The argument of

(d) 240°

$\frac{1-i \sqrt{3}}{1+i \sqrt{3}}$

Rationalising the denominator,

$\frac{1-i \sqrt{3}}{1+i \sqrt{3}} \times \frac{1-i \sqrt{3}}{1-i \sqrt{3}}$

$=\frac{1+3 i^{2}-2 \sqrt{3} i}{1-3 i^{2}}$

$=\frac{-2-2 \sqrt{3} i}{4}$ $\left(\because i^{2}=-1\right)$

$=\frac{-1}{2}-i \frac{\sqrt{3}}{2}$

$\tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$

Then, $\tan \alpha=\left|\frac{\frac{\sqrt{3}}{2}}{\frac{-1}{2}}\right|$

$=\sqrt{3}$

$\Rightarrow \alpha=60^{\circ}$

Since the points $\left(\frac{-1}{2}, \frac{-\sqrt{3}}{2}\right)$ lie in the third quadrant, the argument is given by:

$\theta=180^{\circ}+60^{\circ}$

= 240°