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: 24



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 to obtain: $=\dfrac{5i(2+i)}{(2-i)(2+i)} \\=\dfrac{10i+5i^2}{(2-i)(2+i)}$ Use 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 fact that $i^2=-1$ to obtain: $=\dfrac{10i+5(-1)}{4-(-1)} \\=\dfrac{10i-5}{4+1} \\=\dfrac{-5+10i}{5} \\=\dfrac{-5}{5} + \dfrac{10}{5}i \\=-1+2i$
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