Math 411: Honours Complex Variables - University of Alberta

Example. Let z be a complex number. Then multiplication by z is a R-linear map
from C = R2 into itself and thus uniquely represented by a real 2×2 matrix A of the
form � �
a −b
,
b a
where a = Rez and b = Imz. It follows that A t is the matrix representing ¯z. Hence,
A is orthogonal if and only if |z|= 1.
Theorem 17.1 (Conformality at Nondegenerate Points). Let D1,D2 ⊂ C be open,
and let f: D1 → D2 be holomorphic. Then f is angle preserving at z0 ∈ D1 whenever
f ′ (z0) �= 0.
Proof. Let z0 ∈ D1 be such that f ′ (z0) �= 0. In view of Lemma 17.1 and the example
following it, the claim is clear if |f ′ (z0)|= 1.
For the general case, let
and define
and
1
|f ′ (z0)| D2
�
z
:=
|f ′ �
: z ∈ D2 ,
(z0)|
g: D1 → 1
|f ′ (z0)| D2, z ↦→ f(z)
|f ′ (z0)|
h:
1
|f ′ (z0)| D2 → D2, z ↦→ |f ′ (z0)|z.
Then g is angle preserving at z0 because |g ′ (z0)|= 1, and it is easily seen that h is
angle preserving at g(z0). Consequently, f = h◦g is angle preserving at z0.
Corollary 17.1.1 (Conformality of Biholomorphic Maps). Let D1,D2 ⊂ C be open
and connected, and let f: D1 → D2 be biholomorphic. Then f is angle preserving at
every point of D1.
Theorem 17.2 (Holomorphic Inverses). Let D1,D2 ⊂ C be open and connected,
and let f : D1 → D2 be holomorphic and bijective. Then f is biholomorphic and
Z(f ′ ) = ∅.
Proof. We first show that f −1 is continuous.
Let w0 ∈ D2, and let ǫ > 0 be such that Bǫ(f −1 (w0)) ⊂ D1. By the Open
Mapping Theorem, f(Bǫ(f −1 (w0))) is open. Hence, there exists δ > 0 such that
Bδ(w0) ⊂ f(Bǫ(f −1 (w0))). Hence, if w ∈ Bδ(w0), then f −1 (w) ∈ Bǫ(f −1 (w0)), That
is, f −1 is continuous at w0.
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