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

Solution:

Let $Z=-\sqrt{3} i+1=r(\cos \theta+i \sin \theta)$

Now, separating real and complex part, we get

$1=\operatorname{rcos} \theta \ldots \ldots \ldots .$ eq. 1

$-\sqrt{3}=\operatorname{rsin} \theta \ldots \ldots \ldots \ldots$ 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,

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

$\operatorname{Tan} \theta=-\frac{\sqrt{3}}{1}$

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

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

Representing the complex no. in its polar form will be

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