This document provides solutions to homework problems from a complex analysis course. It analyzes the differentiability of several complex functions by checking if they satisfy the Cauchy-Riemann equations. For differentiable functions, it computes their derivatives according to the Cauchy-Riemann theorem. The solutions involve defining real and imaginary parts of functions, taking partial derivatives, and applying limits.

1. NCU Math, Spring 2014: Complex Analysis Homework Solution 2
Text Book: An Introduction to Complex Analysis
Problem. 7.3
Sol:
(a)
f(z) = u(x, y) + iv(x, y) where we dene u(x, y) = x and v(x, y) = −y2
. It is easy to see that u and v are
smooth. So we have
∂u
∂x = 1, ∂u
∂y = 0,
∂v
∂x = 0, ∂v
∂y = −2y.
Since ∂u
∂y = −∂v
∂x = 0 and ∂u
∂x = ∂v
∂y = 1 i y = −1
2 , we can get that f(z) is dierentiable only at y = −1
2 and is
not dierentiable at z = −1
2 by virtue of Theorem 6.4 and 6.5. Moreover, f (z) = ∂u
∂x (0, −1
2 ) + i∂v
∂x (0, −1
2 ) = 1 for
z = −1
2 by Theorem 6.5.
(b)
f(z) = u(x, y)+iv(x, y) where we dene u(x, y) = x2
and v(x, y) = y2
. It is easy to see that u and v are smooth.
So we have
∂u
∂x = 2x, ∂u
∂y = 0,
∂v
∂x = 0, ∂v
∂y = 2y.
Since ∂u
∂y = −∂v
∂x = 0 and ∂u
∂x = ∂v
∂y = 2x = 2y i x = y, we can get that f(z) is dierentiable only at x = y and is not
dierentiable at z ∈ {x + iy|x = y} by virtue of Theorem 6.4 and 6.5. Moreover, f (z) = ∂u
∂x (x, x) + i∂v
∂x (x, x) = 2x
for z = x + ix and x ∈ R by Theorem 6.5.
(c)
f(z) = u(x, y)+iv(x, y) where we dene u(x, y) = yx and v(x, y) = y2
. It is easy to see that u and v are smooth.
So we have
∂u
∂x = y, ∂u
∂y = x,
∂v
∂x = 0, ∂v
∂y = 2y.
Since ∂u
∂x = ∂v
∂y and ∂u
∂y = −∂v
∂x i x = y = 0, we can get that f(z) is dierentiable only at x = y = 0 and
is not dierentiable at z ∈ {x + iy|x = 0 or y = 0} by virtue of Theorem 6.4 and 6.5. Moreover, f (z) =
∂u
∂x (0, 0) + i∂v
∂x (0, 0) = 0 for z = 0 by Theorem 6.5.
(d)
1

2. f(z) = u(x, y) + iv(x, y) where we dene u(x, y) = x3
and v(x, y) = (1 − y)3
. It is easy to see that u and v are
smooth. So we have
∂u
∂x = 3x2
, ∂u
∂y = 0,
∂v
∂x = 0, ∂v
∂y = −3(1 − y)2
.
Since ∂u
∂y = −∂v
∂x = 0 and ∂u
∂x = ∂v
∂y = 3x2
= −3(1−y)2
= 0 i x = 0 and y = 1, we can get that f(z) is dierentiable
only at x = 0, y = 1 and is not dierentiable at z ∈ {x + iy|x = 0 or y = 1} by virtue of Theorem 6.4 and 6.5.
Moreover, f (z) = ∂u
∂x (0, 1) + i∂v
∂x (0, 1) = 0 for z = i by Theorem 6.5.
Problem. 7.4
Sol:
(a)
We dene u(x, y) = x3
+ 3xy2
− 3x and v(x, y) = y3
+ 3x2
y − 3y. So f(z) = u + iv. It is easy to see that u and
v are smooth. So we have
∂u
∂x = 3x2
+ 3y2
− 3, ∂u
∂y = 6xy,
∂v
∂x = 6xy, ∂v
∂y = 3y2
+ 3x2
− 3.
Since ∂u
∂x = ∂v
∂y = 3x2
+ 3y2
− 3 and ∂u
∂y = −∂v
∂x i xy = 0 , we can get that f(z) is dierentiable only at xy = 0 and
is not dierentiable at z ∈ {x + iy|xy = 0} by virtue of Theorem 6.4 and 6.5. Thus f is not analytic everywhere.
Moreover, f (z) = ∂u
∂x (x, y) + i∂v
∂x (x, y) = 3x2
− 3 or 3y2
− 3 for z = x or iy by Theorem 6.5.
(b)
From some computations,
f(z) = 6(x2
− y2
− 2xyi) − 2(x − yi) − 4i(x2
+ y2
)
= (6x2
− 6y2
− 2x) + i(−4x2
− 4y2
− 12xy + 2y).
We dene u(x, y) = 6x2
− 6y2
− 2x and v(x, y) = −4x2
− 4y2
− 12xy + 2y. So f(z) = u + iv. It is easy to see that
u and v are smooth. So we have
∂u
∂x = 12x − 2, ∂u
∂y = −12y,
∂v
∂x = −8x − 12y, ∂v
∂y = −8y − 12x + 2.
Then
∂u
∂x
=
∂v
∂y
⇔ 6x + 2y = 1,
∂u
∂y
= −
∂v
∂x
⇔ x + 3y = 0.
This implies that ∂u
∂x = ∂v
∂y and ∂u
∂y = −∂v
∂x i z = 3
16 − 1
16 i. Thus we can get that f(z) is dierentiable only at
z = 3
16 − 1
16 i and is not dierentiable at z ∈ C − { 3
16 − 1
16 i} by virtue of Theorem 6.4 and 6.5. Therefore f is not
analytic everywhere. Moreover, f ( 3
16 − 1
16 i) = ∂u
∂x ( 3
16 , − 1
16 ) + i∂v
∂x ( 3
16 , − 1
16 ) = 1
4 − 3
4 i by Theorem 6.5.
2

