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: 21


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

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 to obtain: $=\dfrac{2(3+i)}{(3-i)(3+i)} \\=\dfrac{6+2i}{(3-i)(3+i)}$ Use 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 fact that $i^2=-1$ 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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