## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 7 - Section 7.7 - Complex Numbers - 7.7 Exercises - Page 491: 63

#### Answer

$4$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $(1+i)^2(1-i)^2 ,$ use the laws of exponents to express the factors in a single exponent. Then use the special product on multiplying the sum and difference of like terms. Also, use $i^2=-1.$ $\bf{\text{Solution Details:}}$ Using the extended Power Rule of the laws of exponents which is given by $\left( x^my^n \right)^p=x^{mp}y^{np},$ the expression above is equivalent to \begin{array}{l}\require{cancel} [(1+i)(1-i)]^2 .\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} [(1)^2-(i)^2]^2 \\\\= [1-i^2]^2 .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} [1-(-1)]^2 \\\\= [1+1]^2 \\\\= ^2 \\\\= 4 .\end{array}

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