Explanation

(a)
We have, $z^{2} + pz + q = 0$ and let $p^{2} = 3q$
$\Rightarrow \ \ \ \ \ \ z = \frac{- p \pm \sqrt{p^{2} - 4q}}{2}$
$= \frac{- p \pm \sqrt{3q - 4q}}{2}$
$= \frac{- p \pm i\sqrt{q}}{2}$
Let $z_{1} = \frac{- p + i\sqrt{q}}{2}$
And $z_{2} = \frac{- p - i\sqrt{q}}{2}$
Further, let $z_{1}$ and $z_{2}$ be the affixes of points $A$ and $B$ respectively. Then,
$OA = \left| z_{1} \right| = \sqrt{\left( - \frac{p}{2} \right)^{2} + \left( \frac{\sqrt{q}}{2} \right)^{2}} = \sqrt{\frac{p^{2}}{4} + \frac{q}{4}}$
$= \sqrt{\frac{3q}{4} + \frac{q}{4}} = \sqrt{q}$
$OB = \left| z_{2} \right| = \sqrt{\left( - \frac{p}{2} \right)^{2} + \left( + \frac{\sqrt{q}}{2} \right)^{2}}\$
$= \sqrt{\frac{p^{2}}{4} + \frac{q}{4}} = \sqrt{\frac{3q}{4} + \frac{q}{4}} = \sqrt{q}$
And $AB = \left| z_{1} - z_{2} \right| = \left| i\sqrt{q} \right| = \sqrt{0 + \left( \sqrt{q} \right)^{2}} = \sqrt{q}$
$\therefore\ \ \ \ \ OA = OB = AB$
$\Rightarrow \ \ \ \ \ \mathrm{\Delta}\ AOB$ is an equilateral triangle.
Thus, $p^{2} = 3q$