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.


  • $\begingroup$ Similar problem: Conformal map from a disk onto a disk with a slit $\endgroup$ – user147263 Feb 23 '15 at 2:21
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    $\begingroup$ 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 $\endgroup$ – Dan Feb 23 '15 at 12:48

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.

  • $\begingroup$ The first step (the rotation) wasn't necessary, and I only used it because I already know what happens when the slit is on the right side. An alternate solution would be to apply the inverse Cayley transform without rotation and then see what happens from there. $\endgroup$ – The Substitute Feb 26 '15 at 12:04
  • $\begingroup$ Thanks for the explanation. but, I have question. when you use the map z↦z^2 , why there is slit in [0,∞). $\endgroup$ – Dan Feb 28 '15 at 23:30
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    $\begingroup$ 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. $\endgroup$ – The Substitute Mar 1 '15 at 0:37

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