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

We need to find all $$z \in \mathbb{C}$$ satisfying $$\bar{z} = i(z^2 + \text{Re}(\bar{z}))$$, then compute $$\sum_{z \in S} |z|^2$$.

Then $$\bar{z} = x - iy$$ and $$\text{Re}(\bar{z}) = x$$.

Also, $$z^2 = x^2 - y^2 + 2ixy$$.

$$ x - iy = i(x^2 - y^2 + 2ixy + x) $$

$$ x - iy = i(x^2 - y^2 + x) + i(2ixy) $$

$$ x - iy = -2xy + i(x^2 - y^2 + x) $$

Real part: $$x = -2xy \Rightarrow x(1 + 2y) = 0$$

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

Imaginary part: $$-y = x^2 - y^2 + x \Rightarrow x^2 + x - y^2 + y = 0$$

$$ 0 + 0 - y^2 + y = 0 \Rightarrow y(y - 1) = 0 $$

So $$y = 0$$ or $$y = 1$$.

Solutions: $$z = 0$$ and $$z = i$$.

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

$$ x^2 + x - \frac{3}{4} = 0 \Rightarrow 4x^2 + 4x - 3 = 0 $$

$$ (2x - 1)(2x + 3) = 0 $$

So $$x = \frac{1}{2}$$ or $$x = -\frac{3}{2}$$.

Solutions: $$z = \frac{1}{2} - \frac{i}{2}$$ and $$z = -\frac{3}{2} - \frac{i}{2}$$.

$$S = \left\{0,\; i,\; \frac{1}{2} - \frac{i}{2},\; -\frac{3}{2} - \frac{i}{2}\right\}$$

$$ |0|^2 = 0 $$

$$ |i|^2 = 1 $$

$$ \left|\frac{1}{2} - \frac{i}{2}\right|^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} $$

$$ \left|-\frac{3}{2} - \frac{i}{2}\right|^2 = \frac{9}{4} + \frac{1}{4} = \frac{5}{2} $$

$$ \sum_{z \in S} |z|^2 = 0 + 1 + \frac{1}{2} + \frac{5}{2} = 4 $$

The correct answer is Option B: 4.