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: 56

Answer

$5+12i$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $ (3+2i)^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} (3)^2+2(3)(2i)+(2i)^2 \\\\= 9+12i+4i^2 .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 9+12i+4(-1) \\\\= 9+12i-4 .\end{array} Combining like terms results to \begin{array}{l}\require{cancel} (9-4)+12i \\\\= 5+12i .\end{array}
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