Find the modulus and the argument of the complex number

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

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

On squaring and adding, we obtain

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

$\Rightarrow r^{2}=3+1=4 \quad\left[\cos ^{2} \theta+\sin ^{2} \theta=1\right]$

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

$\therefore$ Modulus $=2$

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

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

$\therefore \theta=\pi-\frac{\pi}{6}=\frac{5 \pi}{6}$ [As $\theta$ lies in the II quadrant]

Thus, the modulus and argument of the complex number $-\sqrt{3}+i$ are 2 and $\frac{5 \pi}{6}$ respectively.