3. (c)
We dene u(x, y) = 3x2
+ 2x − 3y2
− 1 and v(x, y) = 6xy + 2y. So f(z) = u + iv. It is easy to see that u and v
are smooth. So we have
∂u
∂x = 6x + 2, ∂u
∂y = −6y,
∂v
∂x = 6y, ∂v
∂y = 6x + 2.
Since ∂u
∂x = ∂v
∂y = 6x + 2 and ∂u
∂y = −∂v
∂x = −6y for all z ∈ C , we can get that f(z) is dierentiable for all z ∈ C by
virtue of Theorem 6.4 and 6.5. Thus f is analytic in C. Moreover, f (z) = ∂u
∂x (x, y)+i∂v
∂x (x, y) = 6x+2+6yi = 6z+2
for z ∈ C by Theorem 6.5.
(d)
Since f doesn't dene in z = 0, ±2i, we only consider z ∈ C − {0, ±2i}. By virtue of Theorem 6.1, we have
f(z) = 2z2
+6
z(z2+4) is dierentiable for all z ∈ C − {0, ±2i}. f is not analytic in C but is analytic in C − {0, ±2i}. When
we use Theorem 6.1 again, we can get that
f (z) =
4z(z(z2
+ 4)) − (2z2
+ 6)(3z2
+ 4)
(z(z2 + 4))
2
=
−2(z4
+ 5z2
+ 12)
z2(z2 + 4)2
.
for all z ∈ C − {0, ±2i}.
(e)
It is easy to see that
f(z) = ey2
−x2
e−2xyi
= e−(x2
+2xyi−y2
)
= e−z2
.
Since −z2
, ez
are dierentiable in C and Theorem 6.1, we can get that f(z) = e−z2
is dierentiable in C. So f is
analytic in C. Moreover,
f (z) = e−z2
· (−2z)
= −2ze−z2
by virtue of Theorem 6.1 and Example 7.1.
Problem. 7.5
Sol:
3

4. Since
w(z) = (ay3
+ ix3
) + xy(bx + icy)
= (ay3
+ bx2
y) + i(x3
+ cxy2
),
we dene u(x, y) = ay3
+ bx2
y and v(x, y) = x3
+ cxy2
. It is easy to see that u and v are smooth and
∂u
∂x = 2bxy, ∂u
∂y = 3ay2
+ bx2
,
∂v
∂x = 3x2
+ cy2
, ∂v
∂y = 2cxy.
So w is analytic i
∂u
∂x = ∂v
∂y ,
∂u
∂y = −∂v
∂x ,
⇔
2bxy = 2cxy,
3ay2
+ bx2
= −(3x2
+ cy2
),
from Theorem 6.4 and 6.5. This implies that w is analytic i a = 1 and b = c = −3. Moreover, we have
∂w
∂z
=
∂u
∂x
+ i
∂v
∂x
= −6xy + i(3x2
− 3y2
)
= 3i(x2
+ 2ixy − y2
)
= 3i · z2
by Theorem 6.5.
Problem. 7.8
Sol:
(a)
It is easy to see that
f(z) =
z2
z if z = 0,
0 if z = 0,
=
z3
|z|2 if z = 0,
0 if z = 0,
=
x3
−3xy2
x2+y2 + i−3x2
y+y3
x2+y2 if z = 0,
0 if z = 0,
= u(x, y) + iv(x, y)
4

