$z=-1-i \sqrt{3}$

Let $r \cos \theta=-1$ and $r \sin \theta=-\sqrt{3}$

On squaring and adding, we obtain

$(r \cos \theta)^{2}+(r \sin \theta)^{2}=(-1)^{2}+(-\sqrt{3})^{2}$

$\Rightarrow r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=1+3$

$\Rightarrow r^{2}=4 \quad\left[\cos ^{2} \theta+\sin ^{2} \theta=1\right]$

$\Rightarrow r=\sqrt{4}=2 \quad[\text { Conventionally }, r>0]$

$\therefore$ Modulus $=2$

$\therefore 2 \cos \theta=-1$ and $2 \sin \theta=-\sqrt{3}$

$\Rightarrow \cos \theta=\frac{-1}{2}$ and $\sin \theta=\frac{-\sqrt{3}}{2}$

since both the values of $\sin \theta$ and $\cos \theta$ negative and $\sin \theta$ and $\cos \theta$ are negative in $III$ quadrant,

Argument $=-\left(\pi-\frac{\pi}{3}\right)=\frac{-2 \pi}{3}$

Thus, the modulus and argument of the complex number $-1-\sqrt{3} i$ are $2$ and $-\frac{2 \pi}{3}$ respectively.