Question

Determine the locus of the point $z$ such that $\frac{z^2}{z-1}$ is always real.

Answer 1

We know that if $z$ is real, then $z=\overline{z}$
If $\frac{z^2}{z-1}$ is real then $\frac{z^2}{z-1}=\frac{{\overline{z}}^2}{\overline{z}-1}$
Cross multiplying, we get
$(z-\overline{z})\{z\overline{z}-(z+\overline{z})\}=0$
$\therefore z-\overline{z}=0$ or $z=\overline{z}$ or $z$ is real $\therefore y=0$
or $z\overline{z}-(z+\overline{z})=0$ or $x^2+y^2-2x=0$
Hence the locus is either $y=0$ i.e. $x-$axis or a circle $x^2+y^2-2x=0$
Note : You may do it by Cartesian method and put imaginary part equal to zero.
$\therefore y(x^2+y^2-2x)=0$.