## 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: 35

#### Answer

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

#### Work Step by Step

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

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