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

Let $z=x+i y$ <br/><br/> So $2 x=(1+i)\left(x^{2}-y^{2}+2 x y i\right)$ <br/><br/> $\Rightarrow 2 x=x^{2}-y^{2}-2 x y\quad$ ...(i) and <br/><br/> $x^{2}-y^{2}+2 x y=0\quad\dots(ii)$ <br/><br/> From (i) and (ii) we get <br/><br/> $x=0 \text { or } y=-\frac{1}{2}$ <br/><br/> When $x=0$ we get $y=0$ <br/><br/> When $y=-\frac{1}{2}$ we get $x^{2}-x-\frac{1}{4}=0$ <br/><br/> $\Rightarrow \quad x=\frac{-1 \pm \sqrt{2}}{2}$ <br/><br/> So there will be total 3 possible values of $z$, which are $0,\left(\frac{-1+\sqrt{2}}{2}\right)-\frac{1}{2} i$ and $\left(\frac{-1-\sqrt{2}}{2}\right)-\frac{1}{2} i$ <br/><br/> Sum of squares of modulus <br/><br/> $$ \begin{aligned} &=0+\left(\frac{\sqrt{2}-1}{2}\right)^{2}+\frac{1}{4}+\left(\frac{\sqrt{2}+1}{2}\right)^{2}=+\frac{1}{4} \\\\ &=2 \end{aligned} $$