Given two biholomorphic maps such that $f(z_0)=g(z_0)=0$, prove there exists $c$ such that $f(z)=cg(z)$

Question

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?

Why do you think that you can assume that $\lvert f\rvert\leqslant\lvert g\rvert$ without loss of generality? — José Carlos Santos, Dec 19 '18 at 17:25
@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. — Gregoire Rocheteau, Dec 19 '18 at 17:43
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 $\bigl\lvert f(w)\bigr\rvert>\bigl\lvert g(w)\bigr\rvert$. — José Carlos Santos, Dec 19 '18 at 17:47
To be clear, are you saying that for all $z\in\mathbb{C}$ there is a unique $c\in\mathbb{C}$ such that $f(z)=cg(z)$. — R. Burton, Dec 19 '18 at 17:55
@R.Burton The problem itself doesn't state uniqueness of $c$.But it should hold for all $z\in\mathbb{C}$ — Gregoire Rocheteau, Dec 19 '18 at 17:57
@R.Burton No. if $c=i$, $|c|=1$ and that doesn't imply $|f(z)|=|g(z)|$ — Gregoire Rocheteau, Dec 19 '18 at 18:25
@YaG But if $|c|=1$, then $|cg(z)|=|g(z)|$; only the argument changes, unless I'm missing something? — R. Burton, Dec 19 '18 at 18:30
@R.Burton Oh right. You have a good point. May be I should assume one is greater than another and show by contradiction? — Gregoire Rocheteau, Dec 19 '18 at 18:31
I think Schwarz Lemma might be involved in this since $f \circ g^{-1}$ and $g \circ f^{-1}$ map $\mathbb{D} \rightarrow \mathbb{D}$ such that they fix the origin. — Story123, Dec 19 '18 at 21:34
@Story123 I also think that is a great idea. But all I can show is $|f\circ g^{-1}|\leq|z|$ because I cannot find what $[f(g^{-1}(0))]'$ would be — Gregoire Rocheteau, Dec 20 '18 at 03:59

score 3 · Accepted Answer · answered Dec 21 '18 at 18:52

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 $1$.

Given two biholomorphic maps such that $f(z_0)=g(z_0)=0$, prove there exists $c$ such that $f(z)=cg(z)$

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