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: 73

Answer

$\dfrac{-5-12i}{13}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To divide the given expression, $ \dfrac{2-3i}{2+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{2-3i}{2+3i}\cdot\dfrac{2-3i}{2-3i} \\\\= \dfrac{(2-3i)(2-3i)}{(2+3i)(2-3i)} \\\\= \dfrac{(2-3i)^2}{(2+3i)(2-3i)} .\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{(2-3i)^2}{(2)^2-(3i)^2} \\\\= \dfrac{(2-3i)^2}{4-9i^2} .\end{array} Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{(2)^2-2(2)(3i)+(3i)^2}{4-9i^2} \\\\= \dfrac{4-12i+9i^2}{4-9i^2} .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{4-12i+9(-1)}{4-9(-1)} \\\\= \dfrac{4-12i-9}{4+9} \\\\= \dfrac{-5-12i}{13} .\end{array}
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