Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Section 7.7 - Complex Numbers - Exercise Set: 83

Answer

$-\dfrac{7}{3}+\dfrac{4}{3}i$

Work Step by Step

Multiply both the numerator and the denominator by the conjugate of the denominator, which is $3i$, to obtain: $=\dfrac{(4+7i)(3i)}{-3i(3i)} \\=\dfrac{12i+21i^2}{-9i^2}$ Use the fact that $i^2=-1$ to obtain: $=\dfrac{12i+21(-1)}{-9(-1)} \\=\dfrac{12i-21}{9} \\=\dfrac{-21+12i}{9} \\=-\dfrac{21}{9} + \dfrac{12}{9}i \\=-\dfrac{7}{3}+\dfrac{4}{3}i$
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