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

$-1=r \cos \theta \ldots \ldots \ldots . . \mathrm{q} \cdot 1$

$\sqrt{3}=r \sin \theta \ldots \ldots \ldots \ldots .$ 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,

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

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

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

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

Representing the complex no. in its polar form will be

$\mathrm{Z}=2\left\{\cos \left(\frac{2 \pi}{3}\right)+\operatorname{isin}\left(\frac{2 \pi}{3}\right)\right\}$