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: 87

Answer

$-i$

Work Step by Step

Note that: $i^{11} = 1^{10+1}$ RECALL: (1) $a^{mn} = (a^m)^n$ (2) $i^2=-1$ (3) $a^{m+n} = a^m \cdot a^n$ Use rule (3) above to obtain: $=i^{10+1}=i^{10} \cdot i$ Use rule (1) above to obtain: $=(i^2)^5\cdot i$ Use rule (2) to obtain: $=(-1)^{5} \cdot i \\=(-1)(-1)(-1)(-1)(-1)(i) \\=-i$
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