Conformal mapping of an infinite strip onto itself

[DeadOriginal]

Homework Statement

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.

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...

 

[DeadOriginal]

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.