Show existence of holomorphic function $g$ such that $f(g(z)) = g(z^n)$

Question

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?

This is an important result in complex dynamics as it allows us to understand the nature of a function near a super-attracting fixed point. Thus, you should be able to find a proof in any reasonable text on complex dynamics. It appears as theorem 4.1 on page 33 of Complex Dynamics by Carleson and Gamelin, for example. — Mark McClure, Mar 29 '18 at 16:14

score 2 · Accepted Answer · answered Jun 05 '18 at 13:44

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.

Show existence of holomorphic function $g$ such that $f(g(z)) = g(z^n)$

1 Answers1