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: 26


$-\dfrac{12}{13} - \dfrac{18}{13}i$

Work Step by Step

Rationalize the denominator by multiplying the conjugate of the denominator, which is $3-2i$, to both the numerator and the denominator to obtain: $=\dfrac{-6i(3-2i)}{(3+2i)(3-2i)} \\=\dfrac{-18i+12i^2}{(3+2i)(3-2i)}$ Use the rule $(a-b)(a+b) = a^2-b^2$ to obtain: $=\dfrac{-18i+12i^2}{3^2-(2i)^2} \\=\dfrac{-18i+12i^2}{9-4i^2}$ Use the fact that $i^2=-1$ to obtain: $=\dfrac{-18i+12(-1)}{9-4(-1)} \\=\dfrac{-18i-12}{9+4} \\=\dfrac{-12-18i}{13} \\=-\dfrac{12}{13} - \dfrac{18}{13}i$
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