# Conformal mapping of an infinite strip onto itself

1. Apr 6, 2014

1. The problem statement, all variables and given/known data
Find a conformal mapping of the strip $D=\{z:|\Re(z)|<\frac{\pi}{2}\}$ onto itself that transforms the real interval $(-\frac{\pi}{2},\frac{\pi}{2})$ to the full imaginary axis.

3. The attempt at a solution
I tried to map the strip to a unit circle and then map it back to the strip. First I used $f_{1}(z)=2z$ to expand the strip to $\{z:|\Re(z)|<\pi\}$ and then I rotated it by applying $f_{2}(z)=iz$. I then applied $f_{3}(z)=e^z$ to send the region to the unit disk. I then applied $f_{4}(z)=\frac{1+z}{1-z}$ to send the unit disk to the half plane $\{z:\Re(z)>0\}$ and then I applied $f_{5}(z)=Log(z)$ to send that back to $\{z:|\Im(z)|<i\frac{\pi}{2}\}$. I then applied $f_{2}$ once more to rotate it back to the strip $D=\{z:|\Re(z)|<\frac{\pi}{2}\}$.

After composing all these functions, I ended up with something along the lines of $-z$ which might even be $z$ depending on whether or not I messed up a sign somewhere in the calculations. This doesn't map the interval to the imaginary axis. Can someone help me include that part?

EDIT**: I know that the easiest way to map the strip to itself is by using $z,-z$ but that doesn't help me map the interval to the imaginary axis. I was hoping that something magical would happen in the mess up there but nothing happened...

Last edited: Apr 7, 2014
2. Apr 7, 2014

I figured it out for future reference to anybody.

Use $sin(z)$ to take the infinite strip to $\mathbb{C}\sim\{w:|\Re(w)|\geq 1$ and $\Im(w)=0\}$. Then rotate this by multiplying by $i$ and finally use $Arctan(w)$ to take it back to the infinite strip.