Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 8 - Review Exercises - Page 410: 46

Answer

The fifth roots of -32 are: $2~(cos~36^{\circ}+i~sin~36^{\circ})$ $2~(cos~108^{\circ}+i~sin~108^{\circ})$ $2~(cos~180^{\circ}+i~sin~180^{\circ})$ $2~(cos~252^{\circ}+i~sin~252^{\circ})$ $2~(cos~324^{\circ}+i~sin~324^{\circ})$ -32 has one real fifth root.

Work Step by Step

$z = -32 + 0~i$ $z = 32(cos~180^{\circ}+i~sin~180^{\circ})$ $r = 32$ and $\theta = 180^{\circ}$ We can use this equation to find the fifth 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: $z^{1/5} = 32^{1/5}~[cos(\frac{180^{\circ}}{5}+\frac{(360^{\circ})(0)}{5})+i~sin(\frac{180^{\circ}}{5}+\frac{(360^{\circ})(0)}{5})]$ $z^{1/5} = 2~(cos~36^{\circ}+i~sin~36^{\circ})$ When k = 1: $z^{1/5} = 32^{1/5}~[cos(\frac{180^{\circ}}{5}+\frac{(360^{\circ})(1)}{5})+i~sin(\frac{180^{\circ}}{5}+\frac{(360^{\circ})(1)}{5})]$ $z^{1/5} = 2~(cos~108^{\circ}+i~sin~108^{\circ})$ When k = 2: $z^{1/5} = 32^{1/5}~[cos(\frac{180^{\circ}}{5}+\frac{(360^{\circ})(2)}{5})+i~sin(\frac{180^{\circ}}{5}+\frac{(360^{\circ})(2)}{5})]$ $z^{1/5} = 2~(cos~180^{\circ}+i~sin~180^{\circ})$ When k = 3: $z^{1/5} = 32^{1/5}~[cos(\frac{180^{\circ}}{5}+\frac{(360^{\circ})(3)}{5})+i~sin(\frac{180^{\circ}}{5}+\frac{(360^{\circ})(3)}{5})]$ $z^{1/5} = 2~(cos~252^{\circ}+i~sin~252^{\circ})$ When k = 4: $z^{1/5} = 32^{1/5}~[cos(\frac{180^{\circ}}{5}+\frac{(360^{\circ})(4)}{5})+i~sin(\frac{180^{\circ}}{5}+\frac{(360^{\circ})(4)}{5})]$ $z^{1/5} = 2~(cos~324^{\circ}+i~sin~324^{\circ})$ The imaginary part of the root is zero only when the angle is $0^{\circ}$ or $180^{\circ}$. One of the solutions has an angle of $180^{\circ}$. Therefore, there is one real root.
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