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

Solution:

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

Now, separating real and complex part, we get

$-\sqrt{3}=r \cos \theta \ldots \ldots \ldots . . q .1$

$-1=r \sin \theta \ldots \ldots \ldots \ldots$ eq.2

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

$4=r^{2}$

Since r is always a positive no., therefore,

r = 2

Hence its modulus is 2.

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

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

$\tan \theta=\frac{1}{\sqrt{3}}$

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

$\tan \theta=\frac{1}{\sqrt{3}}$, therefore $\theta=-\frac{5 \pi}{6}$.

Representing the complex no. in its polar form will be

$Z=2\left\{\cos \left(-\frac{5 \pi}{6}\right)+i \sin \left(-\frac{5 \pi}{6}\right)\right\}$