Question

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

• A. $2\pi /3$
• B. $\pi /3$
• C. $-\pi /3$
• D. $-2\pi /3$

Answer 1

Answer: (A)

$\frac{1+i\sqrt{3}}{1-i\sqrt{3}}=\frac{1+i\sqrt{3}}{1-i\sqrt{3}}\times \frac{1+i\sqrt{3}}{1+i\sqrt{3}}$
$=\frac{{(1+i\sqrt{3})}^2}{1+3}$
$=\frac{1-3+2i\sqrt{3}}{4}$
$=-\frac{1}{2}+\frac{i\sqrt{3}}{2}$
Here; $x=-\frac{1}{2},y=\frac{\sqrt{3}}{2}$
$\therefore$ $\tan \theta =\frac{y}{x}=\frac{\sqrt{3}}{2}\times \frac{2}{-1}=-\sqrt{3}=-\tan \frac{\pi }{3}$
$\Rightarrow$ $\tan \theta =\tan \left(\pi -\frac{\pi }{3}\right)$
$\Rightarrow$ $\theta =\frac{2\pi }{3}$