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

Question:

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

express each of them in polar form: $\frac{2+6 \sqrt{3} i}{5+\sqrt{3} i}$

Solution:

$=\frac{2+6 \sqrt{3} i}{5+\sqrt{3} i} \times \frac{5-\sqrt{3} i}{5-\sqrt{3} i}$

$=\frac{10+28 \sqrt{3} i-18 i^{2}}{25-3 i^{2}}$

$=\frac{28 \sqrt{3} i+28}{28}$

$=\sqrt{3} \mathrm{i}+1$

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

Now , separating real and complex part , we get

1 = rcosθ ……….eq.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

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

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

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

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

Representing the complex no. in its polar form will be

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