Explanation

(a)
Let $OA,\ OB$ be the sides of an equilateral $\mathrm{\Delta}OAB$ and let $OA,\ OB$ represent the complex numbers $\mathcal{z}_{1},\ \mathcal{z}_{2}$ respectively
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From the equilateral $\mathrm{\Delta}OAB$,
$\overrightarrow{\mathbf{AB}} = \mathcal{z}_{2} - \mathcal{z}_{1}$
$\therefore\arg{\left( \frac{\mathcal{z}_{2} - \mathcal{z}_{1}}{\mathcal{z}_{2}} \right) = \arg{\left( \mathcal{z}_{2} - \mathcal{z}_{1} \right) - \arg{\mathcal{z}_{2} = \frac{\pi}{3}}}}$
and $\arg{\left( \frac{\mathcal{z}_{2}}{\mathcal{z}_{1}} \right) = \arg{\left( \mathcal{z}_{2} \right) - \arg{\left( \mathcal{z}_{1} \right) = \frac{\pi}{3}}}}$
Also, $\left| \frac{\mathcal{z}_{2} - \mathcal{z}_{1}}{\mathcal{z}_{2}} \right| = 1 = \left| \frac{\mathcal{z}_{2}}{\mathcal{z}_{1}} \right|$, since triangle is equilateral. Thus, the complex numbers $\frac{\mathcal{z}_{2} - \mathcal{z}_{1}}{\mathcal{z}_{2}}$and $\frac{\mathcal{z}_{2}}{\mathcal{z}_{1}}$ have same modulus and same argument, which implies that the numbers are equal, that is
$\ \frac{\mathcal{z}_{2} - \mathcal{z}_{1}}{\mathcal{z}_{2}} = \frac{\mathcal{z}_{2}}{\mathcal{z}_{1}}\ \Rightarrow \mathcal{z}_{1}\mathcal{z}_{2} - \mathcal{z}_{1}^{2} = \mathcal{z}_{2}^{2}$
$\Rightarrow \ \mathcal{z}_{1}^{2} + \mathcal{z}_{2}^{2} = \mathcal{z}_{1}\mathcal{z}_{2}$