5. where we dene
u(x, y) =
0 if x = y = 0
x3
−3xy2
x2+y2 o.w.
and v(x, y) =
0 if x = y = 0
−3x2
y+y3
x2+y2 o.w.
So
∂u
∂x
(0, 0) = lim
h→0
h3
−3h·02
h2+02 − 0
h
= lim
h→0
h
h
= 1,
∂u
∂y
(0, 0) = lim
k→0
03
−3·0·k2
02+k2 − 0
k
= lim
h→0
0
k
= 0,
∂v
∂x
(0, 0) = lim
h→0
−3h2
·0+03
h2+02 − 0
h
= lim
h→0
0
h
= 0,
∂v
∂y
(0, 0) = lim
k→0
−3·02
k+k3
02+k2 − 0
k
= lim
h→0
k
h
= 1.
This implies that ∂u
∂x (0, 0) = ∂v
∂y (0, 0) = 1 and ∂u
∂y (0, 0) = −∂v
∂x (0, 0) = 0. Thus f satises Cauchy-Riemann equations
at z = 0.
(b)
From some computations, we have for z = 0
f(z) − f(0)
z
=
z2
z
z
=
z
z
2
.
It suces to show that limz→0
z
z
2
doesn't exist. It is easy to see that
lim
z=x,x→0
z
z
2
= lim
z=x→0
x
x
2
= 1.
Also
lim
z=x+ix,x→0
z
z
2
= lim
z=x+ix,x→0
x − ix
x + ix
2
= lim
z=x+ix,x→0
1 − i
1 + i
2
=
1 − i
1 + i
2
=
(1 − i)4
4
.
Since 1 = (1−i)4
4 , we can get that limz→0
z
z
2
doesn't exist.
5

6. Problem. 7.10
Sol:
Since x = r cos θ and y = r sin θ, we can get that
∂x
∂r
= cos θ,
∂x
∂θ
= −r sin θ,
∂y
∂r
= sin θ,
∂y
∂θ
= r cos θ.
By chain rule, we have
∂u
∂r
=
∂u
∂x
·
∂x
∂r
+
∂u
∂y
·
∂y
∂r
=
∂u
∂x
cos θ +
∂u
∂y
sin θ,
∂u
∂θ
=
∂u
∂x
·
∂x
∂θ
+
∂u
∂y
·
∂y
∂θ
= r(−
∂u
∂x
sin θ +
∂u
∂y
cos θ),
∂v
∂r
=
∂v
∂x
·
∂x
∂r
+
∂v
∂y
·
∂y
∂r
=
∂v
∂x
cos θ +
∂v
∂y
sin θ,
∂v
∂θ
=
∂v
∂x
·
∂x
∂θ
+
∂v
∂y
·
∂y
∂θ
= r(−
∂v
∂x
sin θ +
∂v
∂y
cos θ).
So
∂u
∂r
=
∂v
∂y
cos θ −
∂v
∂x
sin θ =
1
r
∂v
∂θ
,
∂u
∂θ
= r(−
∂v
∂y
sin θ −
∂v
∂x
cos θ) = −r
∂v
∂r
.
by virtue of Cauchy-Riemann conditions.
Conversely, we have
∂r
∂x
=
x
r
,
∂r
∂y
=
y
r
,
∂θ
∂x
=
−y
r2
,
∂θ
∂y
=
x
r2
,
6

7. from r2
= x2
+ y2
and θ = tan−1
(y
x ). Since chain rule, we can get that
∂u
∂x
=
∂u
∂r
·
∂r
∂x
+
∂u
∂θ
·
∂θ
∂x
=
∂u
∂r
x
r
−
∂u
∂θ
y
r2
,
∂u
∂y
=
∂u
∂r
·
∂r
∂y
+
∂u
∂θ
·
∂θ
∂y
=
∂u
∂r
y
r
+
∂u
∂θ
x
r2
,
∂v
∂x
=
∂v
∂r
·
∂r
∂x
+
∂v
∂θ
·
∂θ
∂x
=
∂v
∂r
x
r
−
∂v
∂θ
y
r2
,
∂v
∂y
=
∂v
∂r
·
∂r
∂y
+
∂v
∂θ
·
∂θ
∂y
=
∂v
∂r
y
r
+
∂v
∂θ
x
r2
.
Thus
∂u
∂x
=
∂v
∂θ
x
r2
+
∂v
∂r
y
r
=
∂v
∂y
,
∂u
∂y
=
∂v
∂θ
y
r2
−
∂v
∂r
x
r
= −
∂v
∂x
,
by (7.7).
Moreover,
f (z) =
∂u
∂x
+ i
∂v
∂x
= (
∂u
∂r
x
r
−
∂u
∂θ
y
r2
) + i(
∂v
∂r
x
r
−
∂v
∂θ
y
r2
)
= (
∂u
∂r
x
r
+
∂v
∂r
y
r
) + i(
∂v
∂r
x
r
−
∂u
∂r
y
r
)
= (cos θ − i sin θ)ur + (sin θ + i cos θ)vr
= e−iθ
ur + ie−iθ
vr
= e−iθ
(ur + ivr).
from above, Theorem 6.5, and (7.7).
We dene u(r, θ) =
√
r cos θ
2 and v(r, θ) =
√
r sin θ
2 . Then f(z) =
√
re−iθ/2
=
√
r cos θ
2 + i
√
r sin θ
2 = u(r, θ) +
iv(r, θ). Obviously, u and v are smooth in C − {0}. Also,
∂u
∂r = 1
2
√
r
cos θ
2 , ∂u
∂θ = −
√
r
2 sin θ
2 ,
∂v
∂r = 1
2
√
r
sin θ
2 , ∂v
∂θ =
√
r
2 cos θ
2 ,
for all z ∈ C − {0}. So
∂u
∂r
=
1
2
√
r
cos
θ
2
=
1
r
∂v
∂θ
,
∂u
∂θ
=
−
√
r
2
sin
θ
2
= −r
∂v
∂r
.
7

