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I need to proof that no holomorphic function exists such that $|f(z)| > |z|$. I defined $g(z) = \frac{z}{f(z)}$. It follows that $|g(z)| < 1$ and with Liouville's theorem follows that $g$ has to be constant. How can I show that $f$ is constant? As I see it $f$ must be linear but not constant.

Akut Luna
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    You are almost there: notice that $f$ cannot have a zero, and ask yourself: "what linear functions are never zero?" –  Nov 08 '21 at 15:05

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You're right. From $g$ being constant (and equal to $\alpha$), what we can conclude is that $f(z)=\alpha z$ for some $\alpha\in\Bbb C$. Next, you have to show that $|\alpha z|>|z|$ can't hold everywhere, and you're done.

Arthur
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Simply note that $g(0)=0$ and you're done, because $g(z)\ne 0$ for $z\ne0$.

egreg
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