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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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    $f^{-1} \circ g(z_0)=z_0$ not $0.$2011-12-14

1 Answers 1

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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)|.