Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Section 7.7 - Complex Numbers - Exercise Set - Page 570: 93



Work Step by Step

Note that $i^{17} = i^{16+1}$ RECALL: (1) $a^{mn} = (a^m)^n$ (2) $i^2=-1$ (3) $a^{m+n} = a^m \cdot a^n$ Use rule (1) above to obtain: $=(i^2)^{8} \cdot i$ Use rule (2) to obtain: $=(-1)^{8} \cdot i$ Note that when $-1$ raised to an even power, the result is $1$. Thus, the expression above simplifies to: $= 1 \cdot i \\=i$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After 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.