Find $z$ s.t. $\frac{1+z}{z-1}$ is real I must find all $z$ s.t. $\dfrac{1+z}{z-1}$ is real. 
So, $\dfrac{1+z}{1-z}$ is real when the Imaginary part is $0$. 
I simplified the fraction to $$-1 - \dfrac{2}{a+ib-1}$$
but for what $a,b$ is the RHS $0$?

Hint:
$$\begin{align}\frac{1+z}{z-1} \,\,\, \text{is real} \  &\iff \frac{1+z}{z-1} = \overline{\frac{1+z}{z-1}} = \frac{\overline{1+z}}{\overline {z-1}} \\&\iff (1+z)(\overline {z}-1) = (1+\overline {z})(z-1)\\&\iff \mathfrak {Im} (z) = 0\end{align}$$
