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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. . . . . .
$2$
$3$
$4$
$5$
Solution
Given,
$\bar{z}-z^2=i\left(\bar{z}+z^2\right)$
$\Rightarrow(1-i) \bar{z}=(1+i) z^2$
$\Rightarrow \frac{(1-i)}{(1+i)} \bar{z}=z^2$
$\Rightarrow\left(-\frac{2 i}{2}\right) \bar{z}=z^2$
$\therefore z^2=-i \bar{z}$
Let $z = x +i y$,
$\therefore\left( x ^2- y ^2\right)+i(2 xy )=-i( x -i y )$
$\text { so, } x ^2- y ^2+ y =0$
$\text { and }(2 y +1) x =0$
$\Rightarrow x =0 \text { or } y =-\frac{1}{2}$
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 $4$ .