Q52 / 90 · JEE Main 2012 (Offline) · Mathematics

If $z \neq 1$ and $\frac{z^2}{z-1}$ is real, then the point represented by the complex number $z$ lies

  • Correct answer
    A
    either on the real axis or on a circle passing through the origin
  • B
    on a circle with centre at the origin
  • C
    either on the real axis or on a circle not passing through the origin
  • D
    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.

Hence, the correct answer is option: 'A'.

Relevant Questions