Trigonometry (10th Edition)

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

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


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

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: $[2{cis 0^{\circ},2cis 120^{\circ},2cis 240^{\circ}}]$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.