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(C) Given,$\bar{z}-z^2=i(\bar{z}+z^2)$.Rearranging the terms,we get $(1-i)\bar{z}=(1+i)z^2$.Thus,$\bar{z} = \frac{1+i}{1-i}z^2 = \frac{(1+i)^2}{1^2+1^

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$4$

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(C) Given,$\bar{z}-z^2=i(\bar{z}+z^2)$.Rearranging the terms,we get $(1-i)\bar{z}=(1+i)z^2$.Thus,$\bar{z} = \frac{1+i}{1-i}z^2 = \frac{(1+i)^2}{1^2+1^2}z^2 = \frac{2i}{2}z^2 = iz^2$.Let $z = x+iy$,then $\bar{z} = x-iy$.Substituting into $\bar{z} = iz^2$,we have $x-iy = i(x+iy)^2 = i(x^2-y^2+2ixy) = -2xy + i(x^2-y^2)$.Equating real and imaginary parts:$x = -2xy \Rightarrow x(1+2y) = 0$.$-y = x^2-y^2 \Rightarrow x^2 = y^2-y$.Case $I$: If $x=0$,then $y^2-y=0$,so $y=0$ or $y=1$. This gives roots $z=0$ and $z=i$.Case $II$: If $y=-\frac{1}{2}$,then $x^2 = (-\frac{1}{2})^2 - (-\frac{1}{2}) = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}$,so $x = \pm \frac{\sqrt{3}}{2}$. This gives roots $z = \frac{\sqrt{3}}{2} - \frac{1}{2}i$ and $z = -\frac{\sqrt{3}}{2} - \frac{1}{2}i$.There are $4$ distinct roots.

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 $\bar{z}-z^2=i(\bar{z}+z^2)$ is . . . . . .

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