Given the function $f(z)=\bigl(\frac{z+1}{z-1}\bigr)^{2}$ when $f(1)=\infty,f(\infty)=1$, I need to find the image of the extended real line and the extended imaginray line.
In order to do so, I worked with the Möbius transformation $\psi(z)=\frac{z+1}{z-1}$.
Finding the image of the extended real line was easy- as a Möbius transformation sends extended circles to extended circles in $\overline{\mathbb{C}}$ which are characterized by three points, I found the image of three points on the extended real line to be on the extended real line (from here I'll use e.r.l), so $\psi(z)$ sends the e.r.l to itself. $f(z)=(\psi(z))^{2}$ thus sends the e.r.l to the positive half of the e.r.l in $\overline{\mathbb{C}}$.
Now, for finding the image of the extended imaginary line (e.i.l) in $\overline{\mathbb{C}}$, I used the same reasoning and got:
$\psi(i)=\frac{i+1}{i-1}=\frac{-(i+1)^{2}}{2}, \psi(0)=-1, \psi(-i)=\frac{-i+1}{-i-1}=\frac{-(i-1)^{2}}{2}$.
So the e.i.l is sent by $\psi$ to a circle going through $\frac{-(i+1)^{2}}{2},\frac{-(i-1)^{2}}{2},-1$.
Before I go through the quite tedious work of finding the equation of the circle, a troubling thought came to my mind. Say I have the circle. I need to find the image under $f(z)=(\psi(z))^2$.
Should I square the circle?
Silly me. Looking again at the points on the circle which is the image of the e.i.l under $\psi(z)$, I see they simplify to:
$\psi(i)=\frac{i+1}{i-1}=\frac{-(i+1)^{2}}{2}=\frac{-(-1+1+2i)}{2}=-i$, $\psi(0)=-1$ and $\psi(-i)=\frac{-i+1}{-i-1}=\frac{-(i-1)^{2}}{2}=\frac{-(-1+1-2i)}{2}=i$.
These points determine the unit circle in $\overline{\mathbb{C}}$!
Now, let's see what happens under the given function $f(z)=(\psi(z))^{2}$.
Let $a\in S^1$ be a general point on the unit circle. Then $a=e^{i\theta}$ for some $\theta$.
$f(a)=f(e^{i\theta})=e^{2i\theta}$.
So $f$ transforms the unit circle in $\overline{\mathbb{C}}$ onto the unit circle in $\overline{\mathbb{C}}$.
Finally, we conclude that $f(z)=\bigl(\frac{z+1}{z-1}\bigr)^{2}$ takes the extended imaginary line to the unit circle in $\overline{\mathbb{C}}$.