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. $\frac{9\pi }{10}$
• D. $\frac{2\pi }{5}$

Answer 1

Answer: (C)

$\sin \frac{6\pi }{5}+i\left(1+\cos \frac{6\pi }{5}\right)$
Notice the point $\sin \left(\frac{6\pi }{5}\right)+i\left(1+\cos \frac{6\pi }{5}\right)$
or $-\sin \left(\frac{6\pi }{5}\right)+i\left(1-\cos \frac{\pi }{5}\right)$ lies in the second quadrant of complex plane hence its argument is given as
$\arg (z)=\pi -{\tan }^{-1}|y/x|(\because z=x+iy)$
$(\forall x<0,y\ge 0)$
$\arg (z)=\pi -{\tan }^{-1}\left|\frac{1-\cos \frac{\pi }{5}}{\sin \frac{\pi }{5}}\right|$ $=\pi -{\tan }^{-1}\left|\frac{2{\sin }^2\frac{\pi }{10}}{2\sin \frac{\pi }{10}\cos \frac{\pi }{10}}\right|=\pi -{\tan }^{-1}\left|\frac{\sin \frac{\pi }{10}}{\cos \frac{\pi }{10}}\right|$
$=\pi -{\tan }^{-1}\left|\tan \frac{\pi }{10}\right|\left(\because {\tan }^{-1}\frac{\pi }{10}>1\right)$
$=\pi -\frac{\pi }{10}\left(\because -\frac{\pi }{2}\le {\tan }^{-1}x\le \frac{\pi }{2}\right)$
$\arg (z)=\frac{9\pi }{10}$