16

Suppose $f$ is a holomorphic function defined on a neighborhood of $0$, such that $f(0) = 0$. Let $f^n$ denote the $n$-fold composition of $f$ with itself. If $f^n$ is the identity for some natural number $n$, then is $f$ necessarily conjugate to the map $z \mapsto \zeta z$ for some $n^{\text{th}}$ root of unity $\zeta$? That is, if $f^n(z) = z$ for all $z$ sufficiently near to $0$, is there some biholomorphic function $h$ with $h(0) = 0$ such that $f(z) = h^{-1}(\zeta h(z))$ for all $z$ sufficiently near to $0$?

EthanK
  • 305
  • Chain rule and $f(0)=0$ give that $f'(0)$ is an $n$-th root of unity, presumably the candidate for $\zeta$. Now how to get the $h$, I don't know. Can one prove that if $f'(0)=1$, only $f(z)=z$ does it? – Torsten Schoeneberg Nov 07 '25 at 04:24

1 Answers1

19

Set $\zeta:=f'(0)$ (we have $\zeta^n=1$). Since $f(0)=0$ the function $$ h(z):=\frac{1}{n}\sum_{k=0}^{n-1}\zeta^{-k}f^k(z) $$ is defined and holomorphic in a neighborhood of $0$. Moreover $$ h(f(z))=\frac{1}{n}\sum_{k=0}^{n-1}\zeta^{-k}f^{k+1}(z) = \zeta\frac{1}{n}\sum_{k=0}^{n-1}\zeta^{-k-1}f^{k+1}(z) $$ $$ =\zeta\frac{1}{n} (\zeta^{-1}f^{1}(z) + \dots + \zeta^{-(n-2)-1}f^{(n-2)+1}(z)+ \zeta^{-n}f^{n}(z)) = \zeta h(z) $$ (note that $\zeta^{-n}f^{n}(z)=z=\zeta^{-0}f^{0}(z)$). Moreover $h(0)=0$ and $$ h'(0)=\frac{1}{n}\sum_{k=0}^{n-1}\zeta^{-k} \zeta^k=1. $$ Thus $h$ is one-to-one near $0$ and $h^{-1}$ exists and is holomorphic in a neighborhood of $0$. Thus $f(z)=h^{-1}(\zeta h(z))$ in a neighborhood of $0$.

Gerd
  • 10,510
  • 1
  • 8
  • 29
  • Wonderful! I actually thought of a similar solution, taking the product of all the $f^k(z)$ then taking the $n^{\text{th}}$ root, but I think this is better. – EthanK Nov 07 '25 at 16:17
  • Neat, and almost easier than suspected. You do not really need that scaling factor $1/n$, do you? – Torsten Schoeneberg Nov 07 '25 at 18:04
  • @TorstenSchoeneberg Thanks. And yes, I just scaled $h'(0)$ to $1$, but that is not necessary. – Gerd Nov 07 '25 at 18:29