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

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?

Answer: 
Let $z=x+iy$ where $x,y$ are real

$$\frac{iz-1}{z-i}=\frac{i(x+iy)-1}{x+iy-i}=\frac{ix-(y+1)}{x+i(y-1)}$$

$$=\frac{[ix-(y+1)][x-i(y-1)]}{[x+i(y-1)][x-i(y-1)]}$$

whose imaginary part is
$$=\frac{x^2+y^2-1}{x^2+(y-1)^2}$$ which must be zero

$$\implies x^2+y^2=1$$

So, $z$ must lie on the circle whose centre is $\left(0,0\right)$ and radius $=1$
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

$$\frac{z+i}{z-i}=iy\implies z+i=iy(z-i)=z(iy)+y\implies z(1-iy)=y-i\implies z=\frac{y-i}{1-iy}$$