Question

The argument of the complex number $\sin\left(\frac{6\pi}{5}\right) + i\left(1 +\cos \frac{6\pi}{5}\right)$ is

• A. $\frac{\pi}{10}$
• B. $\frac{5 \pi}{6}$
• C. $\frac{ - \pi}{10}$
• D. $\frac{2 \pi}{5}$

Answer 1

Answer: (C)

$\frac{ - \pi}{10}$

Answer 2

Let $z = \sin\frac{6\pi}{5} + i\left(1 +\cos \frac{6\pi}{5}\right)$
putting $\sin \frac{6\pi}{5} = r \cos \alpha$ ....(i)
and $1 + \cos \frac{6 \pi}{5} = r \sin \alpha$ .....(ii)
Dividing (ii) by (i), we get
$\tan \alpha = \frac{1+\cos \frac{6\pi}{5}}{\sin \frac{6\pi}{5}} = \frac{2 \cos^{2} \frac{3\pi}{5}}{2 \sin \frac{3\pi}{5} \cos \frac{3\pi}{5}}$
$\tan \alpha = \cot \frac{3\pi}{5} =\tan \left(\frac{\pi}{2} - \frac{3\pi}{5}\right)$
$\tan \alpha =\tan \left(\frac{-\pi}{10}\right)$
$\therefore$ argument of z is $\alpha = \frac{-\pi}{10}$