I want to find a Möbius transformation $T$ such that

  • T fixes $i$,
  • Sends the region $\vert z-i\vert < 2$ to the upper half plane,
  • Maps the imaginary axis onto itself

I'm really stuck on this even though it's probably a very basic problem. I know that I have to define $6$ points; $z_1,z_2,z_3,w_1,w_2,w_3 \in \mathbb{C}^{*}$ and solve the cross-ratio equation $(z,z_1,z_2,z_3) =(w,w_1,w_2,w_3)$ where $w = T(z)$. I also know that Möbius transformations maps generalized circles to generalized circles and that Möbius transformations preserves orientation induced by $z_1,z_2,z_3$ if these points happen to live on a generalized circle and that I probably should use these facts to solve this.

Obviously one $z_j = w_j = i$ for some $j \leq 3$ but I'm not sure how to proceed. The imaginary axis should be mapped to itself but does this mean $T(ai)= ai, \ \forall a \in \mathbb{R}$ or something like $T(ai) = bi$? I'm also not sure how to think regarding mapping the region to the upper half plane.

EDIT: I know the answer should be $T(z) = -i\frac{z+i}{z-3i}$ or $T(z) = -i\frac{z-3i}{z+i}$ but I want to know how to get there!


1 Answer 1


Let $$\varphi (z)=e^{i\alpha }\frac{z-i}{2}\quad (\alpha \in\mathbb{R}).$$ This is a general Moebius transformation mapping $\vert z-i\vert < 2$ to the unit disc and sending $i$ to the origin. Now consider $$ g(z)=i\frac{1+z}{1-z}, $$ which maps the unit disc to the upper half plane with $g(0)=i.$ Thus $$ T(z)=(g\circ\varphi )(z)=i\frac{2+e^{i\alpha }(z-i)}{2-e^{i\alpha }(z-i)}\quad(\alpha \in \mathbb{R})$$ is a general form of $T$ such that

  • $T$ fixes $i,$
  • Sends the region $|z-i|<2$ to the upper half plane.

Our task is to determine $\alpha $ so that $T$ maps the imaginary axis onto itself. The condition is $\operatorname{Re} \varphi (it)=0$ for all $-1<t<3$. Easy calculation yields that $\alpha =\frac{\pi}{2}, -\frac{\pi}{2}.$ Thus we get $$ T(z)=-i\frac{z-3i}{z+i},\quad -i\frac{z+i}{z-3i}.$$


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