Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Review Exercises - Page 578: 101



Work Step by Step

Solve $i^{23}$ Since, $i^2=-1$ and $i^2 \cdot i^2=(-1)(-1)=1$ and $i^3=i^2 \cdot i=-i$ Thus, $i^{23}=(i^2)^{11} \cdot i=(-1)^{11} \cdot i$ Hence, $i^{23}=-i$
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