Sum of squares of modulus of all the complex numbers z satisfying z ¯ = i z 2 + z 2 - z is equal to

Given,

$\mathrm{z}+\overline{\mathrm{z}}={\mathrm{izz}}^2+{\mathrm{z}}^2$

Consider $z=x+iy$

$2x=\left(i+1\right)\left(x^2-y^2+2xyi\right)$

On comparing real and imaginary part we get,

$\Rightarrow 2x=x^2-y^2-2xy$ and $x^2-y^2+2xy=0$

$\Rightarrow 2x=-4xy$

$\Rightarrow x=0$ or $y=\frac{-1}{2}$

Case 1: $x=0\Rightarrow 2\times 0=\left(i+1\right)\left(0^2-y^2+2\times 0\times yi\right)y=0$ 

Here $z=0$

Case 2: $y=\frac{-1}{2}$

$\Rightarrow 4x^2-4x-1=0$

${\left(2x-1\right)}^2=2$

$2x-1=\pm \sqrt{2}$

$x=\frac{1\pm \sqrt{2}}{2}$

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 $z$

$=0+\frac{{\left(1+\sqrt{2}\right)}^2+1}{4}+\frac{{\left(1-\sqrt{2}\right)}^2+1}{4}=\frac{8}{4}=2$