I am trying to solve a below problem.
Problem
Let $f$ be a conformal map from upper-half plane to upper-half plane and $f(i)=i$. Show that there exists $\theta \in \mathbb{R}$ such that $f(z) = \frac{(cos{\theta})z + (sin{\theta})}{(-sin{\theta})z + (cos{\theta})}$.
By Schwarz lemmma and Cayley transformation $T(z)=\frac{z-i}{z+i}$, $|T\circ f\circ T^{-1} (z)| = |z|$. It seems that the lemma and the transformation is useful to solve this problem, however I cannot find the $\theta$. How can I find the $\theta$ ? Thank you.
Best Answer
Hint: Let $\phi (z)=\frac {i(i-z)} {i+z}$. $g(z)=\phi^{-1}(f(\phi(z)))$ defines a conformal equivalence of the open unit disk such that $g(0)=0$. All conformal equivalences of the unit disk have the form $e^{i\theta }\frac {z-a} {1-\overline a z}$ and among them the only one's vanishing at $0$ are of the form $z \to e^{i\theta} z$. [This is proved in Rudin's RCA]. Hence $g(z)=e^{i\theta} z$ and you can write down $f(z)=\phi(g(\phi^{-1}(z)))$ from this.