8. Thus we can apply Theorem 6.5 to get that f(z) is dierentiable at all z except z = 0. Moreover, we have
f (z) = e−iθ
(ur + ivr)
= e−iθ
(
1
2
√
r
cos
θ
2
+ i
1
2
√
r
sin
θ
2
)
=
1
2
√
r
e−iθ
e
iθ
2
=
1
2
√
r
e− iθ
2 .
Problem. 7.18
Sol:
(a)
Since ∂2
u
∂x2 = e−x
sin y and ∂2
u
∂y2 = −e−x
sin y are both smooth, we have ∂2
u
∂x2 + ∂2
u
∂y2 = 0. So u is harmonic.
Let v(x, y) satises that ∂v
∂y = ∂u
∂x = −e−x
sin y and ∂v
∂x = −∂u
∂y = −e−x
cos y. Then v = e−x
cos y +h(x) for some
real-valued function h by ∂v
∂y = −e−x
sin y. But ∂v
∂x = −e−x
cos y. So h (x) = 0. This implies that v = e−x
cos y + C
for some real constant C. Thus the corresponding analytic function is e−x
sin y + ie−x
cos y + iC for some real
constant C.
(b)
Since ∂2
v
∂x2 = − cos x cosh y and ∂2
u
∂y2 = cos x cosh y are both smooth, we have ∂2
u
∂x2 + ∂2
u
∂y2 = 0. So u is harmonic.
Let u(x, y) satises that ∂u
∂x = ∂v
∂y = cos x sinh y and ∂u
∂y = −∂v
∂x = sin x cosh y. Then u = sin x sinh y + h(y)
for some real-valued function h by ∂u
∂x = cos x sinh y. But ∂u
∂y = sin x cosh y. So h (y) = 0. This implies that u =
sin x sinh y +C for some real constant C. Thus the corresponding analytic function is sin x sinh y +C +i cos x cosh y
for some real constant C.
Problem. 7.24
Sol:
For convenience, we dene C = {(x, y)|u(x, y) = c} and D = {(x, y)|v(x, y) = d}. Let (a, b) ∈ C ∩ D with
f (a+ib) = 0. Since (a, b) ∈ C and f (a+ib) = 0, the tangent vector to C at (a, b) is parallel to (uy(a, b), −ux(a, b)) =
0. Similarly, we have the tangent vector to D at (a, b) is parallel to (vy(a, b), −vx(a, b)) = 0. It suces to show that
(uy(a, b), −ux(a, b)) · (vy(a, b), −vx(a, b)) = 0. Because of Cauchy-Riemann equations, we can get that
(uy(a, b), −ux(a, b)) · (vy(a, b), −vx(a, b))
=uy(a, b)vy(a, b) + ux(a, b) · vx(a, b)
= − vx(a, b)ux(a, b) + ux(a, b) · vx(a, b)
=0.
8

9. Now, we consider the case f(z) = z2
= (x2
− y2
) + 2xyi. So C = {(x, y)|x2
− y2
= c} and D = {(x, y)|2xy = d}.
Let (a, b) ∈ C ∩ D with f (a + ib) = 0. Thus a2
= b2
and ab = 0. The tangent vector to C at (a, b) is parallel to
(b, a) = 0. Similarly, The tangent vector to D at (a, b) is parallel to (a, −b) = 0. Then C and D are orthogonal by
(b.a) · (a, −b) = 0.
9