I'm looking to investigate how the exponential map acts on the semi inifite strip

$$U:=\{z \in \mathbb{C} : Re(z)<0, 0<Im(z)<\pi \} $$

My understanding is that for the infinite strip, the region will be mapped to the UHP, but I'm not sure how the semi infinite condition affects the mapping.

Ultimately I'm looking for a conformal map from $U$ to the UHP.

Thanks in advance.

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    $\begingroup$ It maps to the uppar half of the unit disc. $\endgroup$ – WimC May 27 '12 at 18:05
  • $\begingroup$ @WimC: Thanks, are you able to briefly explain why? $\endgroup$ – Mathmo May 27 '12 at 18:08
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    $\begingroup$ $e^{x + i y} = e^x(\cos(y) + i \sin(y))$. $\endgroup$ – WimC May 27 '12 at 18:10
  • $\begingroup$ @WimC: Thanks! Should've seen that! $\endgroup$ – Mathmo May 27 '12 at 18:15

Apologies.. you were asking for the exponential mapping. Should have seen that! answer left for reference though. :)

Clearly a semi-infinite strip is still infinite, but only on say X>X0 or Y>Y0 with the "non-infinite" dimension bounded. The mapping for a semi-infinite strip still maps to an upper half plane, it is just slightly different mathematically. The mapping is really quite simple if one is but to apply a Schwarz-Christoffel mapping as the limiting form of a triangle (to quote the wikipedia page on Schwarz-Christoffel transformations).

I know it has been a year since your post, but I stumbled upon this unanswered question in my hunt for my own answers!


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