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

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. – Ya G 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 $z\in\mathbb{C}$ – Ya G Dec 19 '18 at 17:57

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 $$$$f \circ g^{-1} = e^{i\theta} \frac{z - \alpha}{1-\overline{\alpha}z}$$$$ In particular as $$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$$). So in particular, $$$$f \circ g^{-1} = z e^{i \theta}$$$$ So it follows from composing more that $$$$f = cg$$$$ for some $$c$$ with magnitude $$1$$.