Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8.4 De Moivre's Theorem: Powers and Roots of Complex Numbers - 8.4 Exercises - Page 376: 19


(a) The three cube roots are: $4~(cos~60^{\circ}+i~sin~60^{\circ})$ $4~(cos~180^{\circ}+i~sin~180^{\circ})$ $4~(cos~300^{\circ}+i~sin~300^{\circ})$ (b) We can see the three vectors in the complex plane:

Work Step by Step

(a) $z = -64$ $z = 64~(-1+0~i)$ $z = 64~(cos~180^{\circ}+i~sin~180^{\circ})$ $r = 64$ and $\theta = 180^{\circ}$ We can use this equation to find the cube roots: $z^{1/n} = r^{1/n}~[cos(\frac{\theta}{n}+\frac{360^{\circ}~k}{n})+i~sin(\frac{\theta}{n}+\frac{360^{\circ}~k}{n})]$, where $k \in \{0, 1, 2,...,n-1\}$ When k = 0: $z^{1/3} = 64^{1/3}~[cos(\frac{180^{\circ}}{3}+\frac{(360^{\circ})(0)}{3})+i~sin(\frac{180^{\circ}}{3}+\frac{(360^{\circ})(0)}{3})]$ $z^{1/3} = 4~(cos~60^{\circ}+i~sin~60^{\circ})$ When k = 1: $z^{1/3} = 64^{1/3}~[cos(\frac{180^{\circ}}{3}+\frac{(360^{\circ})(1)}{3})+i~sin(\frac{180^{\circ}}{3}+\frac{(360^{\circ})(1)}{3})]$ $z^{1/3} = 4~(cos~180^{\circ}+i~sin~180^{\circ})$ When k = 2: $z^{1/3} = 64^{1/3}~[cos(\frac{180^{\circ}}{3}+\frac{(360^{\circ})(2)}{3})+i~sin(\frac{180^{\circ}}{3}+\frac{(360^{\circ})(2)}{3})]$ $z^{1/3} = 4~(cos~300^{\circ}+i~sin~300^{\circ})$ (b) We can see the three vectors in the complex plane:
Small 1529031326
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.