Further explanation needed :Finding all $z$ such that the modulus of $f(z)=e^{(z+1)/(z-1)}$ is equal to/at most $1$ I was solving a previous exam paper and there I got stuck on the following problem:  

Let $f(z)=e^{\frac{z+1}{z-1}}$. Then find all $z \in \Bbb C$ for which    
  
  
*
  
*$|f(z)|=1$,   
  
*$|f(z)|\le 1$. 
  

Any idea about how to tackle this problem will be highly appreciated. Thanks for your time.
 A: If $w$ is any complex number, $|e^w| =  e^{\operatorname{Re}(w)}$, so you are trying to respectively find the $z$ for which $\operatorname{Re}({z +1 \over z- 1}) = 0$ and $\operatorname{Re}({z +1 \over z- 1}) \leq 0$. To figure out the details of which $z$ satisfy these, knowledge of Möbius transformations are helpful, but you can work them out by hand too.
A: Let $ z = a + bi, (a, b) \in \mathbb{R}^2 $. Hence $$\begin{align*} \left|e^{\frac{z+1}{z-1}}\right| &= \left|e^{\frac{(a + 1 + bi)(a - 1 - bi)}{(a - 1 + bi)(a - 1 - bi)}} \right| \\&= \left|\ e^{\frac{a^2 - 1 + b^2 -2bi}{(a - 1)^2 + b^2}}\ \ \right| \\ &= \left|\ e^{ \frac{a^2 -1 + b^2}{(a - 1)^2 + b^2}}\ \ \ \ \ \right|\cdot\left|e^{-\frac{2bi}{(a-1)^2 +b^2}}\right|\\&=\ \ e ^{\frac{a^2 - 1 + b^2}{(a - 1)^2 + b^2)}}\end{align*}$$
Hence, the modulus is $ 1 $ when $ a^2 + b^2 = 1 $, which is a unit circle. Note that the point $ (1, 0) $ should be excluded because $ f(1) $ is not defined.
Also, the modulus is less than $1 $ on the entirety of the disk that is contained by that circle, i.e. $ a^2 + b^2 < 1 $. 
A: Let $z=r e^{i\theta}$, where $r\ge 0$ and $\theta\in \mathbb{R}$. 
Then 
$$\begin{eqnarray*}
|f(r,\theta)| 
&=& \exp\left(\frac{r^2-1}{r^2-2r\cos\theta+1}\right) \\
&=& \exp\left(\frac{r^2-1}{(r-1)^2+2r(1-\cos\theta)}\right).
\end{eqnarray*}$$
I. If $\cos\theta = 1$, $|f| = \exp\left(\frac{r+1}{r-1}\right)$. 
In this case $|f|\ne 1$ for any $r$. 
However, $|f|<1$ for $r<1$. 
II. If $\cos\theta\ne 1$, $(r-1)^2+2r(1-\cos\theta) > 0$. 
Therefore, $|f| = 1$ for $r=1$ and $|f|<1$ for $r<1$. 
We find, $|f| = 1$ for all
$z = e^{i\theta}$,
where $\theta \ne 2\pi n$ and $n\in\mathbb{Z}$.
This is the unit circle with $(1,0)$ deleted.
In addition, $|f|\le 1$ for all 
$z = r e^{i\theta}$
such that 
(1) $r=1$ and $\theta\ne 2\pi n$ or 
(2) $r<1$ and $\theta\in\mathbb{R}$. 
This is the unit circle and its interior with $(1,0)$ deleted.
The results agree with those of @JonClaus, who used Cartesian coordinates. 
Here is a three dimensional plot of $|f|$ above the complex plane. 
