1. The document provides solutions to homework problems from a complex analysis class. 2. It shows the work to find harmonic conjugates and derivatives of complex functions, evaluate complex expressions, and take logarithms and exponents of complex numbers. 3. Key steps include using the Cauchy-Riemann equations to test if functions are analytic, decomposing complex expressions into polar form, and applying properties of logarithms and exponents to manipulate expressions.

1. Homework #5 Solutions
Math 128, Fall 2013
Instructor: Dr. Doreen De Leon
1 p. 81: 1 (b), (c)
1. Show that u(x, y) is harmonic in some domain and find a harmonic conjugate v(x, y).
(b) u(x, y) = 2x3 + 3xy2
ux = 2 − 3x2
+ 3y2
=⇒ uxx = −6x
uy = 6xy =⇒ uyy = 6x.
Then,
uxx + uyy = −6x + 6x = 0.
So, u(x, y) is harmonic for all (x, y). Now to find a harmonic conjugate v(x, y).
ux = 2 − 3x2
+ 3y2
= vy
=⇒ v = (2 − 3x2
+ 3y2
) dy
= 2y − 3x2
y + y3
+ g(x).
uy = 6xy = −vx
=⇒ −(−6xy + g (x)) = 6xy
6xy − g (x) = 6xy
g (x) = 0 =⇒ g(x) = c.
So, v(x, y) = 2y − 3x2y + y3 + c. If we assume v(0, 0) = 0, then 0 = c. So,
v(x, y) = 2y − 3x2
y + y3
.
(c) u(x, y) = sinh x sin y
ux = cosh x sin y =⇒ uxx = sinh x sin y
uy = sinh x cos y =⇒ uyy = − sinh x sin y.
Then,
uxx + uyy = sinh x sin y + (− sinh x sin y) = 0.
1

2. So, u(x, y) is harmonic for all (x, y). Now, we need to find a harmonic conjugate v(x, y).
ux = cosh x sin y = vy
=⇒ v = (cosh x sin y) dy
= − cosh x cos y + g(x).uy = sinh x cos y = −vx
=⇒ −(− sinh x cos y + g (x)) = sinh x cos y
sinh x cos y − g (x) = sinh x cos y
g (x) = 0 =⇒ g(x) = c.
So, v(x, y) = − cosh x cos y + c. If we asume v(0, 0) = −1, then c = 0, and
v(x, y) = − cosh x cos y .
2 p. 92: 1, 4, 8
1. Show that
(a) exp(2 ± 3πi) = −e2
e2+3πi = e2e3πi e2−3πi = e2e−3πi
= e2(−1) = e2(−1)
= −e2 = −e2.
.
(b) exp
2 + πi
4
=
e
2
(1 + i)
exp
2 + πi
4
= exp
1
2
+
π
4
i
= e
1
2 · e
π
4
i
=
√
e cos
π
4
+ i sin
π
4
=
√
e
1
√
2
+ i
1
√
2
=
e
2
(1 + i).
(c) exp(z + πi) = − exp(z)
exp(z + πi) = ez+πi
= ez
eπi
= −ez
= − exp(z).
4. Show in two ways that the function f(z) = exp(z2) is entire. What is its derivative?
2

3. (I) Let g(z) = ez
and h(z) = z2. g and h are both entire. Therefore, (g ◦ h)(z) is entire.
Therefore, f(z) = (g ◦ h)(z) is entire.
(II)
f(z) = ez2
= e(x+iy)2
= e(x2−y2)+i(2xy)
= ex2−y2
· ei(2xy)
= ex2−y2
(cos(2xy) + i sin(2xy))
= ex2−y2
cos(2xy) + iex2−y2
sin(2xy).
u(x, y) = ex2−y2
cos(2xy) v(x, y) = ex2−y2
sin(2xy)
ux = 2xex2−y2
cos(2xy) − 2yex2−y2
sin(2xy)
uy = −2yex2−y2
cos(2xy) − 2xex2−y2
sin(2xy)
vx = 2xex2−y2
sin(2xy) + 2yex2−y2
cos(2xy)
vy = −2yex2−y2
sin(2xy) + 2xex2−y2
cos(2xy)



