Find conformal mapping from open strip onto open quarter disk.

Question

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.

Answer 1

$e^{\pi z/2}$ 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.