Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8.1 Complex Numbers - 8.1 Exercises - Page 358: 103

Answer

$-i$

Work Step by Step

We know that $i^{2}=-1$. Therefore, $i^{4}=i^{2}\times i^{2}=-1\times-1=1$. We can use the fact that $i^{4}=1$ in order to evaluate higher powers of $i$. $\frac{1}{i^{-11}}=i^{11}=i^{8}\times i^{2}\times i=(i^{4})^{2}\times-1\times i=(1)^{2}\times-1\times i=1\times-1\times i=-i$
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