I'm trying to find a conformal mapping $f:A \rightarrow B$ from the open strip $$A = \{z \in \mathbb{C}| Re(z) < 0,0<Im(z)<1\}$$ onto the open quarter disk in the first quadrant given by $$B=\{z \in \mathbb{C}| Re(z) > 0,Im(z)>0\} \cap\{z \in \mathbb{C}| |z|<1\} $$. First of all, am I right in assuming that $e^{\frac{\pi}{2}z}$ will map the $A$ onto the upper half plane from which I can map onto the unit disk using the Möbius transform? Also, how can I restrict the Möbius transform in a way that I will end up with a quarter disk? Thanks for any answers.

  • $\begingroup$ $z^2/i$ maps the open quarter disc to the open half-disc in the right half-plane, then one maps the unit circle to the upper half-plane so that half-disc becomes a quadrant, so by another squaring we get the full half-plane, so reversing that plus your map from the strip to the upper half plane gives the required $f$ $\endgroup$ – Conrad Jun 1 '20 at 2:58
  • $\begingroup$ No, $e^{\pi z/2}$ maps $A$ to something much more convenient :). Consider the absolute value and the argument of $e^{\pi (x + i y)/2}$ on $A$. $\endgroup$ – Maxim Jun 1 '20 at 19:59

$e^{\pi z/2}$ does not map $A$ to the upper half-plane, but to $B$. It is already the answer.

$z\to\pi z/2$ only changes the strip's height to $\pi/2$; it remains infinite to the left. Then the exponential map turns the strip into the region between radii of $e^{-\infty}=0$ and $e^0=1$, and between angles of $0$ and $\pi/2$ – in other words, $B$ itself.


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