JEE Main 2014 · 9 Apr · Shift 1, Mathematics question 62

Solve for $z$: $z(k-\bar w)=k-w$, so $z=\dfrac{k-w}{k-\bar w}$.

The numerator and denominator are conjugates (as $k$ is real), so $|z|=1$. Since $w\ne\bar w$, $z\ne1$. Every such $z$ arises for some real $k$.

So the set is $\{z:|z|=1,\ z\ne1\}$.