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 $0$?
 A: The idea is good:
$$
\frac{1+z}{z-1}=\frac{z-1+2}{z-1}=1+\frac{2}{z-1}
$$
Since $1$ and $2$ are real, you need that $z-1$ is real, which means…
Another way: write
$$
w=\frac{1+z}{z-1}
$$
Then $zw-w=1+z$ and so $z(w-1)=w+1$, which means
$$
z=\frac{w+1}{w-1}
$$
If $w$ is real, then also $z$ is real. And conversely.
A: Try writing 
$$\frac{2}{a+ib-1}=\frac{2}{a+ib-1}\cdot \frac{a - ib - 1}{a - ib - 1}$$
After you perform the multiplication, it should be pretty clear what values of $a,b$ make this fraction real...
A: Hint:
$$\begin{align}\frac{1+z}{z-1} \,\,\, \text{is real} \  &\iff \frac{1+z}{z-1} = \overline{\frac{1+z}{z-1}} = \frac{\overline{1+z}}{\overline {z-1}} \\&\iff (1+z)(\overline {z}-1) = (1+\overline {z})(z-1)\\&\iff \mathfrak {Im} (z) = 0\end{align}$$
A: Clearly, real $z$ (except $z=1$) make the expression real, and the map $z\mapsto\frac{1+z}{z-1}$ is a bijection from $\mathbb{C}\cup\{\infty\}$ to itself (with $1\leftrightarrow\infty$). The inverse map is the very same map.
Also, it maps $\mathbb{R}\cup\{\infty\}$ to itself. It follows that no other complex $z$ beyond real $z$ will give real values of the expression.
A: 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.
