# Proof for $|f'(z_0)|\geq |g'(z_0)|$

I need some help with this question (I tried to solve it but I'm not so sure that the solution is exact).

Let $\Omega \subset \mathbb C$ open set. Let $f\colon\Omega \to D$ analytic bijective and on function, $D=\{z:|z|<1\}$. Let $g\colon\Omega \to D$ an analytic function.

I want to show that if $z_0 \in \Omega$ and $f(z_0)=g(z_0)=0$ then $|f'(z_0)|\geq |g'(z_0)|$.

My solution was: $h=f^{-1}\circ g\colon D \to D$, and $f^{-1}\circ g(z_0)=0$.

$h'=c\varphi_{z_0}(z)=c\frac{z-z_0}{1-\overline{z_0}z}$ as $|c|=1$ now, I want to say that $h' \leq c$ and therefore: $\frac{|g'(z_0)|}{|f'(z_0)|}\leq |c|$ so we get: $|f'(z_0)| \geq |g'(z_0)|$.

I'm sorry that the question is long but I'll be very happy if someone will order the solution (or even give a better solution).

Thanks.

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We have $h(z_0)=f^{-1}(g(z_0))=f^{-1}(0)=z_0$. By "injective", do you mean bijective? –  Davide Giraudo Dec 13 '11 at 21:23
yes, bijective. I wrote "injective and on" but bijective is the correct terminology. I'll edit it, thanks. –  bond Dec 13 '11 at 21:34
$f^{-1} \circ g(z_0)=z_0$ not $0.$ –  Ehsan M. Kermani Dec 14 '11 at 7:45
Define the holomorphic map $h:=g\circ f^{-1} : D \to D.$ Then, $h(0)=g(f^{-1}(0))=0.$ Now, by Schwarz lemma, we have $|h'(0)| \leq 1$ which means that $|g'(f^{-1}(0)) (f^{-1})'(0)| \leq 1$ or $|g'(z_0) \frac{1}{f'(z_0)}| \leq 1$, hence $|g'(z_0)| \leq |f'(z_0)|.$