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: 98

Answer

$-i$

Work Step by Step

Note that $(-i) = -1(i)$ Thus, the given expression is equivalent to: $=[-1(i)]^{13}$ Use the rule $(ab)^m = a^mb^m$ to obtain: $=(-1)^{13}(i)^{13} \\=(-1)(i^{13}) \\=-i^{13} \\=-i^{12+1}$ 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: $-i^{12+1}=-(i^{12} \cdot i)$ Use rule (1) above to obtain: $=-[(i^2)^6 \cdot i]$ Use rule (2) above to obtain: $\\=-[(-1)^6 \cdot i] \\=-(1 \cdot i) \\=-(i) \\=-i$
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