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: 91

Answer

$0-\frac{2}{3}i$

Work Step by Step

$\frac{2}{3i}=\frac{2}{3i}(\frac{-i}{-i})=\frac{2\times -i}{3i\times -i}=\frac{-2i}{-3i^{2}}=\frac{-2i}{-3(-1)}=\frac{-2i}{3}=-\frac{2}{3}i=0-\frac{2}{3}i$ We know that $i^{2}=-1$.
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