TITLE: Show existence of holomorphic function $g$ such that $f(g(z)) = g(z^n)$
QUESTION [9 upvotes]: Let $f:\mathbb{C} \rightarrow \mathbb{C}$ be a holomorphic function, with $f(0)=0$, and $n\ge 0$ the multiplicity of $0$ as a zero of $f$. 
Show that there exist a holomorphic function $g:\mathbb{C}\rightarrow \mathbb{C}$ with $g(0)=0$ such that $$f(g(z)) = g(z^n).$$ 
I know the existence of $g$ such that $f(g(z)) = z^n$, but the above seems out of reach. Any ideas or possible sources for further research?

REPLY [2 votes]: The statement is only true when $n\geq 2$.
Let $G$ be the inverse of $g$. Equation $G\circ f=G^n$ which is equivalent to your
equation is called the Bottcher equation in complex dynamics. Bottcher theorem says that under your condition there exists a solution $G$ analytic in a neighborhood of $0$ with the properties
that $G(0)=0$ and $G'(0)=1$. (Therefore, $G$ is invertible and the inverse is your $g$.) There are many proofs of this. In the most elementary proof, $G$ is constructed as a limit $(f^{*k})^{n^{-k}}$, where $f^{*k}$ is the $k$-th iterate
of $f$. For a complete proof, see any book on holomorphic dynamics.
The best introduction is J. Milnor, Dynamics in one complex variable. Available on arXiv.