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: 68


$\dfrac{9}{5} + \dfrac{18}{5}i$

Work Step by Step

Multiply both the numerator and the denominator by the conjugate of the denominator, which is $1+2i$, to obtain: $=\dfrac{9(1+2i)}{(1-2i)(1+2i)} \\=\dfrac{9+18i}{(1-2i)(1+2i)}$ Simplify using the rule $(a-b)(a+b)=a^2-b^2$ to obtain: $=\dfrac{9+18i}{1^2-(2i)^2} \\=\dfrac{9+18i}{1-4i^2}$ Use the fact that $i^2=-1$ to obtain: $=\dfrac{9+18i}{1-4(-1)} \\=\dfrac{9+18i}{1+4} \\=\dfrac{9+18i}{5} \\=\dfrac{9}{5} + \dfrac{18}{5}i$
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