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?


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    $\begingroup$ Turn it to become horizontal. What does $\exp$ do now? Adjust the width as necessary. $\endgroup$ – 75064 May 16 '13 at 18:32
  • $\begingroup$ Didn't work out the $z \mapsto iz$. Thanks for your advice. $\endgroup$ – Mathmo May 16 '13 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.

  • $\begingroup$ Thanks! Couldn't make the jump to apply $z \mapsto iz$. $\endgroup$ – Mathmo May 16 '13 at 18:54
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    $\begingroup$ Also known as $i \tan(\pi z/4)$. So $\tan(\pi z/4)$ is a nice option. $\endgroup$ – WimC May 16 '13 at 19:01
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    $\begingroup$ @WimC, yes, but a lot harder to find without the intermediate steps, unless you're really good at complex trigonometric functions. $\endgroup$ – mrf May 16 '13 at 19:08
  • $\begingroup$ Why not using $e^{i\frac{\pi}{2}}$ which is the same as $iz$ ? $\endgroup$ – Richard Clare Jun 21 '17 at 19:29

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