Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.1 - Complex Numbers - Exercise Set - Page 314: 25


$-\dfrac{24}{25} + \dfrac{32}{25}i$

Work Step by Step

Rationalize the denominator by multiplying the conjugate of the denominator, which is $4+3i$, to both the numerator and the denominator to obtain: $=\dfrac{8i(4+3i)}{(4-3i)(4+3i)} \\=\dfrac{32i+24i^2}{(4-3i)(4+3i)}$ Use the rule $(a-b)(a+b) = a^2-b^2$ to obtain: $=\dfrac{32i+24i^2}{4^2-(3i)^2} \\=\dfrac{32i+24i^2}{16-9i^2}$ Use the fact that $i^2=-1$ to obtain: $=\dfrac{32i+24(-1)}{16-9(-1)} \\=\dfrac{32i-24}{16+9} \\=\dfrac{-24+32i}{25} \\=-\dfrac{24}{25} + \dfrac{32}{25}i$
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