COMPLEX GEOMETRY Course notes

Theorem 1.1.3 (Local Structure of f ∈ O(U)). If f ∈ O(U) is non-constant at z 0 ∈ U. Let
m = min{n > 0 / f (n) (z 0 ) ≠ 0}.
Then there exists a bi-holomorphic function ϕ : V −→ W from a neighbourhood of V of z 0 to a neighbourhood
W of 0 ∈ C with ϕ(z 0 ) = 0 such that
f(z) − f(z 0 ) = ϕ(z) m , for every z ∈ V.
Proof: Note that f(z) − f(z 0 ) = (z − z 0 ) m g(z) with g(z 0 ) ≠ 0 and g ∈ O(U). Since the quotient
f(z)−f(z 0)
z−z 0
is bounded on U − z 0 and continuous at z 0 , we have by the Riemann Extension Theorem that
(z − z 0 ) m−1 g(z) = f(z)−f(z0)
z−z 0
is holomorphic on U. Proceeding this way, we have that g(z) ∈ O(U).
We study several cases: If n = 1 then f ′ (z 0 ) ≠ 0 and by the Inverse Function Theorem we can choose
ϕ(z) = f(z) − f(z 0 ). Now assume n ≠ 1. Since g(z 0 ) ≠ 0 then g(z) ≠ 0 on a neighbourhood of z 0 . So
we can write g = h m on a neighbourhood V . We have
f(z) − f(z 0 ) = [h(z)(z − z 0 )] m
with ϕ ′ (z 0 ) = h(z 0 ) ≠ 0. Hence, up to a local change of coordinates, f is locally of the form z ↦→ z m for
some m. Such a number m is called the ramification degree of f at z 0 .
Corollary 1.1.3 (Open Mapping Theorem). If f ∈ O(U) is non-constant and U is connected, then f is an
open mapping.
Corollary 1.1.4. If f ∈ O(U) and |f| has a local maximum at z 0 ∈ U, where U is an open connected set,
then f is constant on U.
Proof: Suppose f is not constant. Then by the Open Mapping Theorem, we have that B ɛ (f(z 0 )) ⊆ f(U)
for some ɛ > 0. In this neighbourhood there are some points of modulus greater that 0. Hence f(z 0 ) is
not a local maximum.
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