I am trying to solve a below 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.


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.

  • $\begingroup$ 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$(z) = \phi(f(\phi^{-1}(z)))$ ? $\endgroup$ – 0721 ubari Nov 29 '20 at 7:49
  • $\begingroup$ @0721ubari First prove that $\phi^{-1}(z)=\frac {i (i-z)} {i+z}$ by solving $\phi(z)=\zeta$ for $z$ in terms of $\zeta$. . $\endgroup$ – Kavi Rama Murthy Nov 29 '20 at 8:03
  • $\begingroup$ You told $f(z)=\phi(f(\phi ^{-1}(z)))$,but is that $f(z)=\phi(g(\phi ^{-1}(z)))$ ? (sry for my poor English.) $\endgroup$ – 0721 ubari Dec 1 '20 at 7:49
  • $\begingroup$ @0721ubari Yes, there was a typo. $\endgroup$ – Kavi Rama Murthy Dec 1 '20 at 7:51
  • $\begingroup$ 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? $\endgroup$ – 0721 ubari Dec 1 '20 at 8:28

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