Question

There exists an analytic univalent function f that maps the infinite strip {z : 0 < Im z < 1} onto the unit disk.

Answer 1

$z_1=\pi z$ maps the horizontal strip $\{z: \ 0<\text{Im } z<1\}$ onto the horizontal strip $\{z_1: \ 0<\text{Im } z_1<\pi\}$ .

The exponential function $z_2=e^{z_1}$ maps the horizontal strip $\{z_1: \ 0<\text{Im } z_1<\pi\}$ to the half plane $\{z_2: \ \text{Im} \ z_2>0\}$ .

$z_3=e^{-i \pi/2}z_2$ rotates the complex plane by $-90^\circ$, $\{z_3:\ \text{Re}\ z_3>0\}$ .

The linear fractional transformation $w=\frac{1-z_3}{1+z_3}$ maps the right half plane $\{z_3:\ \text{Re} \ z_3>0\}$ onto the unit disk.

So, we have function $f=\frac{1-z_3}{1+z_3}= \frac{1-e^{-i \pi/2}z_2}{1+ e^{-i \pi/2}z_2}=\frac{1+i z_2}{1-i z_2}= \frac{1+i e^{z_1}}{1-i e^{z_1}}= \frac{1+i e^{\pi z}}{1-i e^{\pi z}}$ .

Answer: $f(z)= \frac{1+i e^{\pi z}}{1-i e^{\pi z}}.$