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.