Question

The modulus and argument of $\sin \frac{\mathrm{\pi }}{5}+i\left(1-\cos \frac{\mathrm{\pi }}{5}\right)$ are _______ and _______ respectively.

Answer 1

Since complex number is $\sin \frac{\mathrm{\pi }}{5}+i\left(1-\cos \frac{\mathrm{\pi }}{5}\right)$
$$\begin{gathered} \mathrm{Modulus}\mathrm{is}\sqrt{{\sin }^2\frac{\mathrm{\pi }}{5}+{\left(1-\cos \frac{\mathrm{\pi }}{5}\right)}^2} \\ =\sqrt{{\sin }^2\frac{\mathrm{\pi }}{5}+1+{\cos }^2\frac{\mathrm{\pi }}{5}-2\cos \frac{\mathrm{\pi }}{5}} \\ =\sqrt{-2\cos \frac{\mathrm{\pi }}{5}+1+1}\left(\because {\cos }^2\frac{\mathrm{\pi }}{5}+{\sin }^2\frac{\mathrm{\pi }}{5}=1\right) \\ =\sqrt{2-2\cos \frac{\mathrm{\pi }}{5}} \\ =\sqrt{2\left(1-\cos \frac{\mathrm{\pi }}{5}\right)} \\ =\sqrt{2\times 2{\sin }^2\frac{\mathrm{\pi }}{10}}\left(\because 1-\cos 2\theta =2{\sin }^2\theta \right) \\ \mathrm{i}.\mathrm{e}\mathrm{modulus}=2\sin \frac{\mathrm{\pi }}{10} \end{gathered}$$
and argument is ${\tan }^{-1}\left(\frac{1-\cos {\displaystyle \frac{\mathrm{\pi }}{5}}}{\sin {\displaystyle \frac{\mathrm{\pi }}{5}}}\right)$
$$\begin{gathered} ={\tan }^{-1}\left(\frac{2{\sin }^2{\displaystyle \frac{\mathrm{\pi }}{10}}}{2\sin {\displaystyle \frac{\mathrm{\pi }}{10}}\cos {\displaystyle \frac{\mathrm{\pi }}{10}}}\right)\left[\because \sin 2\theta =2\sin \theta \cos \theta 1-\cos 2\theta =2{\cos }^2\theta \right] \\ ={\tan }^1\left(\frac{2\sin {\displaystyle \frac{\mathrm{\pi }}{10}}}{2\cos {\displaystyle \frac{\mathrm{\pi }}{10}}}\right) \\ ={\tan }^1\left(\tan \frac{\mathrm{\pi }}{10}\right) \\ =\frac{\mathrm{\pi }}{10} \end{gathered}$$

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