# Conformal map from upper-half plane to upper-half plane

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.

## 1 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.

• I understood $g(z) = e^{i\theta}z$, but I think $f(z) = \phi(e^{i\theta}\phi^{-1}(z))$ from $g(z) = e^{i\theta}z$. How can I write down f$(z) = \phi(f(\phi^{-1}(z)))$ ? – 0721 ubari Nov 29 '20 at 7:49
• @0721ubari First prove that $\phi^{-1}(z)=\frac {i (i-z)} {i+z}$ by solving $\phi(z)=\zeta$ for $z$ in terms of $\zeta$. . – Kavi Rama Murthy Nov 29 '20 at 8:03
• You told $f(z)=\phi(f(\phi ^{-1}(z)))$,but is that $f(z)=\phi(g(\phi ^{-1}(z)))$ ? (sry for my poor English.) – 0721 ubari Dec 1 '20 at 7:49
• @0721ubari Yes, there was a typo. – Kavi Rama Murthy Dec 1 '20 at 7:51
• ok, then I calculate the equation $f(z)=\phi(g(\phi ^{-1}(z))) = i\frac{(i+e^{i\theta}) z -(ie^{i\theta}+1)}{(i-e^{i\theta})z+(ie^{i\theta}-1)}$ and I stopped. What should I do next? – 0721 ubari Dec 1 '20 at 8:28