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 $$\mathrm{C}$$ be the circle in the complex plane with centre $$\mathrm{z}_{0}=\frac{1}{2}(1+3 i)$$ and radius $$r=1$$. Let $$\mathrm{z}_{1}=1+\mathrm{i}$$ and the complex number $$z_{2}$$ be outside the circle $$C$$ such that $$\left|z_{1}-z_{0}\right|\left|z_{2}-z_{0}\right|=1$$. If $$z_{0}, z_{1}$$ and $$z_{2}$$ are collinear, then the smaller value of $$\left|z_{2}\right|^{2}$$ is equal to :

For $$a \in \mathbb{C}$$, let $$\mathrm{A}=\{z \in \mathbb{C}: \operatorname{Re}(a+\bar{z}) > \operatorname{Im}(\bar{a}+z)\}$$ and $$\mathrm{B}=\{z \in \mathbb{C}: \operatorname{Re}(a+\bar{z})<\operatorname{Im}(\bar{a}+z)\}$$. Then among the two statements :

(S1): If $$\operatorname{Re}(a), \operatorname{Im}(a) > 0$$, then the set A contains all the real numbers

(S2) : If $$\operatorname{Re}(a), \operatorname{Im}(a) < 0$$, then the set B contains all the real numbers,