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?


  • 2
    $\begingroup$ Turn it to become horizontal. What does $\exp$ do now? Adjust the width as necessary. $\endgroup$
    – 75064
    May 16, 2013 at 18:32
  • $\begingroup$ Didn't work out the $z \mapsto iz$. Thanks for your advice. $\endgroup$
    – Mathmo
    May 16, 2013 at 18:54

1 Answer 1


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, 2013 at 18:54
  • 2
    $\begingroup$ Also known as $i \tan(\pi z/4)$. So $\tan(\pi z/4)$ is a nice option. $\endgroup$
    – WimC
    May 16, 2013 at 19:01
  • 1
    $\begingroup$ @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, 2013 at 19:08
  • $\begingroup$ Why not using $e^{i\frac{\pi}{2}}$ which is the same as $iz$ ? $\endgroup$ Jun 21, 2017 at 19:29
  • $\begingroup$ One could mention is that what you get is essentially the tangent function, $i \cdot \tan(\pi z/4)$ to be precise. $\endgroup$
    – Martin R
    Dec 15, 2020 at 13:18

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