exist and continuous for all z.
It is clear that ux = vy and uy = −vx, so the Cauchy-Riemann equations are satisfied for all
z. Therefore, f(z) is differentiable everywhere and, thus, f(z) is entire.
Then,
f (z) =
d
dz
ez2
= 2zez2
.
8. Find all values of z such that
(a) ez = −2
First, exeiy = 2eiπ. So,
ex
= 2 and y = π + 2nπ, n ∈ Z
x = ln 2 and y = π(2n + 1), n ∈ Z.
So, z = ln 2 + (2n + 1)πi, n ∈ Z.
(b) ez = 1 +
√
3i
1 +
√
3i = reiθ
where r = 12 + (
√
3)2 = 2, θ = tan−1
√
3
1
=
π
3
.
So,
ex
eiy
= 2eiπ
3
=⇒ ex
= 2 and y =
π
3
+ 2nπ, n ∈ Z
=⇒ x = ln 2.
So, z = ln 2 +
π
3
+ 2nπ i, n ∈ Z.
(c) exp(2z − 1) = 1
1 = 1ei0
exp(2z − 1) = exp(2(x + iy) − 1)
= exp[(2x − 1) + i(2y)]
= e2x−1
ei(2y)
.
3

4. So,
e2x−1
ei(2y)
= ei(0)
=⇒ e2x−1
= 1 and 2y = 0 + 2nπ, n ∈ Z
=⇒ 2x − 1 = ln 1 and y = nπ, n ∈ Z
=⇒ x =
1
2
.
So, z =
1
2
+ nπi, n ∈ Z.
3 p. 97: 2, 3, 7, 9(a)
2. Show that
(a) log e = 1 + 2nπi, n ∈ Z
e = e · ei0
(so, r = e, θ = 0)
=⇒ log e = ln e + i(0 + 2nπ), n ∈ Z
= 1 + 2nπi, n ∈ Z.
(b) log i = 2n +
1
2
πi, n ∈ Z
i = 1ei(π
2 )
=⇒ log i = ln 1 + iπ
1
2
+ 2nπ , n ∈ Z
= 0 + iπ
1
2
+ 2n , n ∈ Z.So, log i = 2n +
1
2
πi, n ∈ Z.
(c) log(−1 +
√
3i) = ln 2 + 2 1 +
1
3
πi, n ∈ Z
−1 +
√
3i = reiθ
, where r = (−1)2 + (
√
3)2 = 2, θ = tan−1
√
3
−1
=
π
3
So, log(−1 +
√
3i) = log 2ei(2π
3 )
= ln 2 + i
2π
3
+ 2nπ , n ∈ Z
= ln 2 + i2π
1
3
+ n , n ∈ Z.
Then, log(−1 +
√
3i) = ln 2 + 2 n +
1
3
πi, n ∈ Z.
3. Show that
4

5. (a) Log (1 + i)2 = 2Log (1 + i)
Log (1 + i) = Log
√
2ei π
4
= ln
√
2 + i
π
4
=
1
2
ln 2 +
π
4
i.So, 2Log (1 + i) = ln 2 +
π
2
i.
(1 + i)2
= 2i and
Log (2i) = Log 2ei π
2
= ln 2 +
π
2
i = 2Log (1 + i).
(b) Log (−1 + i)2 = 2Log (−1 + i)
Log (−1 + i) = Log
√
2ei(3π
4 )
= ln
√
2 + i
3π
4
=
1
2
ln 2 +
3π
4
i.So, 2Log (−1 + i) = ln 2 +
3π
2
i.
But, Log (−1 + i)2
= Log (−2i)
= Log 2ei(−π
2 )
= ln 2 −
π
2
i = 2Log (−1 + i).
7. Find all roots of the equation log z = i
π
2
.
log z = ln |z| + i(θ + 2nπ), n ∈ Z
=⇒ ln |z| + i(θ + 2nπ) = i
π
2
=⇒ ln |z| = 0 and θ + 2nπ =
π
2
=⇒ |z| = 1 and θ =
π
2
− 2nπ =
π
2
+ 2nπ.
So, z = 1ei(π
2
+2nπ) = i .
9(a) Show that the function f(z) = Log (z − i) is analytic everywhere excepte on the portion x ≤ 0 of
the line y = 1.
Log (z − i) = Log (x + yi − i)
= Log (x + (y − 1)i).
In class, we saw that Log z is analytic for all z except on the set {(x, y)|x ≤ 0 and y = 0} (i.e., for
all z except the portion x ≤ 0 of the x-axis). Therefore, Log (z − i) is analytic for all z except on
the set
{(x, y)|x ≤ 0 and y − 1 = 0} or {(x, y)|x ≤ 0 andy = 1},
i.e., everywhere except the portion x ≤ 0 of the line y = 1.
5

