# Find conformal mapping from open strip onto open quarter disk.

I'm trying to find a conformal mapping $$f:A \rightarrow B$$ from the open strip $$A = \{z \in \mathbb{C}| Re(z) < 0,0 onto the open quarter disk in the first quadrant given by $$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.

• $z^2/i$ 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 $f$ – Conrad Jun 1 '20 at 2:58
• 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 $A$. – 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.