Explanation

(a)
Let the vertices of triangle be $A\left( z_{1} \right),\ B\left( z_{2} \right)\$and$\ C(z_{3})$
Given, $\frac{z_{1} - z_{3}}{z_{2} - z_{3}} = \frac{1 - i\sqrt{3}}{2}$
$\Rightarrow \ \ \ \ \ \left| \frac{z_{1} - z_{3}}{z_{2} - z_{3}} \right| = \frac{|2|}{|2|} = 1$
$\therefore\ \ \ \ \ \ \left| z_{1} - z_{3} \right| = |z_{2} - z_{3}|$
$\Rightarrow \ \ \ \ \ |AC| = |BC|$
Now, $\frac{z_{1} - z_{3}}{z_{2} - z_{3}} = e^{- i\pi/3}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \arg{\left( \frac{z_{1} - z_{3}}{z_{2} - z_{3}} \right) = - \frac{\pi}{3}}$
$\therefore\ \ \ \ \ \ \ \ \ \angle BCA = \frac{\pi}{3}$
$\Rightarrow \ \ \ \ \ \ \ |AC| = |BC|\ \$ and $\angle BCA = 60{^\circ}$
$\Rightarrow \ \ \ \ |AB| = |BC| = |CA|$
$\Rightarrow \ \ \ \ \mathrm{\Delta}ABC$ is an equilateral triangle.