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: 69


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

Work Step by Step

Multiply both the numerator and the denominator by the conjugate of the denominator, which is $3+2i$, to obtain: $=\dfrac{6i(3+2i)}{(3-2i)(3+2i)} \\=\dfrac{18i+12i^2}{(3-2i)(3+2i)}$ Simplify using 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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