Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 8 - Section 8.3 - Products and Quotients in Trigonometric Form - 8.3 Problem Set - Page 439: 37

Answer

$16+16i$

Work Step by Step

$-2+2i$ in trigonometric form is $-2+2i=2\sqrt {2}(\cos \frac{3\pi}{4}+i\sin \frac{3\pi}{4})$ Using de Moivre's theorem ($z^{n}=r^{n} cis \,n\theta$ where $z=r\, cis\,\theta$ and $n$ is an integer), we get $(-2+2i)^{3}=[2\sqrt {2}(\cos \frac{3\pi}{4}+i\sin \frac{3\pi}{4})]^{3}$ $=(2\sqrt {2})^{3}(\cos 3\cdot\frac{3\pi}{4}+i\sin 3\cdot\frac{3\pi}{4})$ $=16\sqrt {2}(\cos \frac{9\pi}{4}+i\sin\frac{9\pi}{4})$ $=16\sqrt {2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})$ ($\frac{\pi}{4}$ and $\frac{9\pi}{4}$ are coterminal.) In standard form, our result is $=16\sqrt {2}(\frac{\sqrt 2}{2}+i\cdot\frac{\sqrt 2}{2})$ $=16+16i$
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.