Answer
$3-4i$
Work Step by Step
$\bf{\text{Solution Outline:}}$
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}
