Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.3 Complex Numbers - 1.3 Exercises - Page 113: 97

Answer

$-i$

Work Step by Step

Apply exponent rule: $\displaystyle \quad a^{-n}=\frac{1}{a^{n}}$ $i^{-13}=\displaystyle \frac{1}{i^{13}}$ Calculate $i^{13}$: $i^{13}=i^{12+1}=i^{12}i \qquad$...Apply: $a^{mn}=(a^{m})^{n}$ $=i(i^{2})^{6}\qquad$...Apply: $\quad i^{2}=-1$ $=(-1)^{6}i \qquad$...Apply: $(-a)^{n}=a^{n},$ if $n$ is even $=1i$ $=i$ Thus, $i^{-13}=\displaystyle \frac{1}{i}$ Multiply by the conjugate $\displaystyle \frac{-i}{-i}$ $=\displaystyle \frac{1\cdot(-i)}{i(-i)}$ $=1\cdot(-i)$ $=-i$
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