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 - Page 809: 41


$[2{cis 0^{\circ},2cis 120^{\circ},2cis 240^{\circ}}]$

Work Step by Step

Given: $x^{3}-8=0$ or $x^{3}=8$ This can be written in trigonometric form as: $x^{3}=8+0.i=8(\cos 180^{\circ}+i \sin 180^{\circ})$ $x=[8(\cos 180^{\circ}+i \sin 180^{\circ})]^{\frac{1}{3}}$ $x=2[(\cos 180^{\circ}+i \sin 180^{\circ})]^{\frac{1}{3}}$ Absolute value of third root is given as $\sqrt[3] 1=1$ Now, the arguments can be wriiten as: $k=0,1,2$ Roots: $2(\cos 0^{\circ}+\sin 0^{\circ})$,$2(\cos 120^{\circ}+\sin 120^{\circ})$,$2(\cos 240^{\circ}+\sin 240^{\circ})$ Solution set of the equation can be written as: $[2{cis 0^{\circ},2cis 120^{\circ},2cis 240^{\circ}}]$
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