Find the modulus of each of the following complex numbers and hence

$=\frac{-16}{1+\sqrt{3} i} \times \frac{1-\sqrt{3} i}{1-\sqrt{3} i}$

$=\frac{-16+16 \sqrt{3} i}{1-3 i^{2}}$

$=\frac{16 \sqrt{3} i-16}{4}$

$=4^{\sqrt{3}} \mathrm{i}-4$

Let $Z=4^{\sqrt{3}} i-4=r(\cos \theta+i \sin \theta)$

Now , separating real and complex part , we get

$-4=r \cos \theta \ldots \ldots \ldots .$ eq. 1

$4 \sqrt{3}=r \sin \theta \ldots \ldots \ldots \ldots$ eq. 2

Squaring and adding eq.1 and eq.2, we get

$64=r^{2}$

Since r is always a positive no., therefore,

r = 8

Hence its modulus is 8.

Now, dividing eq.2 by eq.1 , we get,

$\frac{r \sin \theta}{r \cos \theta}=\frac{4 \sqrt{3}}{-4}$

$\tan \theta=-\sqrt{3}$

Since $\cos \theta=-\frac{1}{2} \sin \theta=\frac{\sqrt{3}}{2}$ and $\tan \theta=-\sqrt{3}$. Therefore the $\theta$ lies in second quadrant.

$\operatorname{Tan} \theta=-\sqrt{3}$, therefore $\theta=\frac{2 \pi}{3}$

Representing the complex no. in its polar form will be

$\mathrm{Z}=8\left\{\cos \left(\frac{2 \pi}{3}\right)+i \sin \left(\frac{2 \pi}{3}\right)\right\}$