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

Solution:

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

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

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

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

$=\frac{(2-2 \sqrt{3})+i(2 \sqrt{3}+2)}{4}$

$=\frac{(1-\sqrt{3})+i(\sqrt{3}+1)}{2}$

Let $Z=\frac{(1-\sqrt{3})+i(\sqrt{3}+1)}{2}=r(\cos \theta+i \sin \theta)$

Now, separating real and complex part , we get

$\frac{1-\sqrt{3}}{2}=r \cos \theta$ ……….eq.1

$\frac{1+\sqrt{3}}{2}=r \sin \theta$ …………eq.2

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

$2=r^{2}$

Since r is always a positive no., therefore,

$r=\sqrt{2}$

Hence its modulus is $\sqrt{2}$.

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

$\frac{r \sin \theta}{r \cos \theta}=\frac{\frac{1+\sqrt{3}}{2}}{\frac{1-\sqrt{3}}{2}}$

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

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

$\operatorname{Tan} \theta=\frac{1+\sqrt{3}}{1-\sqrt{3}}$, therefore $\theta=\frac{7 \pi}{12}$

Representing the complex no. in its polar form will be

$\mathrm{Z}=\sqrt{2}\left\{\cos \left(\frac{7 \pi}{12}\right)+i \sin \left(\frac{7 \pi}{12}\right)\right\}$