Let $z=x+i y$

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

$\Rightarrow 2 x=x^{2}-y^{2}-2 x y\quad$ ...(i) and

$$ x^{2}-y^{2}+2 x y=0\quad\dots(ii) $$

From (i) and (ii) we get

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

When $x=0$ we get $y=0$

When $y=-\frac{1}{2}$ we get $x^{2}-x-\frac{1}{4}=0$

$\Rightarrow \quad x=\frac{-1 \pm \sqrt{2}}{2}$

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 $\left(\frac{-1-\sqrt{2}}{2}\right)-\frac{1}{2} i$

Sum of squares of modulus

$$ \begin{aligned} &=0+\left(\frac{\sqrt{2}-1}{2}\right)^{2}+\frac{1}{4}+\left(\frac{\sqrt{2}+1}{2}\right)^{2}=+\frac{1}{4} \\\\ &=2 \end{aligned} $$