Answer
$-i$
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, $i^{23}=-i$
Intermediate Algebra for College Students (7th Edition)
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-13417-894-7
ISBN 13:
978-0-13417-894-3
Chapter 7 - Review Exercises - Page 578: 101
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