Answer
The required operation in polar form is $ z=\frac{1}{2}\left( \cos \left( \frac{\pi }{6} \right)+i\sin \left( \frac{\pi }{6} \right) \right)$
Work Step by Step
As
$\frac{2\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)}{4\left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right)}$
So, $\begin{align}
& \frac{2\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)}{4\left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right)}=\left( \frac{2}{4} \right)\left( \cos \left( \frac{\pi }{2}-\frac{\pi }{3} \right)+i\sin \left( \frac{\pi }{2}-\frac{\pi }{3} \right) \right) \\
& =\left( \frac{1}{2} \right)\left( \cos \left( \frac{\pi }{6} \right)+i\sin \left( \frac{\pi }{6} \right) \right) \\
& =\frac{1}{2}\left( \cos \left( \frac{\pi }{6} \right)+i\sin \left( \frac{\pi }{6} \right) \right)
\end{align}$
So, $ z=\frac{1}{2}\left( \cos \left( \frac{\pi }{6} \right)+i\sin \left( \frac{\pi }{6} \right) \right)$
Therefore, the required operation in polar form is $ z=\frac{1}{2}\left( \cos \left( \frac{\pi }{6} \right)+i\sin \left( \frac{\pi }{6} \right) \right)$.
