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  4. The argument of the complex number sin((6π/5)) + i(1 + cos (6π/5)) is

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

COMEDKCOMEDK 2009Complex Numbers and Quadratic Equations

Solution:

$ \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
$\arg ( z )=\pi-\tan ^{-1}| y / x | (\because z = x + iy )$
$(\forall x<0, y \geq 0)$
$\arg ( z )=\pi-\tan ^{-1}\left|\frac{1-\cos \frac{\pi}{5}}{\sin \frac{\pi}{5}}\right|$ $=\pi-\tan ^{-1}\left|\frac{2 \sin ^{2} \frac{\pi}{10}}{2 \sin \frac{\pi}{10} \cos \frac{\pi}{10}}\right|=\pi-\tan ^{-1}\left|\frac{\sin \frac{\pi}{10}}{\cos \frac{\pi}{10}}\right|$
$=\pi-\tan ^{-1}\left|\tan \frac{\pi}{10}\right|\left(\because \tan ^{-1} \frac{\pi}{10}>1\right)$
$=\pi-\frac{\pi}{10}\left(\because-\frac{\pi}{2} \leq \tan ^{-1} x \leq \frac{\pi}{2}\right)$
$\arg ( z )=\frac{9 \pi}{10}$