# conformal map of unit disk slit

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

• Similar problem: Conformal map from a disk onto a disk with a slit – user147263 Feb 23 '15 at 2:21
• Thanks, but the problem that I don't understand how functions transfers a region to another which prevents me from coming up with a suitable transfer myself. I was hoping that someone can give me a clear idea in how the function works – Dan Feb 23 '15 at 12:48

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.
• You have a slit at $(0,\infty)$ since no points are mapped to them under the map $z\mapsto z^2$. The only possible points that could get mapped to $(0,\infty)$ under this map would be $(-\infty,0)$ but this is part of the boundary of the upper half plane and is therefore not part of our graph. – The Substitute Mar 1 '15 at 0:37