Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.6 - Complex Numbers - 1.6 Exercises - Page 64: 43



Work Step by Step

$\dfrac{4+6i}{3i}$ Multiply the fraction by $\dfrac{-3i}{-3i}$: $\Big(\dfrac{4+6i}{3i}\Big)\Big(\dfrac{-3i}{-3i}\Big)=\dfrac{-12i-18i^{2}}{-9i^{2}}=...$ Substitute $i^{2}$ with $-1$: $...=\dfrac{-12i-18(-1)}{-9(-1)}=\dfrac{-12i+18}{9}=\dfrac{18}{9}-\dfrac{12}{9}i=2-\dfrac{4}{3}i$
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