Question

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

Answer 1

Answer: 2

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