Trigonometry (11th Edition) Clone

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

Chapter 8 - Test - Page 412: 8

Answer

The four complex fourth roots of $-16~i$ are: $2~(cos~67.5^{\circ}+i~sin~67.5^{\circ})$ $2~(cos~157.5^{\circ}+i~sin~157.5^{\circ})$ $2~(cos~247.5^{\circ}+i~sin~247.5^{\circ})$ $2~(cos~337.5^{\circ}+i~sin~337.5^{\circ})$

Work Step by Step

$z = -16~i$ $z = 16~(0 - i)$ $z = 16(cos~270^{\circ}+i~sin~270^{\circ})$ $r = 16$ and $\theta = 270^{\circ}$ We can use this equation to find the fourth 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: $z^{1/4} = 16^{1/4}~[cos(\frac{270^{\circ}}{4}+\frac{(360^{\circ})(0)}{4})+i~sin(\frac{270^{\circ}}{4}+\frac{(360^{\circ})(0)}{4})]$ $z^{1/4} = 2~(cos~67.5^{\circ}+i~sin~67.5^{\circ})$ When k = 1: $z^{1/4} = 16^{1/4}~[cos(\frac{270^{\circ}}{4}+\frac{(360^{\circ})(1)}{4})+i~sin(\frac{270^{\circ}}{4}+\frac{(360^{\circ})(1)}{4})]$ $z^{1/4} = 2~(cos~157.5^{\circ}+i~sin~157.5^{\circ})$ When k = 2: $z^{1/4} = 16^{1/4}~[cos(\frac{270^{\circ}}{4}+\frac{(360^{\circ})(2)}{4})+i~sin(\frac{270^{\circ}}{4}+\frac{(360^{\circ})(2)}{4})]$ $z^{1/4} = 2~(cos~247.5^{\circ}+i~sin~247.5^{\circ})$ When k = 3: $z^{1/4} = 16^{1/4}~[cos(\frac{270^{\circ}}{4}+\frac{(360^{\circ})(3)}{4})+i~sin(\frac{270^{\circ}}{4}+\frac{(360^{\circ})(3)}{4})]$ $z^{1/4} = 2~(cos~337.5^{\circ}+i~sin~337.5^{\circ})$
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