Conformal Map from Vertical Strip to Unit Disc

I haven't found a similar question on here, though I suspect the question may be rather well-covered.

I want to find a conformal map from the vertical strip $\{z:-1<Re(z)<1\}$ onto the unit disc.

Under the exponential map the region is taken to the annulus with radii $e$ and $e^{-1}$, but I'm not sure how useful this will be.

Can anyone advise on what the conformal map may be?

Thanks.

• Turn it to become horizontal. What does $\exp$ do now? Adjust the width as necessary. May 16, 2013 at 18:32
• Didn't work out the $z \mapsto iz$. Thanks for your advice. May 16, 2013 at 18:54

First apply $z \mapsto iz$ to map the strip onto the corresponding horizontal strip $-1 < \operatorname{Im} z < 1$.

Next step, apply $z \mapsto \exp(\frac{\pi z}2)$. This will give you the right half-plane $\operatorname{Re} z > 0$.

To finish off, the Möbius transformation $z \mapsto \dfrac{z-1}{z+1}$ takes the half-plane onto the unit disc.

• Thanks! Couldn't make the jump to apply $z \mapsto iz$. May 16, 2013 at 18:54
• Also known as $i \tan(\pi z/4)$. So $\tan(\pi z/4)$ is a nice option.
– WimC
May 16, 2013 at 19:01
• @WimC, yes, but a lot harder to find without the intermediate steps, unless you're really good at complex trigonometric functions.
– mrf
May 16, 2013 at 19:08
• Why not using $e^{i\frac{\pi}{2}}$ which is the same as $iz$ ? Jun 21, 2017 at 19:29
• One could mention is that what you get is essentially the tangent function, $i \cdot \tan(\pi z/4)$ to be precise. Dec 15, 2020 at 13:18