The modulus and argument of sin

Solution:

Since complex number is $\sin \frac{\pi}{5}+i\left(1-\cos \frac{\pi}{5}\right)$

Modulus is $\sqrt{\sin ^{2} \frac{\pi}{5}+\left(1-\cos \frac{\pi}{5}\right)^{2}}$

$=\sqrt{\sin ^{2} \frac{\pi}{5}+1+\cos ^{2} \frac{\pi}{5}-2 \cos \frac{\pi}{5}}$

$=\sqrt{-2 \cos \frac{\pi}{5}+1+1} \quad\left(\because \cos ^{2} \frac{\pi}{5}+\sin ^{2} \frac{\pi}{5}=1\right)$

$=\sqrt{2-2 \cos \frac{\pi}{5}}$

$=\sqrt{2\left(1-\cos \frac{\pi}{5}\right)}$

$=\sqrt{2 \times 2 \sin ^{2} \frac{\pi}{10}} \quad\left(\because 1-\cos 2 \theta=2 \sin ^{2} \theta\right)$

i. e modulus $=2 \sin \frac{\pi}{10}$

and argument is $\tan ^{-1}\left(\frac{1-\cos \frac{\pi}{5}}{\sin \frac{\pi}{5}}\right)$

$=\tan ^{-1}\left(\frac{2 \sin ^{2} \frac{\pi}{10}}{2 \sin \frac{\pi}{10} \cos \frac{\pi}{10}}\right) \quad\left[\because \sin 2 \theta=2 \sin \theta \cos \theta 1-\cos 2 \theta=2 \cos ^{2} \theta\right]$

$=\tan ^{1}\left(\frac{2 \sin \frac{\pi}{10}}{2 \cos \frac{\pi}{10}}\right)$

$=\tan ^{1}\left(\tan \frac{\pi}{10}\right)$

$=\frac{\pi}{10}$

$\therefore$ argument is given by $\frac{\pi}{10}$

$\therefore$ modulus of $\sin \frac{\pi}{5}+i\left(1-\cos \frac{\pi}{5}\right)$ is $2 \sin \frac{\pi}{10}$ and argument is $\frac{\pi}{10}$.