Explanation

(b)
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$\therefore\ \ \overline{AC} = \overline{AB}\ e^{i\pi/3}$
By rotating $\frac{\pi}{3}\$in clockwise sense
$\Rightarrow \left( \mathcal{z}_{3} - \mathcal{z}_{1} \right) = \left( \mathcal{z}_{2} - \mathcal{z}_{1} \right)e^{i\pi/3}$ …(i)
Also, $\left( \mathcal{z}_{1} - \mathcal{z}_{2} \right) = \left( \mathcal{z}_{3} - \mathcal{z}_{2} \right)e^{i\pi/3}$ …(ii)
On dividing Eq.(i) by Eq. (ii), we get
$\frac{(\mathcal{z}_{3} - \mathcal{z}_{1})}{(\mathcal{z}_{1} - \mathcal{z}_{2})} = \frac{(\mathcal{z}_{2} - \mathcal{z}_{1})}{(\mathcal{z}_{3} - \mathcal{z}_{2})}$
$\Rightarrow \ \mathcal{z}_{1}^{2} + \mathcal{z}_{2}^{2} + \mathcal{z}_{3}^{2} = \mathcal{z}_{1}\mathcal{z}_{2} + \mathcal{z}_{2}\mathcal{z}_{3} + \mathcal{z}_{3}\mathcal{z}_{1}$