I am trying to map the region G, {|z|<1, |z+i|> (2)^0.5} to the infinite vertical strip at x = +/- pi.

I have started by using a Mobius Map which sends the two common points of the circles to 0 and infinity (z=+/- 1), namely, using the map f(z)=(z-1)/(z+1). This maps the unit circle to the positive imaginary axis and should map the other circle to a circle. However, when I calculate the image of the other circle, I don't get a region that is one of the axes as one would expect...

If Mobius maps take circlines to circlines, why is this the case?

  • $\begingroup$ Such a Mobius transform doesn't exist: the two boundary circles have two common points, while the two boundary lines have one common point ($\infty$). You can find a mapping between the two given regions in the form $a \ln((1 + z)/(1 - z)) + b$. $\endgroup$ – Maxim Mar 26 at 18:29
  • $\begingroup$ the two boundary lines can have 2 common points though - 0 and ∞? $\endgroup$ – MathematicianP Mar 26 at 19:42
  • $\begingroup$ Which straight lines are we talking about? The vertical strip $-\pi < \operatorname{Re} z < \pi$ is bounded by the lines $\operatorname{Re} z = \pm \pi$. They do not go through zero. $\endgroup$ – Maxim Mar 26 at 20:03
  • $\begingroup$ Oh i see, but shouldn't the mobius map (z-1)/(z+1) take the two circles to circles? I don't think it does for |z+i|=(2)^0.5 $\endgroup$ – MathematicianP Mar 26 at 20:07
  • $\begingroup$ It does. $(z - 1)/(z + 1)$ maps $|z + i| = \sqrt 2$ to the straight line that goes through zero at the same angle with the imaginary axis as the angle between the two circles at $1$. $\endgroup$ – Maxim Mar 26 at 20:18

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