Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 8 - Review Exercises: 44

Answer

$1.231cis27^\circ$, $1.231cis99^\circ$, $1.231cis171^\circ$, $1.231cis243^\circ$ and $1.231cis315^\circ$ are the five fifth roots of $-2 + 2i$.

Work Step by Step

$-2 + 2i$ is at $135^\circ$ with absolute value $\sqrt{(-2)^2 + 2^2} = \sqrt{8}$. The equivalent trigonometric form of $-2 + 2i$ is $\sqrt{8}cis135^\circ$. Suppose the fifth root of $-2 + 2i$ is represented by $rcis\alpha$, $(rcis\alpha)^5$ = $-2 + 2i$ = $\sqrt{8}cis135^\circ$ By De Moivre’s theorem, this equation becomes $r^5cis5\alpha$ = $\sqrt{8}cis135^\circ$ By equating, $r^5 = \sqrt{8}, r=\sqrt[10]{8}$, and $cos5\alpha = cos135^\circ, sin5\alpha = sin135^\circ$ For these equations to be satisfied, $5\alpha$ must represent an angle that is coterminal with $135^\circ$. Therefore, we must have $\alpha = \frac{135^\circ + 360^\circ\cdot k}{5}$ and k take on the integer values 0, 1, 2, 3 and 4. When $k = 0, \alpha = 27^\circ$, $k = 1, \alpha = 99^\circ$, $k = 2, \alpha = 171^\circ$, $k = 3, \alpha = 243^\circ$, $k = 4, \alpha = 315^\circ$ Hence, when $k = 0$, the root is $\sqrt[10]{8}cis27^\circ$, $k = 1$, the root is $\sqrt[10]{8}cis99^\circ$, $k = 2$, the root is $\sqrt[10]{8}cis171^\circ$, $k = 3$, the root is $\sqrt[10]{8}cis243^\circ$, $k = 4$, the root is $\sqrt[10]{8}cis315^\circ$, $\sqrt[10]{8}cis27^\circ (1.231cis27^\circ)$, $\sqrt[10]{8}cis99^\circ (1.231cis99^\circ)$, $\sqrt[10]{8}cis171^\circ (1.231cis171^\circ)$, $\sqrt[10]{8}cis243^\circ (1.231cis243^\circ)$ and $\sqrt[10]{8}cis315^\circ (1.231cis315^\circ)$ are the five fifth roots of $-2 + 2i$.
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