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Given two biholomorphic maps $f:\Omega\rightarrow\mathbb{D}$ and $g:\Omega\rightarrow\mathbb{D}$ such that $f(z_0)=g(z_0)=0$, prove that there exists $c\in\mathbb{C}$ with $|c|=1$ such that $f(z)=cg(z)$

If $f$ or $g$ is identically zero, it is trivial as $0=c0$, so assume they are both not identically zero. Assume WLOG,$|f|\leq|g|$. Then, $f(z)=(z-z_0)^mk(z)$ and $g(z)=(z-z_0)^nh(z)$ where $k(z_0)$ and $h(z_0)$ are both not zero. Then, for $z\neq z_0$, $$\left|\frac{(z-z_0)^{m-n}k(z)}{h(z)}\right|\leq1$$ and there exists some constant $k$ such that $\left|\frac{k(z)}{h(z)}\right|\geq \frac{1}{k}$ so we have $$\frac{|z-z_0|^{m-n}}{k}\leq\left|\frac{(z-z_0)^{m-n}k(z)}{h(z)}\right|\leq 1\Rightarrow|z-z_0|^{m-n}\leq k$$

How do I proceed further to show that there is a constant $c$? or am I totally wrong?

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  • $\begingroup$ Why do you think that you can assume that $\lvert f\rvert\leqslant\lvert g\rvert$ without loss of generality? $\endgroup$ – José Carlos Santos Dec 19 '18 at 17:25
  • $\begingroup$ @JoséCarlosSantos because if they are equal to each other, it's very trivial and always true for some $c$ where $|c|=1$. Maybe I'm not interpreting the problem correctly. $\endgroup$ – Ya G Dec 19 '18 at 17:43
  • $\begingroup$ Indeed, if $\lvert f\rvert=\lvert g\rvert$, then the problem is trivial. But is is perfectly possible that for some $z$ you have $\bigl\lvert f(z)\bigr\rvert<\bigl\lvert g(z)\bigr\rvert$, where as for some $w$ you have $\bigl\lvert f(w)\bigr\rvert>\bigl\lvert g(w)\bigr\rvert$. $\endgroup$ – José Carlos Santos Dec 19 '18 at 17:47
  • $\begingroup$ To be clear, are you saying that for all $z\in\mathbb{C}$ there is a unique $c\in\mathbb{C}$ such that $f(z)=cg(z)$. $\endgroup$ – R. Burton Dec 19 '18 at 17:55
  • $\begingroup$ @R.Burton The problem itself doesn't state uniqueness of $c$.But it should hold for all $z\in\mathbb{C}$ $\endgroup$ – Ya G Dec 19 '18 at 17:57
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Consider $f \circ g^{-1}$ this is an automorphisms of $\mathbb{D}$ since $f$ and $g$ are biholomorphic. In particular, we know all automorphisms of the unit disk are given by Blaschke factors (http://mathworld.wolfram.com/BlaschkeFactor.html). Therefore, one has \begin{equation} f \circ g^{-1} = e^{i\theta} \frac{z - \alpha}{1-\overline{\alpha}z} \end{equation} In particular as $f \circ g^{-1} (0) = f(z_0) = 0$, one sees $\alpha = 0$ (this can also be seen by the Blaschke Factor inter swaps $0$ and $\alpha$). So in particular, \begin{equation} f \circ g^{-1} = z e^{i \theta} \end{equation} So it follows from composing more that \begin{equation} f = cg \end{equation} for some $c$ with magnitude $1$.

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