Complex Numbers and Quadratic Equations — MATHS STD 11 Science — Question

4. $240^\circ$

Solution:

$\frac{1-\text{i}\sqrt{3}}{1+\text{i}\sqrt{3}}$

Rationalising the denominator,

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

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

$=\frac{-2-2\sqrt{3}\text{i}}{4} \ (\because\text{i}^2=-1)$

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

Then, $\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$

$=\sqrt{3}$

$\Rightarrow\alpha=60^\circ$

Since the points $(\frac{-1}{2},-\frac{-\sqrt{3}}{2})$ lie in the third quadrant, the argument is given by:

$\theta=180^\circ+60^\circ$

$=240^\circ$