## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 7 - Section 7.7 - Complex Numbers - 7.7 Exercises - Page 491: 55

#### Answer

$-9+40i$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $(4+5i)^2 ,$ use the special product on squaring binomials. Then use $i^2=-1$ and combine like terms. $\bf{\text{Solution Details:}}$ 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} (4)^2+2(4)(5i)+(5i)^2 \\\\= 16+40i+25i^2 .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 16+40i+25(-1) \\\\= 16+40i-25 .\end{array} Combining like terms results to \begin{array}{l}\require{cancel} (16-25)+40i \\\\= -9+40i .\end{array}

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