Q180 / 270 · NDA 2020 · Mathematics

What is the argument of the complex number $\frac{1-i \sqrt{3}}{1+i \sqrt{3}}$, where $i=\sqrt{-1}$ ?

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

Solution

$\begin{aligned} & \text { Let } x+i y=\frac{1-i \sqrt{3}}{1+i \sqrt{3}} \\ & =\frac{(1-i \sqrt{3})(1-i \sqrt{3})}{(1+i \sqrt{3})(1-i \sqrt{3})}=\frac{(1-i \sqrt{3})^2}{1^2-(i \sqrt{3})^2} \end{aligned}$ $\begin{aligned} & =\frac{1-3-i 2 \sqrt{3}}{1+3}=-\left(\frac{1+i \sqrt{3}}{2}\right) \\ & =-\frac{1}{2}-i \frac{\sqrt{3}}{2} \\ & \therefore x=-\frac{1}{2}, y=-\frac{\sqrt{3}}{2} \\ & \text {Argument }=\tan ^{-1}\left(\frac{y}{x}\right)=\tan ^{-1}\left(\frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}}\right) \\ & =\pi+\tan ^{-1}(\sqrt{3})=180^{\circ}+60^{\circ}=240^{\circ} .\end{aligned}$

Hence, the correct answer is option: 'A'.

Relevant Questions