ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...

Then
Γ (j)
k
ANALYSIS QUALS 11
f(z)dx = 0 for all k, j = 1, 2. But Γ (j)
k converges uniformly
to Γ (j) with
Γ (1) f(z)dz +
Γ (2) f(z)dz =
f(z)dz. Hence, f(z)dz = 0.
Γ Γ
Thus, by Morera’s theorem, f is entire.
Problem 9: Let
f(z) =
∞
anz n
n=0
be a holomorphic function in D. Show that if
∞
n|an| ≤ |a1|
n=2
with a1 = 0 then f is injective.
Solution. Note that f ′ (z) = ∞ n=1 nanzn−1 . Then
|f ′
∞
(z) − a1| = nanz
n−1
≤
∞
n|an||z| n−1 ∞
< n|an| ≤ |a1|.
n=2
n=2
Hence, |f ′ (z) − a1| < |a1|. Thus, we may choose θ so that Re(e iθ f ′ (z)) > 0
for all z ∈ D. Then if z1 = z2,
z1
|f(z1) − f(z2)| =
f ′
(z)dz
> |z2 − z1|
z2
1
0
n=2
e iθ Re(f ′ (z1 + t(z2 − z1)))dt > 0.
Thus, f(z2) = f(z1), so f is injective.
Problem 10: Prove that the punctured disc {z ∈ C : 0 < |z| < 1} and the
annulus given by {z ∈ C : 1 < |z| < 2} are not conformally equivalent.