Describe all the complex numbers $z$ for which $(iz − 1 )/(z − i)$ is real. 
Describe all the complex numbers $z$ for which $(iz − 1 )/(z − i)$ is real.
Your answer should be expressed as a set of the form $S =  \{z \in\mathbb C : \text{conditions satisfied by }z\}$.

I started solving for $((iz − 1 )/(z − i)) = \overline{ ((iz − 1 )/(z − i))}$.
I got stuck. Is this the right way to go about this?

We need $i\cdot\dfrac{z+i}{z-i}$ purely real  
$\iff\dfrac{z+i}{z-i}$ purely imaginary $=iy$(say) where $y$ is real
$$\frac{z+i}{z-i}=iy\implies z+i=iy(z-i)=z(iy)+y\implies z(1-iy)=y-i\implies z=\frac{y-i}{1-iy}$$
