Let S = {z in mathbb{C} : bar{z} = i(z^2 + op | vedclass

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(B) Let $z = x + iy$, where $x, y \in \mathbb{R}$. Then $\bar{z} = x - iy$ and $\operatorname{Re}(\bar{z}) = x$.Given equation: $x - iy = i(x^2 - y^2

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$4$

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(B) Let $z = x + iy$, where $x, y \in \mathbb{R}$. Then $\bar{z} = x - iy$ and $\operatorname{Re}(\bar{z}) = x$.Given equation: $x - iy = i(x^2 - y^2 + 2ixy + x) = i(x^2 - y^2 + x) - 2xy$.Equating real and imaginary parts:$x = -2xy \implies x(1 + 2y) = 0$$-y = x^2 - y^2 + x$Case $1$: $x = 0$. Substituting into the second equation: $-y = -y^2 \implies y^2 - y = 0 \implies y(y - 1) = 0$. So $y = 0$ or $y = 1$.Solutions: $z_1 = 0 + 0i = 0$, $z_2 = 0 + i = i$.Case $2$: $y = -\frac{1}{2}$. Substituting into the second equation: $\frac{1}{2} = x^2 - \frac{1}{4} + x \implies x^2 + x - \frac{3}{4} = 0 \implies 4x^2 + 4x - 3 = 0$.Solving for $x$: $(2x - 1)(2x + 3) = 0 \implies x = \frac{1}{2}$ or $x = -\frac{3}{2}$.Solutions: $z_3 = \frac{1}{2} - \frac{1}{2}i$, $z_4 = -\frac{3}{2} - \frac{1}{2}i$.Calculating $|z|^2$ for each:$|z_1|^2 = 0$, $|z_2|^2 = 1^2 = 1$, $|z_3|^2 = (\frac{1}{2})^2 + (-\frac{1}{2})^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$, $|z_4|^2 = (-\frac{3}{2})^2 + (-\frac{1}{2})^2 = \frac{9}{4} + \frac{1}{4} = \frac{10}{4} = \frac{5}{2}$.Sum $= 0 + 1 + \frac{1}{2} + \frac{5}{2} = 1 + 3 = 4$.

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

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