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: 71

Answer

$-1+2i$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To divide the given expression, $ \dfrac{-7+4i}{3+2i} ,$ 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{-7+4i}{3+2i}\cdot\dfrac{3-2i}{3-2i} \\\\= \dfrac{(-7+4i)(3-2i)}{(3+2i)(3-2i)} .\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{(-7+4i)(3-2i)}{(3)^2-(2i)^2} \\\\= \dfrac{(-7+4i)(3-2i)}{9-4i^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{-7(3)-7(-2i)+4i(3)+4i(-2i)}{9-4i^2} \\\\= \dfrac{-21+14i+12i-8i^2}{9-4i^2} \\\\= \dfrac{-21+(14+12)i-8i^2}{9-4i^2} \\\\= \dfrac{-21+26i-8i^2}{9-4i^2} .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{-21+26i-8(-1)}{9-4(-1)} \\\\= \dfrac{-21+26i+8}{9+4} \\\\= \dfrac{-13+26i}{13} \\\\= \dfrac{13(-1+2i)}{13} \\\\= \dfrac{\cancel{13}(-1+2i)}{\cancel{13}} \\\\= -1+2i .\end{array}
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