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: 72

Answer

$-5+i$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To divide the given expression, $ \dfrac{-38-8i}{7+3i} ,$ multiply both the numerator and the denominator by the complex conjugate of the denominator. Use special products to multiply the resulting expression and 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{-38-8i}{7+3i}\cdot\dfrac{7-3i}{7-3i} \\\\= \dfrac{(-38-8i)(7-3i)}{(7+3i)(7-3i)} .\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{(-38-8i)(7-3i)}{(7)^2-(3i)^2} \\\\= \dfrac{(-38-8i)(7-3i)}{49-9i^2} .\end{array} Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel} \dfrac{-38(7)-38(-3i)-8i(7)-8i(-3i)}{49-9i^2} \\\\= \dfrac{-266+114i-56i+24i^2}{49-9i^2} \\\\= \dfrac{-266+(114-56)i+24i^2}{49-9i^2} \\\\= \dfrac{-266+58i+24i^2}{49-9i^2} .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{-266+58i+24(-1)}{49-9(-1)} \\\\= \dfrac{-266+58i-24}{49+9} \\\\= \dfrac{-290+58i}{58} \\\\= \dfrac{58(-5+i)}{58} \\\\= \dfrac{\cancel{58}(-5+i)}{\cancel{58}} \\\\= -5+i .\end{array}
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