6. 4 p. 100: 2
2. Show that for any two nonzero complex numbers z1 and z2
Log (z1z2) = Log z1 + Log z2 + 2Nπi,
where N has one of the values 0, ± 1.
Let z1 = r1eiθ1
and z2 = r2eiθ2
, where θ1 = Arg z1 and θ2 = Arg z2. Then, z1z2 = r1r2ei(θ1+θ2)
. So,
Arg (z1z2) =



θ1 + θ2 if − π < θ1 + θ2 ≤ π
θ1 + θ2 − 2π if θ1 + θ2 > π
θ1 + θ2 + 2π if θ1 + θ2 ≤ −π
.
=⇒ Arg (z1z2) = θ1 + θ2 + 2Nπ, N ∈ {0, ±1}
= Arg z1 + Arg z2 + 2Nπ, N ∈ {0, ±1}.
So,
Log (z1z2) = Log r1r2ei(Arg (z1z2))
= ln(r1r2) + iArg (z1z2)
= ln r1 + ln r2 + i(Arg z1 + Arg z2 + 2Nπ), N ∈ {0, ±1}
= (ln r1 + Arg z1) + (ln r2 + Arg z2) + 2Nπ, N ∈ {0, ±1}
= Log z1 + Log z2 + 2Nπ, N ∈ {0, ±1}.
5 p. 104: 1, 2
1. Show that
(a) (1 + i)i
= exp −
π
4
+ 2nπ exp i
ln 2
2
, n ∈ Z
(1 + i)i
= ei log(1+i)
.
log(1 + i) = log(
√
2ei π
4 )
= ln
√
2 + i
π
4
+ 2nπ , n ∈ Z
=
1
2
ln 2 + i
π
4
+ 2nπ , n ∈ Z.
So, (1 + i)i
= ei[1
2
ln 2+i(π
4
+2nπ)], n ∈ Z
= ei1
2
ln 2−(π
4
+2nπ), n ∈ Z
= e−π
4
−2nπ
ei(ln 2
2 ), n ∈ Z
= e−π
4
−2nπ
ei(ln 2
2 ), n ∈ Z.
6

7. (b) (−1)
1
π = e(2n+1)i
, n ∈ Z
(−1)
1
π = e
1
π
log(−1)
.
log(−1) = log(1eiπ
)
= ln 1 + i(π + 2nπ), n ∈ Z
= (2n + 1)πi, n ∈ Z.
So, (−1)
1
π = e
1
π
(2n+1)πi
, n ∈ Z
= e(2n+1)i
, n ∈ Z.
2. Find the principal value of
(a) ii
ii
= eiLog i
.
Log i = Log (eiπ
2 )
= ln 1 + i
π
2
= i
π
2
.
So, ii
= ei(i π
2 )
= e−π
2 .
(b)
e
2
(−1 −
√
3i)
3πi
−1 −
√
3i = reiθ
, −π < θ ≤ π.
r = (−1)2 + (
√
3)2 = 2,
θ = tan−1 −
√
3
−1
= −
2π
3
.
So, − 1 −
√
3i = 2e−i(2π
3 ).
e
2
(−1 −
√
3i) =
e
2
e−i(2π
3 )
= e · e−i(2π
3 ).
Log
e
2
(−1 −
√
3i) = Log e · e−i(2π
3 )
= ln e −
2π
3
i
= 1 −
2π
3
i.
So,
e
2
(−1 −
√
3i)
3πi
= exp 3πi 1 −
2π
3
= exp[3πi + 2π2
]
= −e2π2
.
7

8. (c) (1 − i)4i
(1 − i)4i
= e4iLog (1−i)
1 − i = reiθ
, −π < θ ≤ π.
r = 12 + (−1)2 =
√
2,
θ = tan−1 −1
1
= −
π
4
.
So, 1 − i =
√
2ei(−π
4 ).Log (1 − i) = ln
√
2 −
π
4
i
=
1
2
ln 2 −
π
4
. =⇒ (1 − i)4i
= e4i(1
2
ln 2−π
4
i)
= eπ+2 ln 2i
= eπ
e2 ln 2i
= eπ
(cos(2 ln 2) + i sin(2 ln 2)) .
8