## Precalculus (6th Edition) Blitzer

The required operation in polar form is $z=\frac{1}{2}\left( \cos \left( \frac{\pi }{6} \right)+i\sin \left( \frac{\pi }{6} \right) \right)$
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 $z=\frac{1}{2}\left( \cos \left( \frac{\pi }{6} \right)+i\sin \left( \frac{\pi }{6} \right) \right)$.