Intermediate Algebra (12th Edition)

$5+12i$
$\bf{\text{Solution Outline:}}$ 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}