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 - Page 570: 64


$\dfrac{12}{17} - \dfrac{3}{17}i$

Work Step by Step

Rationalize the denominator by multiplying the conjugate of the denominator, which is $4-i$, to both the numerator and the denominator: $=\dfrac{3(4-i)}{(4+i)(4-i)} \\=\dfrac{12-3i}{(4+i)(4-i)}$ Simplify using the rule $(a+b)(a-b) = a^2-b^2$ to obtain: $=\dfrac{12-3i}{4^2-i^2} \\=\dfrac{12-3i}{16-i^2}$ Use the rule $i^2=-1$ to obtain: $=\dfrac{12-3i}{16-(-1)} \\=\dfrac{12-3i}{16+1} \\=\dfrac{12-3i}{17} \\=\dfrac{12}{17} - \dfrac{3}{17}i$
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