Math 411: Honours Complex Variables - University of Alberta

By the Open Mapping Theorem, there exists r > 0 with Br(g(z0)) ⊂ g(D).
Assume that there exists a point z ∈ D with g(z) ∈ Br(−g(z0)):
r > |g(z)+g(z0)|= |−g(z)−g(z0)|;
this means that −g(z) ∈ Br(g(z0)) ⊂ g(D). Hence, there exists ˜z ∈ D with g(˜z) =
−g(z) and thus ˜z = z, which in turn yields that 0 = [g(z)] 2 = z−w. This contradicts
w /∈ D. Hence, g(D)∩Br(−g(z0)) = ∅ must hold.
Define
˜g: D → D, z ↦→
r
g(z)+g(z0) ,
Then ˜g : D → D is holomorphic and injective. Let a := ˜g(z0). Then φa ◦ ˜g :
D → D is holomorphic and injective with (φa ◦ ˜g)(z0) = 0 and (φa ◦ ˜g) ′ (z0) �= 0
by Corollary 17.2.1. Let c ∈ C with |c|= 1 be such that c(φa ◦ ˜g) ′ (z0) > 0. Then
c(φa ◦ ˜g) ∈ F, so that indeed F �= ∅.
Claim 2. Let (fn) ∞ n=1
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be a sequence in F converging compactly to f: D → C. Then
either f ≡ 0 or f ∈ F.
It is straightforward that f(z0) = 0, f ′ (z0) ≥ 0, and f(D) ⊂ D. By Corollary
13.4.1, f ≡ 0 or f is injective. If f is injective, then f ′ (z0) �= 0 must hold by
Corollary17.2.1, so that f ′ (z0) > 0. Also, since f(D) ⊂ D isopen, we have f(D) ⊂ D,
and hence f ∈ F.
Claim 3. There exists f ∈ F such that f(D) = D.
Choose a sequence (fn) ∞ n=1
in F such that
lim
n→∞ f′ n (z0) = sup{ ˜ f ′ (z0) : ˜ f ∈ F} ∈ (0,∞].
Noticethat|fn(z)|≤ 1forallz ∈ D. ByMontel’s Theorem, thereexists asubsequence
that converges compactly to some f: D → D. In particular,
(fnk )∞ k=1
f ′ (z0) = sup{ ˜ f ′ (z0) : ˜ f ∈ F} > 0 (∗)
holds, so that f ∈ F by Claim 2.
Assume that there exists w ∈ D\f(D). Since D admits holomorphic square roots,
there is a holomorphic function h: D → C such that
[h(z)] 2 = −(φw ◦f)(z) = f(z)−w
1− ¯wf(z) .
for z ∈ D. In particular, h(D) ⊂ D, h is injective and hence h ′ (z0) �= 0. We then
evaluate the derivative of each side of the above equation at z = z0, noting that
f(z0) = 0:
2h(z0)h ′ (z0) = f ′ (z0)+ ¯wf ′ (z0)(−w) = (1−|w| 2 )f ′ (z0).