Question

Sum of squares of modulus of all the complex numbers $z$ satisfying $\overline{ Z }= iz ^{2}+ z ^{2}- z$ is equal to

Answer 1

$z +\overline{ z }= iz ^{2}+ z ^{2}$
Consider $z = x + iy$
$2 x=(i+1)\left(x^{2}-y^{2}+2 x y i\right) $
$\Rightarrow 2 x=x^{2}-y^{2}-2 x y \text { and } x^{2}-y^{2}+2 x y=0$
$\Rightarrow 2 x=-4 x y$
$\Rightarrow x =0$ or $y =\frac{-1}{2}$
Case $1: x =0 $
$\Rightarrow y =0$ here $z =0$
Case $2: y =\frac{-1}{2}$
$\Rightarrow 4 x^{2}-4 x-1=0$
$(2 x-1)^{2}=2 $
$2 x-1=\pm \sqrt{2}$
$x=\frac{1 \pm \sqrt{2}}{2}$
Here $z=\frac{1+\sqrt{2}}{2}-\frac{i}{2}$ or $z=\frac{1-\sqrt{2}}{2}-\frac{i}{2}$
Sum of squares of modulus of $z$
$=0+\frac{(1+\sqrt{2})^{2}+1}{4}+\frac{(1-\sqrt{2})^{2}+1}{4}=\frac{8}{4}=2$