#### Answer

$1$

#### Work Step by Step

Using $i^2=-1$ and the laws of exponents, the given expression, $
i^{32}
,$ is equivalent to
\begin{array}{l}\require{cancel}
\left( i^2 \right)^{16}
\\\\=
\left( -1 \right)^{16}
\\\\=
1
\end{array}

Published by
Pearson

ISBN 10:
0-32184-874-8

ISBN 13:
978-0-32184-874-1

$1$

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