$z +\overline{ z }=i z ^{2}+ z ^{2}$

Consider $z=x+i y$

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

$\Rightarrow 2 x=x^{2}-y^{2}-2 x y \text { and } x^{2}-y^{2}+2 x y=0$

$\Rightarrow 2 x=-4 x y$

$\Rightarrow x=0 \text { or } y=\frac{-1}{2}$

Case $1: x=0 \Rightarrow y=0$ here $z=0$

$\text { Case } 2: y=\frac{-1}{2}$

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

$(2 x-1)^{2}=2$

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

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

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 $z$

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