Question

Express $\sin \frac{\mathrm{\pi }}{5}+i\left(1-\cos \frac{\mathrm{\pi }}{5}\right)$ in polar form.

Answer 1

$$\begin{gathered} \mathrm{Let}z=\sin \frac{\mathrm{\pi }}{5}+i\left(1-\cos \frac{\mathrm{\pi }}{5}\right) \\ \Rightarrow \left|z\right|=\sqrt{{\left(\sin \frac{\mathrm{\pi }}{5}\right)}^2+{\left(1-\cos \frac{\mathrm{\pi }}{5}\right)}^2} \\ =\sqrt{{\sin }^2\frac{\mathrm{\pi }}{5}+1+{\cos }^2\frac{\mathrm{\pi }}{5}-2\cos \frac{\mathrm{\pi }}{5}} \\ =\sqrt{2-2\cos \frac{\mathrm{\pi }}{5}} \\ =\sqrt{2}\left(\sqrt{1-\cos \frac{\mathrm{\pi }}{5}}\right) \\ =\sqrt{2}\left(\sqrt{2{\sin }^2\frac{\mathrm{\pi }}{10}}\right) \\ =2\sin \frac{\mathrm{\pi }}{10} \\ \mathrm{Let}\beta \mathrm{be}\mathrm{an}\mathrm{acute}\mathrm{angle}\mathrm{given}\mathrm{by}\tan \beta =\frac{\left|\mathrm{Im}\left(z\right)\right|}{\left|\mathrm{Re}\left(z\right)\right|}.\mathrm{Then}, \\ \tan \beta =\frac{\left|{\displaystyle 1-\cos \frac{\mathrm{\pi }}{5}}\right|}{\left|{\displaystyle \sin \frac{\mathrm{\pi }}{5}}\right|}=\left|\frac{2{\sin }^2{\displaystyle \frac{\mathrm{\pi }}{10}}}{2\sin {\displaystyle \frac{\mathrm{\pi }}{10}}\cos {\displaystyle \frac{\mathrm{\pi }}{10}}}\right|=\left|\tan \frac{\mathrm{\pi }}{10}\right| \\ \Rightarrow \beta =\frac{\mathrm{\pi }}{10} \\ \mathrm{Clearly},z\mathrm{lies}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{quadrant}.\mathrm{Therefore},\arg \left(z\right)=\frac{\mathrm{\pi }}{10} \\ \mathrm{Hence},\mathrm{the}\mathrm{polar}\mathrm{form}\mathrm{of}z\mathrm{is} \\ 2\sin \frac{\mathrm{\pi }}{10}\left(\cos \frac{\mathrm{\pi }}{10}+i\sin \frac{\mathrm{\pi }}{10}\right) \end{gathered}$$