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: 29

Answer

The cube roots of $i$ are: $cos~30^{\circ}+i~sin~30^{\circ}$ $cos~150^{\circ}+i~sin~150^{\circ}$ $cos~270^{\circ}+i~sin~270^{\circ}$ We can graph the cube roots in the complex plane:

Work Step by Step

$z = i$ $z = 0+i$ $z = cos~90^{\circ}+i~sin~90^{\circ}$ $r = 1$ and $\theta = 90^{\circ}$ 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: $z^{1/3} = 1^{1/3}~[cos(\frac{90^{\circ}}{3}+\frac{(360^{\circ})(0)}{3})+i~sin(\frac{90^{\circ}}{3}+\frac{(360^{\circ})(0)}{3})]$ $z^{1/3} = cos~30^{\circ}+i~sin~30^{\circ}$ When k = 1: $z^{1/3} = 1^{1/3}~[cos(\frac{90^{\circ}}{3}+\frac{(360^{\circ})(1)}{3})+i~sin(\frac{90^{\circ}}{3}+\frac{(360^{\circ})(1)}{3})]$ $z^{1/3} = cos~150^{\circ}+i~sin~150^{\circ}$ When k = 2: $z^{1/3} = 1^{1/3}~[cos(\frac{90^{\circ}}{3}+\frac{(360^{\circ})(2)}{3})+i~sin(\frac{90^{\circ}}{3}+\frac{(360^{\circ})(2)}{3})]$ $z^{1/3} = cos~270^{\circ}+i~sin~270^{\circ}$ We can graph the cube roots in the complex plane:
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.