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