Find the modulus and the argument of the complex number $z = -1 - i \sqrt{3}$.

(N/A) Given $z = -1 - i \sqrt{3}$.
Let $z = r(\cos \theta + i \sin \theta)$,where $r$ is the modulus and $\theta$ is the argument.
Here,$r \cos \theta = -1$ and $r \sin \theta = -\sqrt{3}$.
Squaring and adding both equations:
$(r \cos \theta)^{2} + (r \sin \theta)^{2} = (-1)^{2} + (-\sqrt{3})^{2}$
$r^{2}(\cos^{2} \theta + \sin^{2} \theta) = 1 + 3$
$r^{2} = 4 \Rightarrow r = 2$ (since $r > 0$).
Thus,the modulus is $2$.
Now,$\cos \theta = -\frac{1}{2}$ and $\sin \theta = -\frac{\sqrt{3}}{2}$.
Since both $\sin \theta$ and $\cos \theta$ are negative,the complex number lies in the $III$ quadrant.
The reference angle $\alpha$ is given by $\tan \alpha = |\frac{-\sqrt{3}}{-1}| = \sqrt{3}$,so $\alpha = \frac{\pi}{3}$.
In the $III$ quadrant,the argument $\theta = -(\pi - \alpha) = -(\pi - \frac{\pi}{3}) = -\frac{2\pi}{3}$.
Therefore,the modulus is $2$ and the argument is $-\frac{2\pi}{3}$.