If $z \neq 1$ and $\frac{z^2}{z-1}$ is real, then the point represented by the complex number $z$ lies
If $z \neq 1$ and $\frac{z^2}{z-1}$ is real, then the point represented by the complex number $z$ lies
either on the real axis or on a circle passing through the origin
on a circle with centre at the origin
either on the real axis or on a circle not passing through the origin
on the imaginary axis
Solution
Let $z=x+$ iy $(\because x \neq 1$ as $z \neq 1)$
$z^2=\left(x^2-y^2\right)+i(2 x y)$
$\frac{z^2}{z-1}$ is real
$\Rightarrow$ its imaginary part $=0$
$\Rightarrow 2 x y(x-1)-y\left(x^2-y^2\right)=0$
$\Rightarrow y\left(x^2+y^2-2 x\right)=0$
$\Rightarrow y=0 ; x^2+y^2-2 x=0$
$\therefore \mathrm{z}$ lies either on real axis or on a circle through origin.