How does this complex function map πi/2 to => 0 So I'm working on a question that asks
Find a conformal map that maps the horizontal strip $Im(z) ∈ (0,π)$ onto the unit disk $|z| < 1$
and maps $πi/2$ to 0
.
It says the solutions involve:
$f_1(z)=z-\frac{\pi}{2}$
I'm confused as to why this would even be included arrived as the first one doesn't work for mapping π/2 to 0:
$f_1(\frac{\pi i}{2})=\frac{\pi}{2}(i-1)≠0$
Shouldn't the proper mapping be
$f_1(z)=z-\frac{\pi i}{2}$
 A: A solution involves $f_1(z) = z - \frac{\pi}{2}$, meaning that you probably have to find additional conformal maps $f_2,f_3,...$ in order to solve the exercise. So it is no surprise that $f_1(\frac{\pi}{2})\neq0$. Even if you were to define $f_1(z) = z-\frac{\pi i}{2}$, $f_1$ still wouldn't be a conformal map $A := \{z \in \mathbb{C} \mid \text{Im}(z) \in (0,\pi)\} \to \mathbb{D}$, so that doesn't really help.
However, I can give you a hint on how to solve the exercise differently (in my opinion the easiest way): Try to find conformal maps $f_1 : A \to \mathbb{H}$ and $f_2 : \mathbb{H} \to \mathbb{D}$. Here, $\mathbb{H}$ denotes the upper half plane. Then, you can define $f : A \to \mathbb{D}$ as $f := f_2 \circ f_1$ and you'll see that indeed $f(\frac{\pi i}{2})=0$, as needed.
$f_1$ is a map that I'm sure you've used many times before, and if you're having trouble coming up with $f_2$ you can look up "Cayley transform/map". If you need additional hints, feel free to let me know.
I hope this was helpful!
