Explanation

(a)
$\frac{1 + i\sqrt{3}}{1 - i\sqrt{3}} = \frac{1 + i\sqrt{3}}{1 - i\sqrt{3}} \times \frac{1 + i\sqrt{3}}{1 + i\sqrt{3}}$
$= \frac{\left( 1 + i\sqrt{3} \right)^{2}}{1 + 3} = - \frac{1}{2} + \frac{i\sqrt{3}}{2}$
$\therefore\ \ \ \tan{\theta = \frac{\sqrt{3}}{2} \times \frac{2}{- 1}} = - \sqrt{3} = - \tan\frac{\pi}{3}$
$\Rightarrow \ \ \ \ \ \ \tan{\theta = \tan\left( \pi - \frac{\pi}{3} \right)}$
$\Rightarrow \ \ \ \ \theta = \frac{2\pi}{3}$