0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.