JEE Main · 2012 · 29 Apr 2012 · Offline · Complex Numbers

Put $z=x+iy$. The quotient $\frac{z^2}{z-1}$ is real exactly when its imaginary part vanishes; carrying out the division, the imaginary part is proportional to $y(x^2+y^2-2x)$. This is zero either when $y=0$, the real axis, or when $x^2+y^2-2x=0$, which rearranges to $(x-1{)}^2+y^2=1$, a circle that passes through the origin. So the point $z$ lies either on the real axis or on a circle passing through the origin.