ln(
√
5 − 2) + π
2 i + 2kπi)
=
−i ln(
√
5 + 2) + (π
2 + 2kπ)
−i ln(
√
5 − 2) + (π
2 + 2kπ)
for all k ∈ Z.
(b)
From the denition, we have
sinh−1
i = log(i + (1 + i2
)
1/2
)
= log(i + (0)
1/2
)
= log i
= ln 1 + i
π
2
+ 2kπi
= (
π
2
+ 2kπ)i
for all k ∈ Z.
4

5. Problem. 13.1
Sol:
(b)
It is easy to see that
ˆ π
0
(sin 2t + i cos 2t)dt =
ˆ π
0
sin 2tdt + i
ˆ π
0
cos 2tdt
=
1
2
(− cos 2t|
π
0 ) + i
1
2
(sin 2t|
π
0 )
= 0.
(d)
Since for all t ∈ [1, 2]
Log(1 + it) = (ln |1 + it| + iArg(1 + it))
=
ln(1 + t2
)
2
+ i arctan t,
we have
ˆ 2
1
Log(1 + it)dt =
ˆ 2
1
ln(1 + t2
)
2
+ i arctan t dt
=
t ln(1 + t2
)
2
2
1
−
ˆ 2
1
t2
1 + t2
dt + i t arctan t|
2
1 −
ˆ 2
1
t
1 + t2
dt
= ln 5 −
ln 2
2
− t − arctan t|
2
1 + i 2 arctan 2 − arctan 1 −
1
2
ln(1 + t2
)
2
1
= ln 5 −
ln 2
2
− (2 − arctan 2 − 1 + arctan 1) + i 2 arctan 2 −
π
4
−
1
2
ln 5 +
1
2
ln 2
= ln 5 −
ln 2
2
− 1 + arctan 2 −
π
4
+ i 2 arctan 2 −
π
4
−
1
2
ln 5 +
1
2
ln 2 .
Problem. 13.2
Sol:
5

6. The length of this curve is
ˆ 2π
0
|z (t)|dt =
ˆ 2π
0
|a(1 − cos t) + ai sin t|dt
=
ˆ 2π
0
a2(1 − cos t)2 + a2 sin2
tdt
=
ˆ 2π
0
a 1 − 2 cos t + cos2 t + sin2
tdt
= a
ˆ 2π
0
√
2 − 2 cos tdt
= a
ˆ 2π
0
4 sin2 t
2
dt
= 2a
ˆ 2π
0
| sin
t
2
|dt
= 2a
ˆ 2π
0
sin
t
2
dt
= −4a cos
t
2
2π
0
= 8a.
Problem. 13.3
Sol:
From the denition, we have
ˆ
γ
|z|2
dz =
ˆ 2π
0
|z(t)|2
· z (t)dt
=
ˆ 2π
0
|t + it2
|2
· (1 + 2ti)dt
=
ˆ 2π
0
(t2
+ t4
) · (1 + 2ti)dt
=
ˆ 2π
0
(t2
+ t4
)dt + 2i
ˆ 2π
0
(t3
+ t5
)dt
=
8π3
3
+
32π5
5
+ 2i(4π4
+
32π6
3
)
=
40π3
+ 96π5
15
+ s(
24π4
+ 64π6
3
)i.
6

7. Problem. 13.4
Sol:
Let z(t) = (1 + i)t for t ∈ [0, 1] be the curve γ. Then we have
ˆ
γ
f(z)dz =
ˆ 1
0
f(z(t)) · z (t)dt
=
ˆ 1
0
(t − t − 3it2
) · (1 + i)dt
= −3i(1 + i)
ˆ 1
0
t2
dt
= −i(1 + i)
= 1 − i.
Problem. 13.5
Sol:
From the denition, we have
ˆ
γ
z − 2
z
dz =
ˆ π
0
z(θ) − 2
z(θ)
· z (θ)dθ
=
ˆ π
0
2eiθ
− 2
2eiθ
· 2ieiθ
dθ
=
ˆ π
0
2i(cos θ + i sin θ − 1)dθ
= 2i (sin θ − θ − i cos θ|
π
0 )
= 2i(−π + 2i)
= −4 − 2πi.
For z = 2eiθ
and 0 ≤ θ ≤ π, we have
exp (iArgz) = exp iArg(2eiθ
)
= exp(iθ)
and
|z
1/2
| = exp
1
2
log(2eiθ
)
= exp
1
2
(ln 2 + iθ + 2kπi)
= exp
ln 2
2
+ i
θ
2
+ kπi)
=
√
2.
7

