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?
 A: 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 $=1$
A: Here are two methods: The first is essentially the same as lab's quite nice first answer but doesn't pass to the real and imaginary parts and so gives immediately a characterization of the complex elements $z \in S$; it is clean but I do not claim it is any better than lab's answer. The second uses the fact that the quantity is the rule of a Mobius transformation $z \mapsto \frac{az + b}{cz + d}$, and so enjoys some nice geometric properties.
Method 1 First note that $iz - 1 = i(z + i)$. Then, multiplying by the conjugate gives that the expression is
$$
\frac{i(z + i)}{z - i} \cdot \frac{\bar{z} + i}{\bar{z} + i}
 = \frac{i(z \bar{z} + i(z + \bar{z}) - 1)}{z \bar{z} + 1}
 = \frac{-(z + \bar{z}) + i(z \bar{z} - 1)}{z \bar{z} + 1}.
$$
Now, the denominator of the right hand-size is real, so the quantity is real when the numerator is. Since both $z + \bar{z}$ and $z \bar{z} - 1$ are real, this happens exactly when $z \bar{z} - 1 = 0$, that is, we may write the indicated set as
$$S := \{z : z \bar{z} = 1\},$$
which is just the unit circle.
Method 2 Alternatively, we can view
$$f: z \mapsto \frac{iz - 1}{z - i}$$
as the rule for a Mobius transformation, and recall the fact that Mobius transformations send circles and lines to circles and lines. In particular, the set $f^{-1}(\mathbb{R})$ for which the $f(z)$ is real is, by definition, sent to the real line, and so it must itself be a line or circle. Checking directly shows that if $f(z)$ is real then so is $f(iz)$ (and hence so is $f(-z)$ and $f(-iz)$), so $f^{-1}(\mathbb{R})$ is a circle centered at the origin. Now, $f(1) = - 1 \in \mathbb{R}$, so $f^{-1}(\mathbb{R})$ is the unit circle.
A: 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
$$\frac{z+i}{z-i}=iy\implies z+i=iy(z-i)=z(iy)+y\implies z(1-iy)=y-i\implies z=\frac{y-i}{1-iy}$$
