Let $\overline{z}$  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  $\overline{z}-z^2=i(\overline{z}+z^2)$ is 

 $\begin{array}{l}Let z=x+iy\\ x–iy–x^2+y^2–2ixy=\text{ i}(x–iy+x^2–y^2)+2ixy)\\ (x–x^2+y^2)–i(y+2xy)=(y–2xy)+i(x+x^2–y^2)\\ \Rightarrow x–x^2+y^2=y–2xy \left(1\right)\\ x+x^2–y^2=–y–2xy \left(2\right)\\ \left(1\right)+\left(2\right)\text{⇒ }2x=-4xy\end{array}$

$\Rightarrow x=-2xy\Rightarrow x(1+2y)=0\Rightarrow x=0ory=-\frac{1}{2}$

$\begin{array}{l}Putx=0in\left(1\right)or\left(2\right)weget\\ y^2=y \Rightarrow y=0, 1 \\ \therefore 2complexnumbersarepossible0+0iand0+i\end{array}$

put $\text{y=-}\frac{\text{1}}{\text{2}}{\text{in\hspace{0.17em}\hspace{0.17em}(1)\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}(2)\hspace{0.17em}\hspace{0.17em}x-x}}^{\text{2}}\text{+}\frac{1}{4}=-\frac{1}{2}+x$

$\Rightarrow x^2=\frac{3}{4}\Rightarrow x=\pm \frac{\sqrt{3}}{2}$

$\therefore \frac{\sqrt{\text{3}}}{\text{2}}-\frac{\text{i}}{\text{2}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\frac{\sqrt{\text{3}}}{\text{2}}-\frac{\text{i}}{\text{2}}\hspace{0.17em}$  are possible

$\therefore$ 4 solutions are possible.