Question

The argument of $\frac{1-i\sqrt{3}}{1+i\sqrt{3}}$ is
(a) 60°
(b) 120°
(c) 210°
(d) 240°

Answer 1

(d) 240°

$$\begin{gathered} \frac{1-i\sqrt{3}}{1+i\sqrt{3}} \\ \mathrm{Rationalising}\mathrm{the}\mathrm{denominator}, \\ \frac{1-i\sqrt{3}}{1+i\sqrt{3}}\times \frac{1-i\sqrt{3}}{1-i\sqrt{3}} \\ =\frac{1+3i^2\mathit{-}2\sqrt{3}\mathit{}i}{1-3i^2} \\ =\frac{-2-2\sqrt{3}\mathit{}\mathrm{i}}{4}\left(\because \mathit{}{i}^2=-1\right) \\ =\frac{-1}{2}-\mathrm{i}\frac{\sqrt{3}}{2} \\ \tan \mathrm{\alpha }=\left|\frac{\mathrm{Im}\left(z\right)}{\mathrm{Re}\left(z\right)}\right| \\ \mathrm{Then},\tan \mathrm{\alpha }=\left|\frac{{\displaystyle \frac{-\sqrt{3}}{2}}}{{\displaystyle \frac{-1}{2}}}\right| \\ =\sqrt{3} \\ \Rightarrow \mathrm{\alpha }=60° \\ \mathrm{Since}\mathrm{the}\mathrm{points}\left(\frac{-1}{2},\frac{-\sqrt{3}}{2}\right)\mathrm{lie}\mathrm{in}\mathrm{the}\mathrm{third}\mathrm{quadrant},\mathrm{the}\mathrm{argument}\mathrm{is}\mathrm{given}\mathrm{by}: \\ \mathrm{\theta }=180°+60° \\ =240° \end{gathered}$$