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: 85

Answer

$w_0=\frac{\sqrt 2}{2}+\frac{\sqrt x}{2}i$ $w_1=-\frac{\sqrt 2}{2}+\frac{\sqrt x}{2}i$ $w_2=-\frac{\sqrt 2}{2}-\frac{\sqrt x}{2}i$ $w_3=\frac{\sqrt 2}{2}-\frac{\sqrt x}{2}i$

Work Step by Step

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