## 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: 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:

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