## Trigonometry (10th Edition)

Published by Pearson

# 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: 17

#### Answer

(a) The three cube roots are: $2~(cos~90^{\circ}+i~sin~90^{\circ})$ $2~(cos~210^{\circ}+i~sin~210^{\circ})$ $2~(cos~330^{\circ}+i~sin~330^{\circ})$ (b) We can see the three vectors in the complex plane:

#### Work Step by Step

(a) $z = -8i$ $z = 8~(0-i)$ $z = 8~(cos~270^{\circ}+i~sin~270^{\circ})$ $r = 8$ and $\theta = 270^{\circ}$ 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: $z^{1/3} = 8^{1/3}~[cos(\frac{270^{\circ}}{3}+\frac{(360^{\circ})(0)}{3})+i~sin(\frac{270^{\circ}}{3}+\frac{(360^{\circ})(0)}{3})]$ $z^{1/3} = 2~(cos~90^{\circ}+i~sin~90^{\circ})$ When k = 1: $z^{1/3} = 8^{1/3}~[cos(\frac{270^{\circ}}{3}+\frac{(360^{\circ})(1)}{3})+i~sin(\frac{270^{\circ}}{3}+\frac{(360^{\circ})(1)}{3})]$ $z^{1/3} = 2~(cos~210^{\circ}+i~sin~210^{\circ})$ When k = 2: $z^{1/3} = 8^{1/3}~[cos(\frac{270^{\circ}}{3}+\frac{(360^{\circ})(2)}{3})+i~sin(\frac{270^{\circ}}{3}+\frac{(360^{\circ})(2)}{3})]$ $z^{1/3} = 2~(cos~330^{\circ}+i~sin~330^{\circ})$ (b) We can see the three vectors in the complex plane:

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