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Math 116 - Homework 1 Solutions
January 18, 2017
Problem 1 Prove that if both f and f¯ are holomorphic on a connected open set Ω ⊂ C, then f is
constant.
Solution Check the CR condition. Then we have
ux = vy ux = −vy
uy = −vx uy = vx
From the first line we have ux = vy = 0, so u = u(y) and v = v(x). Similarly apply the same reasoning
to the second line, and we have that u, v are constant on any connected component. Hence, so is f .
Problem 2 Suppose f is a C1 complexed valued function on an open set Ω ⊂ C. Prove that
∂f
∂z
=
∂f
∂z
Solution Recall that
∂
∂z
=
1
2
(
∂
∂x
+
1
i
∂
∂y
)
∂
∂z¯
=
1
2
(
∂
∂x
− 1
i
∂
∂y
)
Therefore, using that complex conjugation is additive and multiplicative:
∂f
∂z
=
1
2
(
∂
∂x
+
1
i
∂
∂y
)
(u+ iv)
=
1
2
(
∂
∂x
− 1
i
∂
∂y
)
u+
1
2
(
∂
∂x
+
1
i
∂
∂y
)
iv
=
1
2
(
∂
∂x
− 1
i
∂
∂y
)
u− 1
2
(
∂
∂x
− 1
i
∂
∂y
)
iv
=
∂f
∂z
Problem 3 Let a be a nonzero complex number. For any integer n > 1 show that the sum of n
solutions to zn = a equals to zero.
Solution Let a = reiθ where r > 0. Write zn − a = (z − x1)(z − x2) · · · (z − xn) where xm are
the roots of zn − a. We know all of those xm exist since xm = r1/neiθ/ne2pim/n.
Now consider the polynomial P (z) = zn − a, the coefficient of z, call it c1, is 0. If calculate the
coefficient from expanding (z − x1)(z − x2) · · · (z − xn), we know that
0 = c1 =
n∑
m=1
xm
1
Problem 4 Show that in polar coordinate, the Cauchy-Riemann equations take the form
∂u
∂r
=
1
r
∂v
∂θ
and
1
r
∂u
∂θ
= −∂v
∂r
Use these equations to show that the logarithm function defined by
log z = log r + iθ where z = reiθ with − pi < θ < pi
is holomorphic in the region r > 0 and −pi < θ < pi.
Solution Write x = r cos θ and y = r sin θ. Consider f = u+iv. Write U(r, θ) = u(x(r, θ), y(r, θ)),
similar for V
Ur = uxxr + uyyr
Uθ = uxxθ + uyyθ
In terms of matrix we have [
Ur
Uθ
]
=
[
cos θ sin θ
−r sin θ r cos θ
] [
ux
uy
]
Take the inverse matrix, which is possible since θ and r are specified, we got[
ux
uy
]
=
1
r
[
r cos θ − sin θ
r sin θ cos θ
] [
Ur,
Uθ
]
We will got the same matrix for Vr and Vθ.
CR⇔
[
vx
vy
]
=
[
0 −1
1 0
] [
ux
uy
]
⇔
[
Vr
Vθ
]
=
1
r
[
cos θ sin θ
−r sin θ r cos θ
] [
0 −1
1 0
] [
r cos θ − sin θ
r sin θ cos θ
] [
Ur
Uθ
]
⇔
[
Vr
Vθ
]
=
[
0 −1r
r 0
] [
Ur
Uθ
]
For the logarithm function, U(r, θ) = log r and V (r, θ) = θ, so Ur = 1r , Vθ = 1, and the other partials
are 0. Thus the polar CR equations hold. The partials with respect to r and θ are continuous, so by
the formulae for the transformations between the partials above, the partials with respect to x and y
are also continuous. So the logarithm is holomorphic.
Problem 5 Show that 4 ∂∂z
∂
∂z¯ = 4
∂
∂z¯
∂
∂z =
∂2
∂x2 +
∂2
∂y2 = ∆. Show that if f = u+ iv is holomorphic on
an open set Ω then u and v are harmonic, i.e ∆u = ∆v = 0.
