AIEEE 2012 (Offline) 29th April Code AQuestion 1
If $z\ne1$ and $\frac{z^2}{z-1}$ is real, then the point represented by the complex number $z$ lies :
Complex Numbers
Solution
Let $z=x+iy$. $\operatorname{Im}\frac{z^2}{z-1}=0\iff\operatorname{Im}(z^2(\bar z-1))=0\iff y(x^2+y^2-2x)=0$. So $z$ lies on the real axis or on the circle $x^2+y^2=2x$ through the origin.