Question

The argument of $\frac{1-i\sqrt{3}}{1+i\sqrt{3}}$ is

• A. ${60}^{\circ }$
• B. ${120}^{\circ }$
• C. ${210}^{\circ }$
• D. ${240}^{\circ }$

Answer 1

The correct option is D

${240}^{\circ }$

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

Rationalising the denominator,

$\frac{1-i\sqrt{3}}{1+i\sqrt{3}}\times \frac{i-i\sqrt{3}}{1-i\sqrt{3}}=\frac{1+3i^2-2\sqrt{3}i}{1-3i^2}=\frac{-2-2\sqrt{3}i}{4}(\because i^2=-1)=\frac{-1}{2}-i\frac{\sqrt{3}}{2}tan\alpha =\left|\frac{Im\left(z\right)}{Re\left(z\right)}\right|\text{Then, tan~}\alpha =\left|\frac{\frac{-\sqrt{3}}{2}}{\frac{-1}{2}}\right|=\sqrt{3}\Rightarrow \alpha ={60}^{\circ }$

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

$\theta ={180}^{\circ }+{60}^{\circ }={240}^{\circ }$