# Proving a function satisfying a condition is polynomial of degree 1

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.

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 $$|a|=\frac{r_2}{r_1}$$