Mathematics - Complex Number Question with Solution | TestHub

Let $z=x+$ iy $(\because x\ne 1$ as $z\ne 1)$ $z^2=\left(x^2-y^2\right)+i(2xy)$ $\frac{z^2}{z-1}$ is real $\Rightarrow$ its imaginary part $=0$ $\Rightarrow 2xy(x-1)-y\left(x^2-y^2\right)=0$ $\Rightarrow y\left(x^2+y^2-2x\right)=0$ $\Rightarrow y=0;x^2+y^2-2x=0$ $\therefore \mathrm{z}$ lies either on real axis or on a circle through origin.