What is the holomorphic automorphism group of $\mathbb{C}\backslash\left\{0\right\}$

Question

I already know that $Aut(\mathbb{C})$ are all the $az+b,a\neq 0,a,b\in \mathbb{C}$.It is easy to know that $az,1/az\in Aut(\mathbb{C}\backslash\left\{0\right\}),a\neq 0$.I want to know that if $Aut(\mathbb{C}\backslash\left\{0\right\})=\left\{az,1/az\mid a\neq 0\right\}$ is corrcect.

Answer 1

The idea is that a univalent functions in some domain can have at most one non-removable isolated singularity and that must be a simple pole; this follows from the local form of an analytic function at an isolated singularity (higher-order poles are like $1/z^k, k \ge 2$ so we get large values $k$ times, two simple poles are both like $1/z$ so we get large values twice, while for essential singularities, most values are taken infinitely many times)

So if $f \in Aut(\mathbb{C}\backslash\left\{0\right\})$ we ave two cases:

1: $0$ is a removable singularity, then (by Liouville) infinity is a pole, hence $f$ is a polynomial so it is linear $az, a \ne 0$ since it is injective and fixes $0$

2: $0$ is a simple pole and infinity is removable (so $f$ is bounded away from zero), hence $zf$ is an entire function with at most a simple pole at infinity so it is a linear polynomial; hence $f=a/z+b, a\ne 0$; but if $b \ne 0, f(-a/b)=0$ which is not allowed, hence $b=0$ and we get $f=a/z, a \ne 0$ so we are done!