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.

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 $. 
