## Intermediate Algebra (12th Edition)

$\dfrac{7+17i}{13}$
$\bf{\text{Solution Outline:}}$ To divide the given expression, $\dfrac{-1+5i}{3+2i} ,$ multiply both the numerator and the denominator by the complex conjugate of the denominator. Use special products to multiply the resulting expression and use $i^2=-1.$ $\bf{\text{Solution Details:}}$ Multiplying both the numerator and the denominator by the complex conjugate of the denominator, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{-1+5i}{3+2i}\cdot\dfrac{3-2i}{3-2i} \\\\= \dfrac{(-1+5i)(3-2i)}{(3+2i)(3-2i)} .\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} \dfrac{(-1+5i)(3-2i)}{(3)^2-(2i)^2} \\\\= \dfrac{(-1+5i)(3-2i)}{9-4i^2} .\end{array} Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel} \dfrac{-1(3)-1(-2i)+5i(3)+5i(-2i)}{9-4i^2} \\\\= \dfrac{-3+2i+15i-10i^2}{9-4i^2} \\\\= \dfrac{-3+(2+15)i-10i^2}{9-4i^2} \\\\= \dfrac{-3+17i-10i^2}{9-4i^2} .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{-3+17i-10(-1)}{9-4(-1)} \\\\= \dfrac{-3+17i+10}{9+4} \\\\= \dfrac{7+17i}{13} .\end{array}