Answer
$-j$
Work Step by Step
Since $j=\sqrt{-1}$, $j^2=-1$, and $j^4=1$, we have
$$ j^{27}=(j^{4} )^6j^2j=-j$$
where we used the fact that $27=4\times 6+2+1$.
Elementary Technical Mathematics
by Nelson, Robert; Eren, Dale;
Published by
Brooks Cole
ISBN 10:
1285199197
ISBN 13:
978-1-28519-919-1
Chapter 11 - Review - Page 380: 22
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