Let $$S=\left\{z \in \mathbb{C}: \bar{z}=i\left(z^{2}+\operatorname{Re}(\bar{z})\right)\right\}$$. Then $\sum_\limits{z \in \mathrm{S}}|z|^{2}$ is equal to :

Let $z=x+i y$ <br/><br/>$$ \begin{aligned} &\bar{z}=i\left(z^2+\operatorname{Re}(\bar{z})\right) \\\\ &\Rightarrow x-i y=i\left(x^2-y^2+2 i x y+x\right) \\\\ & x-i y=i\left(x^2-y^2+x\right)-2 x y \\\\ &x=-2 x y \Rightarrow x(2 y+1)=0 \\\\ &\Rightarrow x=0, y=\frac{-1}{2} ........(i)\\\\ &-y=x^2-y^2+x ...........(ii) \end{aligned} $$ <br/><br/><b>Case (I)</b> $x=0$ <br/><br/>$\Rightarrow-y=-y^2 \Rightarrow y^2-y=0 \Rightarrow y=0,1$ <br/><br/>$\therefore$ $z=0, i$ <br/><br/><b>Case (II)</b> $y=\frac{-1}{2}$ <br/><br/>$\Rightarrow \frac{1}{2}=x^2-\frac{1}{4}+x \Rightarrow x^2+x-\frac{3}{4}=0$ <br/><br/>$\Rightarrow$ $4 x+4 x-3=0 \Rightarrow(2 x-1)(2 x+3)=0$ <br/><br/>$\Rightarrow$ $x=\frac{1}{2}, \frac{-3}{2}$ <br/><br/>$$ \begin{aligned} & z=\frac{1}{2}-\frac{1}{2} i, \frac{-3}{2}-\frac{1}{2} i \\\\ & \sum|z|^2=0+1+\frac{1}{2}+\frac{5}{2}=4 \end{aligned} $$