Let S = {z in mathbb{C} : z^{2} + bar{z} = 0} | vedclass

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(D) Given the equation $z^{2} + \bar{z} = 0$. Let $z = x + iy$,where $x, y \in \mathbb{R}$.Then $z^{2} = x^{2} - y^{2} + 2ixy$ and $\bar{z} = x - iy$.

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(D) Given the equation $z^{2} + \bar{z} = 0$. Let $z = x + iy$,where $x, y \in \mathbb{R}$.Then $z^{2} = x^{2} - y^{2} + 2ixy$ and $\bar{z} = x - iy$.Substituting these into the equation: $(x^{2} - y^{2} + x) + i(2xy - y) = 0$.Equating the real and imaginary parts to zero,we get:$1) x^{2} + x - y^{2} = 0$$2) y(2x - 1) = 0$From equation $(2)$,either $y = 0$ or $x = \frac{1}{2}$.Case $1$: If $y = 0$,then $x^{2} + x = 0 \implies x(x + 1) = 0$,so $x = 0$ or $x = -1$. The roots are $z_{1} = 0$ and $z_{2} = -1$.Case $2$: If $x = \frac{1}{2}$,then $(\frac{1}{2})^{2} + \frac{1}{2} - y^{2} = 0 \implies \frac{1}{4} + \frac{1}{2} = y^{2} \implies y^{2} = \frac{3}{4} \implies y = \pm \frac{\sqrt{3}}{2}$. The roots are $z_{3} = \frac{1}{2} + i\frac{\sqrt{3}}{2}$ and $z_{4} = \frac{1}{2} - i\frac{\sqrt{3}}{2}$.The set $S = \{0, -1, \frac{1}{2} + i\frac{\sqrt{3}}{2}, \frac{1}{2} - i\frac{\sqrt{3}}{2}\}$.Sum of $(\operatorname{Re}(z) + \operatorname{Im}(z))$ for all $z \in S$ is:$(0 + 0) + (-1 + 0) + (\frac{1}{2} + \frac{\sqrt{3}}{2}) + (\frac{1}{2} - \frac{\sqrt{3}}{2}) = 0 - 1 + \frac{1}{2} + \frac{1}{2} = 0$.

Let $S = \{z \in \mathbb{C} : z^{2} + \bar{z} = 0\}$. Then $\sum_{z \in S} (\operatorname{Re}(z) + \operatorname{Im}(z))$ is equal to $......$

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