Bijective holomorphic map with a fixed point

Question

$f$ be a holomorphic on a bounded domain $D$ with fixed point $z_0$. Could any one give a hint how to show the following:

$f$ is bijective iff $|f'(z_0)|=1$.

Well, I was thinking like to compose $f,f^2,\dots,f^n$ and apply some how $f^n$ also has $z_0$ as fixed point. Thank you for help.

the question as stated isn't true: consider f(z) = z^2; then f(1) = 1, , f'(1) = 2 but f is not bijective. do you need your fixed point to be unique? — Kris, Aug 14 '12 at 17:50

score 6 · Accepted Answer · answered Aug 14 '12 at 18:51

6

It's not true. Consider $f(z) = z + z^2$ on $D = \mathbb C$. The unique fixed point is $0$, $f'(0) = 1$, but $f(-1-z) = f(z)$.

answered Aug 14 '12 at 18:51

Robert Israel

if the uniqueness of fixed point withdrawn? – Myshkin Aug 17 '12 at 00:31
and is your map bijective? – Myshkin Aug 17 '12 at 00:41
No, the fact that $f(-1-z) = f(z)$ shows that $f$ is not bijective on any domain that contains both $z$ and $-1-z$ for some $z$. – Robert Israel Aug 17 '12 at 05:25
I am extremely sorry Dear Sir, I have edited my question. – Myshkin Aug 18 '12 at 14:34
Adding the requirement of "bounded domain" does not fix the question. As I said, my example is not bijective on any domain that contains both $z$ and $-1-z$ for some $z$. – Robert Israel Aug 19 '12 at 05:05
It's not stated in the question, but from considering the iterates of $f$, it seems that the intention is that $f \colon D \to D$ is a holomorphic self-map of the bounded domain $D$, and in that case, the conclusion holds. If $D \subset \mathbb{C}$ is a bounded domain, $f\colon D\to D$ holomorphic with a fixed point $z_0\in D$, then $f$ is bijective if and only if $\lvert f'(z_0)\rvert = 1$. (cc @BunuelianTrick) – Daniel Fischer Apr 29 '14 at 22:27

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