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: 88

Answer

$z_0=cos~\frac{\pi}{16}+i~sin~\frac{\pi}{16}$ $z_1=cos~\frac{3\pi}{16}+i~sin~\frac{3\pi}{16}$ $z_2=cos~\frac{5\pi}{16}+i~sin~\frac{5\pi}{16}$ $z_3=cos~\frac{7\pi}{16}+i~sin~\frac{7\pi}{16}$ $z_4=cos~\frac{9\pi}{16}+i~sin~\frac{9\pi}{16}$ $z_5=cos~\frac{11\pi}{16}+i~sin~\frac{11\pi}{16}$ $z_6=cos~\frac{13\pi}{16}+i~sin~\frac{13\pi}{16}$ $z_7=cos~\frac{15\pi}{16}+i~sin~\frac{15\pi}{16}$

Work Step by Step

$z^8-i=0$ $z^8=i$ $r=|i|=1$. Also $i$ lies in the positive real axis. Polar form of $i$: $1(cos~\frac{\pi}{2}+i~sin~\frac{\pi}{2})$ $z_k=\sqrt[8]{1}[cos(\frac{\frac{\pi}{2}+2k\pi}{8})+i~sin(\frac{\frac{\pi}{2}+2k\pi}{8})]$ $z_0=1(cos~\frac{\pi}{16}+i~sin~\frac{\pi}{16})$ $z_1=1(cos~\frac{3\pi}{16}+i~sin~\frac{3\pi}{16})$ $z_2=1(cos~\frac{5\pi}{16}+i~sin~\frac{5\pi}{16})$ $z_3=1(cos~\frac{7\pi}{16}+i~sin~\frac{7\pi}{16})$ $z_4=1(cos~\frac{9\pi}{16}+i~sin~\frac{9\pi}{16})$ $z_5=1(cos~\frac{11\pi}{16}+i~sin~\frac{11\pi}{16})$ $z_6=1(cos~\frac{13\pi}{16}+i~sin~\frac{13\pi}{16})$ $z_7=1(cos~\frac{15\pi}{16}+i~sin~\frac{15\pi}{16})$
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