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

Answer

(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:
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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:
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