## College Algebra (11th Edition)

$7-24i$
$\bf{\text{Solution Outline:}}$ To simplify the given expression, $(4-3i)^2 ,$ use the special product on squaring binomials. Then use the equivalence $i^2=-1.$ Finally, combine combine the real parts and the imaginary parts. $\bf{\text{Solution Details:}}$ 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)(3i)+(3i)^2 \\\\= 16-24i+9i^2 .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 16-24i+9(-1) \\\\= 16-24i-9 .\end{array} Combining the real parts and the imaginary parts results to \begin{array}{l}\require{cancel} (16-9)-24i \\\\= 7-24i .\end{array}