Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Section 7.7 - Complex Numbers - Exercise Set - Page 463: 56



Work Step by Step

Multiplying both the numerator and the denominator by $i$ and using $i^2=-1$, the given expression, $ \dfrac{6+8i}{3i} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{6+8i}{3i}\cdot\dfrac{i}{i} \\\\= \dfrac{6i+8i^2}{3i^2} \\\\= \dfrac{6i+8(-1)}{3(-1)} \\\\= \dfrac{6i-8}{-3} \\\\= -\dfrac{6i-8}{3} \\\\= \dfrac{8-6i}{3} \\\\= \dfrac{8}{3}-\dfrac{6i}{3} \\\\= \dfrac{8}{3}-2i .\end{array}
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