#### 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, $i^{23}=-i$

Published by
Pearson

ISBN 10:
0-13417-894-7

ISBN 13:
978-0-13417-894-3

$-i$

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