## College Algebra (11th Edition)

$3-4i$
$\bf{\text{Solution Outline:}}$ To simplify the given expression, $\dfrac{-7+i}{-1-i} ,$ multiply the numerator and the denominator by the conjugate of the denomniator. Use special products to multiply the resulting expression. Then use the equivalence $i^2=-1.$ Finally, combine the real parts and the imaginary parts. $\bf{\text{Solution Details:}}$ Reversing the operator between the terms of the expression, $-1-i ,$ then its conjugate is $-1+i .$ Multiplying the numerator and the denominator of the given expression by the conjugate of the denominator, the given expression is equivalent to \begin{array}{l}\require{cancel} \dfrac{-7+i}{-1-i}\cdot\dfrac{-1+i}{-1+i} \\\\= \dfrac{(-7+i)(-1+i)}{(-1-i)(-1+i)} .\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{-7(-1)-7(i)+i(-1)+i(i)}{(-1-i)(-1+i)} \\\\= \dfrac{7-7i-i+i^2}{(-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{7-7i-i+i^2}{(-1)^2-(i)^2} \\\\= \dfrac{7-7i-i+i^2}{1-i^2} .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{7-7i-i+(-1)}{1-(-1)} \\\\= \dfrac{7-7i-i-1}{1+1} .\end{array} Combining the real parts and the imaginary parts results to \begin{array}{l}\require{cancel} \dfrac{(7-1)+(-7i-i)}{1+1} \\\\= \dfrac{6-8i}{2} \\\\= \dfrac{\cancel2(3)-\cancel2(4)i}{\cancel2} \\\\= 3-4i .\end{array}