Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 8 - Application of Trigonometry - 8.6 De Moivre's Theorem; Powers and Roots of Complex Numbers - 8.6 Exercises: 42

Answer

$[3{cis 60^{\circ},3cis 180^{\circ},3cis 300^{\circ}}]$

Work Step by Step

Given: $x^{3}+27=0$ or $x^{3}=-27$ This can be written in trigonometric form as: $x^{3}=27(\cos 180^{\circ}+i \sin 180^{\circ})$ $x=[27(\cos 180^{\circ}+i \sin 180^{\circ})]^{\frac{1}{3}}$ $x=3[(\cos 180^{\circ}+i \sin 180^{\circ})]^{\frac{1}{3}}$ $x=3[(\cos (180^{\circ}+2k\pi)+i \sin (180^{\circ}+2k\pi))]^{\frac{1}{3}}$ Apply De-Moivre's Theorem $x=3[(\cos (180^{\circ}+2k\pi)/3+i \sin (180^{\circ}+2k\pi)/3)]$ Now, the arguments can be wriiten as: $\frac{180+2k\pi}{3} $ and $k=0,1,2$ Roots: $3(\cos 60^{\circ}+\sin 60^{\circ})$,$3(\cos 180^{\circ}+\sin 180^{\circ})$,$3(\cos 300^{\circ}+\sin 300^{\circ})$ Solution set of the equation can be written as: $[3{cis 60^{\circ},3cis 180^{\circ},3cis 300^{\circ}}]$
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