Solve the following

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

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

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

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

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

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

$=2 \sin \frac{\pi}{10}$

Let $\beta$ be an acute angle given by $\tan \beta=\frac{|\operatorname{Im}(z)|}{|\operatorname{Re}(z)|}$. Then,

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

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

Clearly, $z$ lies in the first quadrant. Therefore, $\arg (z)=\frac{\pi}{10}$

Hence, the polar form of $z$ is

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