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

$ \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{2 i}{2}\right) \overline{ z }= z ^2 $
$ \therefore z ^2=-i \overline{ z }$
Let $z = x +i y$,
$\therefore\left( x ^2- y ^2\right)+i(2 xy )=-i( x -i y )$
so, $x ^2- y ^2+ y =0$....(1)
and $(2 y+1) x=0$...(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.