Define $U_r=\{z:|z|>r\}$ and let $f:U_{r_1}\to U_{r_2}$ be bijective and holomorphic. Prove that $f(z)=az$ for some $|a|=\frac{r_2}{r_1}$.

My attempt: I tried defining $g(z)=\frac{1}{\overline{f(\frac{1}{\bar{z}})}}$. I believe that since $f$ is bijective and holomorphic, we can show that $\infty$ is necessarily a pole, and so $g(0)=0$, where $g:\mathbb{D}{\frac{1}{r_1}}(0)\to\mathbb{D}{\frac{1}{r_2}}(0)$. Now by Riemann's mapping theorem, I can find $\varphi_{r_1},\varphi_{r_2}$ that fix $0$ and s.t $\varphi_{r_2}\circ g\circ\varphi_{r_1}:\mathbb{D}\to\mathbb{D}$. Now I want to use Schwartz lemma somehow, but don't know how exactly.

Any help would be appreciated.


1 Answer 1


Following your idea, you can take $\varphi_{r_1}=\frac{1}{r_1}z, \varphi_{r_2}=r_2z$. Since $$\varphi_{r_2}\circ g\circ\varphi_{r_1}:\mathbb{D}\to\mathbb{D}$$ is an automorphism of the disk, it is of the form $e^{i\theta}\frac{z-w}{1-z\bar{w}}$. Since it fixed $0$, it's actually of the form $e^{i\theta}z$. So: $$\varphi_{r_2}\circ g\circ\varphi_{r_1}=e^{i\theta}z=h(z)$$ So $g(z)=\varphi^{-1}_{r_2}\circ h\circ\varphi^{-1}_{r_1}$. But we know what those functions are: $$g(z)=\varphi^{-1}_{r_2}\circ h\circ\varphi^{-1}_{r_1}(z)=\varphi^{-1}_{r_2}(h(r_1z))=\varphi_{r_2}^{-1}(r_1e^{i\theta}z)=\frac{r_1e^{i\theta}}{r_2}z$$ So:$$\frac{1}{\overline{f(\frac{1}{\bar{z}})}}=\frac{r_1e^{i\theta}}{r_2}z$$ Hence: $$\overline{f(\frac{1}{\bar{z}})}=\frac{r_2}{r_1e^{i\theta}z}\Rightarrow f(z)=\frac{r_2}{r_1e^{-i\theta}}z$$ So $f(z)=az$ with $a=\frac{r_2}{r_1e^{-i\theta}}$, and $|a|=\frac{r_2}{r_1}$


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