On my complex analysis prelim this morning I was asked to give a conformal map from the region $L=\{z\in\mathbb{C}:|z-i|<\sqrt{2},|z+i|<\sqrt{2}\}$, a lune with vertices at $-1$ and $1$ to the unit disc $\mathbb{D}=\{z:|z|<1\}$. I tried to send $L$ to the upper half plane by the Möbius transform sending $(-1,0,1)$ to $(0,i,\infty)$. Then I composed with the Cayley transformation to get to the unit disc.
My question is: does my first map do what I want it to(presuming I calculated it correctly)?
To be brief, does the Möbius transform which takes $(-1,0,1)$ to $(0,i,\infty)$ send $L$ to the upper half plane?
Best Answer
Your map sending $(-1,0,1)$ to $(0,i,\infty)$ is given by $i = e^{i\pi/2}$ times the cross ratio $[z,0,-1,1] = \frac{z+1}{z - 1} : \frac{1}{-1}$, so it is $$\tilde{\phi}(z) = e^{i\pi/2}\cdot \frac{1+z}{1-z}.$$ It is easy to see geometrically (or by a direct calculation) that $\tilde\phi(L)$ is the quadrant $\{z:\,\operatorname{Im}{z} \gt |\operatorname{Re}{z}|\}$ since the circles $|z-i|=\sqrt{2}$ and $|z+i|=\sqrt{2}$ are sent to the lines $\{\operatorname{Im}{z} = \operatorname{Re}{z}\}$ and $\{\operatorname{Im}{z} = -\operatorname{Re}{z}\}$. Note that the angles at the vertices of the lune are equal to $\pi/2$.
It is more convenient to work with $\displaystyle\phi(z) = e^{i\pi/4}\cdot \frac{1+z}{1-z}$ which sends $L$ to the quadrant $\{z:\operatorname{Re}{z}, \, \operatorname{Im}{z}\gt 0\}$, then square to get to the upper half plane and apply the Cayley-transfom $\displaystyle\kappa(z) = \frac{z-i}{z+i}$ to get to the unit disk.
The solution to your problem then is $\kappa((\phi(z))^2)$, which you can compute yourself if needed.
The general procedure is succinctly explained in GEdgar's answer.