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: 69

Answer

$2+2i$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To divide the given expression, $ \dfrac{8i}{2+2i} ,$ multiply both the numerator and the denominator by the complex conjugate of the denominator. $\bf{\text{Solution Details:}}$ 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}{2+2i}\cdot\dfrac{2-2i}{2-2i} \\\\= \dfrac{8i(2-2i)}{(2+2i)(2-2i)} .\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{8i(2-2i)}{(2)^2-(2i)^2} \\\\= \dfrac{8i(2-2i)}{4-4i^2} .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{8i(2-2i)}{4-4(-1)} \\\\= \dfrac{8i(2-2i)}{4+4} \\\\= \dfrac{8i(2-2i)}{8} \\\\= \dfrac{\cancel8i(2-2i)}{\cancel8} \\\\= i(2-2i) .\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} i(2)+i(-2i) \\\\= 2i-2i^2 .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 2i-2(-1) \\\\= 2i+2 \\\\= 2+2i .\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.