Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 8 - Section 8.3 - Polar Form of Complex Numbers; De Moivre's Theorem - 8.3 Exercises - Page 611: 81

Answer

$w_0=1$ $w_1=\frac{\sqrt 2}{2}+\frac{\sqrt 2}{2}i$ $w_2=i$ $w_3=-\frac{\sqrt 2}{2}+\frac{\sqrt 2}{2}i$ $w_4=-1$ $w_5=-\frac{\sqrt 2}{2}-\frac{\sqrt 2}{2}i$ $w_6=-i$ $w_7=\frac{\sqrt 2}{2}-\frac{\sqrt 2}{2}i$

Work Step by Step

$r=|z|=1$ $θ=0~~$ (Positive real axis) Polar form: $z=1(cos~0+i~sin~0)$ $w_k=\sqrt[8] {1}[cos(\frac{0+2k\pi}{8})+i~sin(\frac{0+2k\pi}{8})]$ $w_0=cos~0+i~sin~0=1$ $w_1=cos(\frac{\pi}{4})+i~sin(\frac{\pi}{4})=\frac{\sqrt 2}{2}+\frac{\sqrt 2}{2}i$ $w_2=cos(\frac{\pi}{2})+i~sin(\frac{\pi}{2})=i$ $w_3=cos~\frac{3\pi}{4}+i~sin(\frac{3\pi}{4})=-\frac{\sqrt 2}{2}+\frac{\sqrt 2}{2}i$ $w_4=cos~\pi+i~sin~\pi=-1$ $w_5=cos(\frac{5\pi}{4})+i~sin(\frac{5\pi}{4})=-\frac{\sqrt 2}{2}-\frac{\sqrt 2}{2}i$ $w_6=cos(\frac{3\pi}{2})+i~sin(\frac{3\pi}{2})=-i$ $w_7=cos(\frac{7\pi}{4})+i~sin(\frac{7\pi}{4})=\frac{\sqrt 2}{2}-\frac{\sqrt 2}{2}i$
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