Answer
-1
Work Step by Step
Since $j=\sqrt{-1}$, $j^2=-1$, $j^3=-j$, $j^4=1$, then by multiplying both the denominator and numerator by $j^2$, we have
$$ \frac{1}{j^6}= \frac{j^2}{j^8}=j^2=-1.$$
Elementary Technical Mathematics
by Nelson, Robert; Eren, Dale;
Published by
Brooks Cole
ISBN 10:
1285199197
ISBN 13:
978-1-28519-919-1
Chapter 11 - Section 11.5 - Imaginary Numbers - Exercise - Page 378: 24
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
