#### Answer

$i$

#### Work Step by Step

$i^{33}=i^{32+1}=i^{32}\times i=(i^{2})^{16}\times i=(-1)^{16}\times i=1\times i=i$

Published by
Pearson

ISBN 10:
978-0-13-421743-7

ISBN 13:
978-0-13421-743-7

$i$

$i^{33}=i^{32+1}=i^{32}\times i=(i^{2})^{16}\times i=(-1)^{16}\times i=1\times i=i$

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