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$f$ 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.

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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
yes then, I hope, edited. – La Belle Noiseuse Aug 14 '12 at 17:51
up vote 6 down vote accepted

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

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if the uniqueness of fixed point withdrawn? – La Belle Noiseuse Aug 17 '12 at 0:31
and is your map bijective? – La Belle Noiseuse Aug 17 '12 at 0:41
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 5:25
I am extremely sorry Dear Sir, I have edited my question. – La Belle Noiseuse 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 5:05

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