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 $0$?

If $z+1 = r(z-1)$, then the complex numbers $0, z-1, z+1$ lie on a line.  But the line through $z-1$ and $z+1$ can only contain $0$ if $z$ is real.