8. This implies that
ˆ
−γ
|z
1/2
| exp (iArgz) dz =
ˆ 0
π
|z
1/2
(θ)| exp (iArgz(θ)) · z (θ)dθ
=
ˆ 0
π
√
2eiθ
· 2ieiθ
dθ
= −2
√
2i
ˆ π
0
e2θi
dθ
= −
√
2e2θi
π
0
= 0.
Problem. 13.11
Sol:
By virtue of Theorem 15.2 (p97) and ez
is analytic, we have
´
γ
ez
dz = 0. Since |z| ≤ 4 for all z ∈ γ and the
length of γ is 12, we can get that
ˆ
γ
(ez
− z)dz =
ˆ
γ
zdz
≤ 4 · 12
= 48
by Theorem 13.1.
Problem. 13.13
Sol:
Let z(t) = Reit
where 0 ≤ t ≤ 2π. So we have
Logz2
(t) = Log|z2
(t)| + iArg(z(t))
= Log|R2
ei2t
| + iArg(Reit
)
=
2LogR + it if t ∈ [0, π],
2LogR + i(t − 2π) if t ∈ (π, 2π].
Since R 2 and 0 ≤ t ≤ 2π, we can get that
|Logz2
(t)| ≤ 2LogR + π.
8

9. For z ∈ γR, we have
|z2
+ z + 1| ≥ |z2
| − |z + 1|
= R2
− |z + 1|
Because
|z + 1| ≤ |z| + 1
= R + 1
and R 2, we can get that
R2
− |z + 1| ≥ R2
− R − 1
= R(R − 1) − 1
2(2 − 1) − 1
= 0.
This implies that |z2
+ z + 1| ≥ R2
− R − 1 for all z ∈ γR.
Thus
Logz2
z2 + z + 1
≤
2LogR + π
R2 − R − 1
for all z ∈ γR. By virtue of Theorem 13.1 and the lehgth of γR is 2πR, we can get
ˆ
γR
Logz2
z2 + z + 1
dz ≤ 2πR
π + 2LogR
R2 − R − 1
.
Problem. 13.14
Sol:
Let f(z) = u(x, y) + iv(x, y) and z(t) = x(t) + iy(t) : [a, b] → S be the smooth and counterclockwise curve γ
with z(a) = z(b) where z = x + iy and u, v, x, y ∈ R. Then f (z) = ux + ivx and f(z) = u − iv. From the denition
of the integral along γ, we have
I =
ˆ b
a
f(z(t))f (z(t))z (t)dt
=
ˆ b
a
(u(t) − iv(t))(ux(t) + ivx(t))(x (t) + iy (t))dt
where we dene u(t) = u(x(t), y(t)), v(t) = v(x(t), y(t)), ux(t) = ux(x(t), y(t)), and vx(t) = vx(x(t), y(t)). Since
(u(t) − iv(t))(ux(t) + ivx(t))
= (u(t)ux(t) + v(t)vx(t)) + i (u(x)vx(t) − v(t)ux(t))
= (u(t)ux(t) + v(t)vx(t)) − i (u(x)uy(t) + v(t)vy(t))
9

10. by virtue of Cauchy-Riemann equation, we can get that
Re(I)
=Re
ˆ b
a
(u(t) − iv(t))(ux(t) + ivx(t))(x (t) + iy (t))dt
=Re
ˆ b
a
((u(t)ux(t) + v(t)vx(t)) − i (u(x)uy(t) + v(t)vy(t))) (x (t) + iy (t))dt
=
ˆ b
a
((u(t)ux(t) + v(t)vx(t)) x (t) + (u(x)uy(t) + v(t)vy(t)) y (t)) dt
It is easy to see that
(u(t)ux(t) + v(t)vx(t)) x (t) + (u(x)uy(t) + v(t)vy(t)) y (t)
=
d
dt
1
2
(u(x(t), y(t)))
2
+ (v(x(t), y(t)))
2
.
This implies that
Re(I)
=
ˆ b
a
d
dt
1
2
(u(x(t), y(t)))
2
+ (v(x(t), y(t)))
2
dt
=
1
2
(u(x(b), y(b)))
2
+ (v(x(b), y(b)))
2
−
1
2
(u(x(a), y(a)))
2
+ (v(x(a), y(a)))
2
=0
by z(a) = z(b). Thus I is purely imaginary.
10