Explanation

(b)
Given, $z^{2} + \overline{z} = 0$
$\therefore\ \ \ \ (x + iy)^{2} + (x - iy) = 0$
$\Rightarrow \ \ \ \ \ x^{2} - y^{2} + x + i(2xy - y) = 0$
$\Rightarrow \ \ \ x^{2} - y^{2} + x = 0$ and $2xy - y = 0$
Now, $2xy - y = 0\ \ \ \ \ \Rightarrow \ \ \ \ \ y = 0,\ \ \ x = \frac{1}{2}$
When $y = 0,\ \ \ x^{2} - 0 + x = 0\ \ \ \Rightarrow \ \ \ \ \ \ x = 0,\ - 1$
When $x = \frac{1}{2}$,
$\left( \frac{1}{2} \right)^{2} - y^{2} + \frac{1}{2} = 0\ \ \ \ \Rightarrow \ \ \ y^{2} = \frac{1}{4} + \frac{1}{2}\ \ \ \ \Rightarrow \ \ \ y = \pm \frac{\sqrt{3}}{2}$
$\therefore$ Solutions are $(0,\ 0),\ ( - 1,\ 0),\ \left( \frac{1}{2},\ \frac{\sqrt{3}}{2} \right),\ \left( \frac{1}{2},\ \ \frac{- \sqrt{3}}{2} \right)$