Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 7 - Section 7.7 - Complex Numbers - 7.7 Exercises: 70

Answer

$-4i-4$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To divide the given expression, $ \dfrac{-8i}{1+i} ,$ multiply both the numerator and the denominator by the complex conjugate of the denominator. Use $i^2=-1.$ $\bf{\text{Solution Details:}}$ Multiplying both the numerator and the denominator by the complex conjugate of the denominator, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{-8i}{1+i}\cdot\dfrac{1-i}{1-i} \\\\= \dfrac{-8i(1-i)}{(1+i)(1-i)} .\end{array} Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent \begin{array}{l}\require{cancel} \dfrac{-8i(1-i)}{(1)^2-(i)^2} \\\\= \dfrac{-8i(1-i)}{1-i^2} .\end{array} Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{-8i(1)-8i(-i)}{1-i^2} \\\\= \dfrac{-8i+8i^2}{1-i^2} .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{-8i+8(-1)}{1-(-1)} \\\\= \dfrac{-8i-8}{1+1} \\\\= \dfrac{-8i-8}{2} \\\\= \dfrac{2(-4i-4)}{2} \\\\= \dfrac{\cancel{2}(-4i-4)}{\cancel{2}} \\\\= -4i-4 .\end{array}
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