## Intermediate Algebra (12th Edition)

$-4i-4$
$\bf{\text{Solution Outline:}}$ To divide the given expression, $\dfrac{-8i}{1+i} ,$ multiply both the numerator and the denominator by the complex conjugate of the denominator. 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{-8i}{1+i}\cdot\dfrac{1-i}{1-i} \\\\= \dfrac{-8i(1-i)}{(1+i)(1-i)} .\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{-8i(1-i)}{(1)^2-(i)^2} \\\\= \dfrac{-8i(1-i)}{1-i^2} .\end{array} Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{-8i(1)-8i(-i)}{1-i^2} \\\\= \dfrac{-8i+8i^2}{1-i^2} .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{-8i+8(-1)}{1-(-1)} \\\\= \dfrac{-8i-8}{1+1} \\\\= \dfrac{-8i-8}{2} \\\\= \dfrac{2(-4i-4)}{2} \\\\= \dfrac{\cancel{2}(-4i-4)}{\cancel{2}} \\\\= -4i-4 .\end{array}