TITLE: Conformal map from a lune to the unit disc in $\mathbb{C}$
QUESTION [13 upvotes]: 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?

REPLY [8 votes]: Just to put everyone's answers together and finish things off, $f:z\mapsto
\frac{1+z}{1-z}$ sends $-1$ to $0$ and $1$ to $\infty$.  Because $f:(-1,1)\to(0,+\infty)$, the sector is symmetric about $\mathbb{R}^+$.  Since $f$ is conformal, it preserves the right angle at which the circles cross, so the sector has a right angle. Therefore, $g:z\mapsto z^2$ maps the sector to the right half-plane.  Then, $h:z\mapsto\frac{z-1}{z+1}$ maps the right half-plane to the unit disk. Thus,
$$
h\circ g\circ f:z\mapsto\frac{2z}{1+z^2}
$$
should map the given lune to the unit disk.