Question

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

Answer 1

\[\text{Let} z = \sin\frac{\pi}{5} + i\left( 1 - \cos\frac{\pi}{5} \right)\]

\[ \Rightarrow \left| z \right| = \sqrt{\left( \sin\frac{\pi}{5} \right)^2 + \left( 1 - \cos\frac{\pi}{5} \right)^2}\]

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

\[ = \sqrt{2 - 2\cos\frac{\pi}{5}}\]

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

\[ = \sqrt{2}\left( \sqrt{2 \sin^2 \frac{\pi}{10}} \right)\]

\[ = 2\sin\frac{\pi}{10}\]

\[\text { Let } \beta \text { be an acute angle given by } \tan\beta = \frac{\left| Im\left( z \right) \right|}{\left| Re\left( z \right) \right|} . \text { Then }, \]

\[\tan\beta = \frac{\left| 1 - \cos\frac{\pi}{5} \right|}{\left| \sin\frac{\pi}{5} \right|} = \left| \frac{2 \sin^2 \frac{\pi}{10}}{2\sin\frac{\pi}{10}\cos\frac{\pi}{10}} \right| = \left| \tan\frac{\pi}{10} \right|\]

\[ \Rightarrow \beta = \frac{\pi}{10}\]

\[\text { Clearly, z lies in the first quadrant . Therefore }, \arg\left( z \right) = \frac{\pi}{10}\]

\[\text {Hence, the polar form of z is } \]

\[2\sin\frac{\pi}{10}\left( \cos\frac{\pi}{10} + i\sin\frac{\pi}{10} \right)\]