Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 1 - Section 1.6 - Complex Numbers - 1.6 Exercises - Page 64: 51



Work Step by Step

$i^{1000}$ Break it down into $(i^{2})^{x}$ since $(i^{2} = -1)$ we start with $(-1)*i^{500}$ $(-1)*i^{500}$ $(-1)(-1)*i^{250}$ $(-1)(-1)(-1)*i^{125}$ now 125 is not divisible by 2 so lets set aside an i and make it $(i*i^{124})$ $i*(-1)(-1)(-1)*i^{124}$ $i*(-1)(-1)(-1)(-1)*i^{62}$ $i*(-1)(-1)(-1)(-1)(-1)*i^{31}$ $i*i*(-1)(-1)(-1)(-1)(-1)*i^{30}$ $i*i*(-1)(-1)(-1)(-1)(-1)(-1)*i^{15}$ $i*i*i*(-1)(-1)(-1)(-1)(-1)(-1)*i^{15}$ $i*i*i*i*(-1)(-1)(-1)(-1)(-1)(-1)*i^{14}$ $i*i*i*i*(-1)(-1)(-1)(-1)(-1)(-1)(-1)*i^{7}$ $i*i*i*i*i*(-1)(-1)(-1)(-1)(-1)(-1)(-1)*i^{6}$ $i*i*i*i*i*(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)*i^{3}$ $i*i*i*i*i*i*(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)*i^{2}$ $i*i*i*i*i*i*(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)$ Remember (i*i) = (-1) so we can combine the remaining i's $(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)$ Then combined the (-1)'s and you end up with the answer of positive 1.
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