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Local normal form of a nonconstant holomorphic map

Statement

Let f:ΩC be nonconstant and holomorphic on a complex domain Ω, let aΩ, and put m=degaf. Then near a there is a biholomorphic coordinate ϕ with ϕ(a)=0 and f(z)f(a)=ϕ(z)m.

Precisely, there is a complex domain V with aVΩ such that ϕ:Vϕ[V] is biholomorphic, ϕ(a)=0, and the displayed identity holds for every zV.

Facts & Assumptions

Given: A nonconstant holomorphic function f on a complex domain Ω, a point aΩ, and the positive natural m=degaf (Local degree of a nonconstant holomorphic map). Holomorphic products obey the product rule (Linearity, product, reciprocal, and quotient rules for complex derivatives).

[L1]

A holomorphic function has finite order m at a exactly when, near a, it has the form (za)mh(z) with h holomorphic and h(a)0 (The order of a zero is the exponent in its local holomorphic factorization).

[L2]

For every positive natural m, a nowhere-zero holomorphic function on a disc has a holomorphic mth root (Holomorphic roots of a nonvanishing function on a disc).

[L3]

If a function is holomorphic near a and has nonzero derivative at a, then it is biholomorphic between neighbourhoods of a and its value (A nonzero complex derivative gives a local biholomorphism).

[L4]

A complex differentiable function is continuous at every point of complex differentiability (Complex differentiability at a point implies continuity there).

Proof

technique · direct
1.1

Apply [L1] to ff(a): on a neighbourhood of a one has f(z)f(a)=(za)mh(z), where h is holomorphic and h(a)0.

L1given
2.1

Since h(a)0, [L4] permits shrinking to a disc on which h is nowhere zero. By [L2] there is a holomorphic q on that disc with qm=h.

step 1.1L2L4
3.1

Define ϕ(z):=(za)q(z). Then ϕ(a)=0 and the product rule gives ϕ(a)=q(a)0, so [L3] makes ϕ biholomorphic after one further shrinking around a.

step 2.1L3givenalgebra
4.1

On that final neighbourhood, steps 1.1 and 2.1 give f(z)f(a)=(za)mq(z)m=ϕ(z)m, which is the required normal form.

step 1.1step 2.1step 3.1algebra

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