Precalculus (6th Edition) Blitzer

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

Chapter 6 - Section 6.7 - The Dot Product - Exercise Set - Page 799: 69


$\frac{1}{2}\left( \cos \pi +i\sin \pi \right)$

Work Step by Step

Division of two complex numbers: The division of two complex numbers ${{z}_{1}}={{r}_{1}}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right)$ and ${{z}_{2}}={{r}_{2}}\left( \cos {{\theta }_{2}}+i\sin {{\theta }_{2}} \right)$, is given by $\frac{{{z}_{1}}}{{{z}_{2}}}=\frac{{{r}_{1}}}{{{r}_{2}}}\left[ \cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)+i\sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right) \right]$ So, the product of the given complex numbers ${{z}_{1}}=5\left( \cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3} \right)\text{ and }{{z}_{2}}=10\left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right)$ is $\begin{align} & \frac{{{z}_{1}}}{{{z}_{2}}}=\frac{5}{10}\left[ \cos \left( \frac{4\pi }{3}-\frac{\pi }{3} \right)+i\sin \left( \frac{4\pi }{3}-\frac{\pi }{3} \right) \right] \\ & =\frac{1}{2}\left( \cos \pi +i\sin \pi \right) \end{align}$ Hence, the product of the complex numbers is $\frac{1}{2}\left( \cos \pi +i\sin \pi \right)$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.