Explanation

(d)
Let $z = \frac{1 - i\sqrt{3}}{1 + i\sqrt{3}} = \frac{1 - i\sqrt{3}}{1 + i\sqrt{3}} \times \frac{1 - i\sqrt{3}}{1 - i\sqrt{3}} = - \frac{1}{2} - \frac{i\sqrt{3}}{2}$
$\Rightarrow \ \ \arg{(z) = \theta = \tan^{- 1}{\left( \frac{\sqrt{3}/2}{1/2} \right) = \tan^{- 1}{(\sqrt{3})}}}$
$\Rightarrow \ \ \ \ \theta = 60{^\circ}$
Since, given number lies in IIIrd quadrant
$\therefore\ \ \ \ \ \theta = 180{^\circ} + 60{^\circ} = 240{^\circ}$