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: 63


$\dfrac{3}{5} - \dfrac{1}{5}i$

Work Step by Step

Rationalize the denominator by multiplying the conjugate of the denominator, which is $3-i$, to both the numerator and the denominator: $=\dfrac{2(3-i)}{(3+i)(3-i)}$ Simplify using the rule $(a+b)(a-b) = a^2-b^2$ to obtain: $=\dfrac{6-2i}{3^2-i^2} \\=\dfrac{6-2i}{9-i^2}$ Use the rule $i^2=-1$ to obtain: $=\dfrac{6-2i}{9-(-1)} \\=\dfrac{6-2i}{9+1} \\=\dfrac{6-2i}{10} \\=\dfrac{6}{10} - \dfrac{2}{10}i \\=\dfrac{3}{5} - \dfrac{1}{5}i$
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