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: 66

Answer

$1+2i$

Work Step by Step

Rationalize the denominator by multiplying the conjugate of the denominator, which is $2-i$, to both the numerator and the denominator: $=\dfrac{5i(2-i)}{(2+i)(2-i)} \\=\dfrac{10i-5i^2}{(2+i)(2-i)}$ Simplify using the rule $(a+b)(a-b) = a^2-b^2$ to obtain: $=\dfrac{10i-5i^2}{2^2-i^2} \\=\dfrac{10i-5i^2}{4-i^2}$ Use the rule $i^2=-1$ to obtain: $=\dfrac{10i-5(-1)}{4-(-1)} \\=\dfrac{10i-(-5)}{4+1} \\=\dfrac{10i+5}{5} \\=\dfrac{5+10i}{5} \\=\dfrac{5}{5} + \dfrac{10}{5}i \\=1+2i$
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