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: 67

Answer

$\dfrac{28}{25} + \dfrac{21}{25}i$

Work Step by Step

Multiply both the numerator and the denominator by the conjugate of the denominator, which is $4+3i$, to obtain: $=\dfrac{7(4+3i)}{(4-3i)(4+3i)} \\=\dfrac{28+21i}{(4-3i)(4+3i)}$ Simplify using the rule $(a-b)(a+b)=a^2-b^2$ to obtain: $=\dfrac{28+21i}{4^2-(3i)^2} \\=\dfrac{28+21i}{16-9i^2}$ Use the fact that $i^2=-1$ to obtain: $=\dfrac{28+21i}{16-9(-1)} \\=\dfrac{28+21i}{16+9} \\=\dfrac{28+21i}{25} \\=\dfrac{28}{25} + \dfrac{21}{25}i$
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