Solution ∂
2
∂x2 +
∂2
∂y2 =
4
4
(
∂
∂x + i
∂
∂y
)(
∂
∂x − i ∂∂y
)
= 4 ∂∂z¯
∂
∂z = 4
∂
∂z
∂
∂z¯ . Now, ∆f = ∆u + i∆v. On
the other hand, ∆f = 4 ∂∂z
∂
∂z¯ f = 0 since f is holomorphic and
∂
∂z¯ f = 0.
Problem 6 Suppose that f is holomorphic in a region Ω. Prove that in any one of the following
cases:
1. Ref is constant.
2. Imf is constant.
3. |f | is constant.
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Solution Note that the problem should have stated that the region is connected. The conclusion
is actually that f is constant on each connected component. (a) and (b) are straightforward by just
checking the Cauchy Riemann condition. For (c), note that for f = u + iv,|f |2 = u2 + v2. If |f | = 0
then we are done using the previous parts. If not we will take the partial derivatives in x and y. We
get: {
2uux + 2vvx = 0
2uuy + 2vvy = 0
Together with CR: {
2uux + 2vvx = 0
−2uvx + 2vux = 0
We have: 2
(
u v
v −u
)(
ux
vx
)
=
(
0
0
)
. Since det
(
u v
v −u
)
= −|f |2 6= 0 this matrix is
invertible and the only solution is the zero solution. Using again the CR equations we get ux = vx =
uy = vy = 0, so f is constant.
Problem 7 Determine the radius of convergence of the series
∑∞
n=1 anz
n when
1. an = (log n)2
2. an = n
2
4n+3n
Solution
1. Consider the limit lim sup |an|1/n. Since the limit of 2n ln(log n) goes to 0, the limit we study
goes to 1. So is our R.
2.
1
4
= lim sup
(
1
4n
) 1
n
≤ lim sup
(
n2
4n + 3n
) 1
n
≤ lim sup
(
n2
4n
) 1
n
≤ (lim sup (n2) 1n )(lim sup( 1
4n
) 1
n
) =
1
4
So R = 4.
Problem 8 Show that if {an} is a sequence of non zero complex numbers such that
lim
n→∞
|an+1|
|an| = L
then
lim
n→∞ |an|
1/n = L
Solution Let bn = |an|, bn is a positive non-zero number. Let ε > 0. Then, there is N so that
for every n > N
∣∣∣ bn+1bn − L∣∣∣ < ε, or
(L− ε)bn < bn+1 < (L+ ε)bn
Using induction, we get that
(L− ε)n−NbN < bn < (L+ ε)n−NbN
Now
(L− ε)n−Nn b1/nN < b1/nn < (L+ ε)
n−N
n b
1/n
N
3
Now take n→∞ and get that
(L− ε) < lim
n→∞ b
1/n
n < (L+ ε)
Since ε was arbitrary, limn→∞ b
1/n
n = L.
Problem 9 Prove the following:
1. The power series
∑
nzn does not converge on any point of the unit circle.
2. The power series
∑
zn/n2 converges at every point of the unit circle.
Solution
1. The series
∑
nzn0 converges only if nzn0 goes to zero. nzn does not goes to zero if |z| = 1
2. |∑Mn=m zn/n| ≤ ∑Mn=m |zn/n2| = ∑Mn=m 1/n2. This means BM = ∑Mn=0 zn/n2 is a Cauchy
sequence since
∑
1/n2 converges.
Problem 10 Consider the function f defined by
f(x) =
{
0 x ≤ 0
e−1/x
2
x > 0
Prove that f is indefinitely differentiable on R and that f (n)(0) = 0 for any n ≥ 1. Conclude that f
does not have a converging power series expansion around origin.
Solution To check f is smooth, one only has to check that it is smooth at 0, which reduces to
taking the right derivative of f and making sure it is zero. To do that, simply check
lim
x→0+
f (n)(x)
x
is zero for n ≥ 0. Note that we can compute f (n)(x) when x > 0 and will obtain, in all cases, a rational
function of x times e−1/x
2
. As x → 0, the exponential will go to 0 faster than any rational function
can go to infinity. For example,
lim
x→0+
e−1/x
2
x
= lim
t→∞
t
et2
= lim
t→∞
1
2et2t
= 0
Using L’Hospital’s rule. Finally, since all the derivatives are 0 at origin, so are all the coefficients
of the putative Taylor series, which means around 0 f(x) would be constantly 0, which is absurd.
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