## Trigonometry (10th Edition)

Published by Pearson

# Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8.4 De Moivre's Theorem: Powers and Roots of Complex Numbers - 8.4 Exercises - Page 376: 36

#### Answer

$3(\cos 60^{\circ}+i\sin 60^{\circ})$,$3(\cos 180^{\circ}+i\sin 180^{\circ})$,$3(\cos 300^{\circ}+i\sin 300^{\circ})$ or $[3{cis 60^{\circ},3cis 180^{\circ},3cis 300^{\circ}}]$

#### 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 $x=3[(\cos \frac{(180^{\circ}+2k\pi)}{3}+i \sin\frac{(180^{\circ}+2k\pi)}{3})]$ Now, the arguments can be written as: $\frac{180^{\circ}+2k\pi}{3}$ and $k=0,1,2$ Roots: $3(\cos 60^{\circ}+i\sin 60^{\circ})$,$3(\cos 180^{\circ}+i\sin 180^{\circ})$,$3(\cos 300^{\circ}+i\sin 300^{\circ})$ Solution set of the equation can be written as: $[3{cis 60^{\circ},3cis 180^{\circ},3cis 300^{\circ}}]$

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