conformal map of unit disk slit

Question

Map the unit disk slit along $(-1,-r ]$, $r \in (0, 1)$, onto the unit disk.

Can anyone explain how to do the conformal map thoroughly since I have difficulty understanding it.

Thanks

Answer 1

Note there are many ways to do this.

Let's denote by $U$ our original disc slit along $(-1,r]$ (along the $x$-axis). Let's rotate by $\pi$ radians counterclockwise by applying the map $z\mapsto e^{\pi i}z=-z$ and denote our new region by $-U$. This is the unit disc slit along $[r,1)$ (along the real axis). Apply the inverse Cayley transform $z\mapsto i\frac{1+z}{1-z}$ on $-U$ to obtain the upper half plane slit along $\Big[i\frac{1+r}{1-r},i\infty\Big)$ (along the $y$-axis ), so denote this region by $\mathbb{H}-\Big[i\frac{1+r}{1-r},i\infty\Big).$ Apply the map $z\mapsto z^2$ to obtain the plane with slits from $\Big[-\infty,-\Big(\frac{1+r}{1-r}\Big)^2\Big]$ and $[0,\infty)$, both on the real axis. Hence, we apply the map $z\mapsto\frac{z}{z+\bigg(\frac{1+r}{1-r} \bigg)^2}$ to obtain the plane slit along $(0,\infty]$ on the real axis. Apply a branch of $\sqrt z$ to map to $\mathbb{H}$, and then apply Cayley's transform into the unit disc.