Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 6 - Test - Page 800: 12

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)$.
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