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: 68

Answer

$1-i$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To divide the given expression, $ \dfrac{2}{1+i} ,$ 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{2}{1+i}\cdot\dfrac{1-i}{1-i} \\\\= \dfrac{2(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{2(1-i)}{(1)^2-(i)^2} \\\\= \dfrac{2(1-i)}{1-i^2} .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{2(1-i)}{1-(-1)} \\\\= \dfrac{2(1-i)}{1+1} \\\\= \dfrac{2(1-i)}{2} \\\\= \dfrac{\cancel{2}(1-i)}{\cancel{2}} \\\\= 1-i .\end{array}
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