Let S = \ z ∈ ℂ: z ^ 2 + ¯ z = 0\ . Then ∑ _ z ∈ S (Re (z) + Im...

Given,

$S = \left\{z \in C: {z}^{2} + \overset{¯}{z} = 0\right\}$

Let $z = x + iy$

${z}^{2} = {x}^{2} - {y}^{2} + 2ixy$

$\overset{¯}{z} = x - iy$

Now putting the value in ${z}^{2} + \overset{¯}{z}$ we get,

${z}^{2} + \overset{¯}{z} = {x}^{2} - {y}^{2} + x + i\left(2xy - y\right) = 0$

Now on comparing both side we get,

$⇒ {x}^{2} + x - {y}^{2} = 0$ and $2xy - y = 0$

$⇒ y = 0$ or $x = \dfrac{1}{2}$

Now if $y = 0; x = 0, - 1$

If $x = \dfrac{1}{2}\text{then}y = \dfrac{\sqrt{3}}{2}, \dfrac{- \sqrt{3}}{2}$

So, $\mathop{∑}_{z \in S}\left(Re \left(z\right) + Im \left(z\right) = \left(0 - 1 + \dfrac{1}{2} + \dfrac{1}{2}\right) + 0 + 0 + \dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{3}}{2}\right) = 0$