Question

Let $\bar{z}$ denote the complex conjugate of a complex number $z$ and let $i=\sqrt{-1}$. In the set of complex numbers, the number of distinct roots of the equation
$\bar{z}-z^2=i\left(\bar{z}+z^2\right)$ is ___

Answer 1

Answer: 4.00

$\overline{Z}-z^2=i\left(\overline{Z}+z^2\right)$
$\Rightarrow (1-i)\overline{z}=(1+i)z^2$
$\Rightarrow \frac{(1-i)}{(1+i)}\overline{z}=z^2$
$\Rightarrow \left(-\frac{2i}{2}\right)\overline{z}=z^2$
$\therefore z^2=-i\overline{z}$
Let $z=x+iy$,
$\therefore \left(x^2-y^2\right)+i(2xy)=-i(x-iy)$
so, $x^2-y^2+y=0$....(1)
and $(2y+1)x=0$...(2)
$\Rightarrow x=0\text{ or }y=-\frac{1}{2}$
Case I: When $x=0$
$\therefore (1)\Rightarrow y(1-y)=0\Rightarrow y=0,1$
$\therefore (0,0),(0,1)$
Case II : When $y=-\frac{1}{2}$
$\therefore (1)\Rightarrow x^2-\frac{1}{4}-\frac{1}{2}=0\Rightarrow x^2=\frac{3}{4}\Rightarrow x=\pm \frac{\sqrt{3}}{2}$
$\therefore \left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right),\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right)$
$\Rightarrow$ Number of distinct ' $z$ ' is equal to 4.