#### Answer

$2+2i$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To divide the given expression, $
\dfrac{8i}{2+2i}
,$ multiply both the numerator and the denominator by the complex conjugate of the denominator.
$\bf{\text{Solution Details:}}$
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}{2+2i}\cdot\dfrac{2-2i}{2-2i}
\\\\=
\dfrac{8i(2-2i)}{(2+2i)(2-2i)}
.\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{8i(2-2i)}{(2)^2-(2i)^2}
\\\\=
\dfrac{8i(2-2i)}{4-4i^2}
.\end{array}
Since $i^2=-1,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{8i(2-2i)}{4-4(-1)}
\\\\=
\dfrac{8i(2-2i)}{4+4}
\\\\=
\dfrac{8i(2-2i)}{8}
\\\\=
\dfrac{\cancel8i(2-2i)}{\cancel8}
\\\\=
i(2-2i)
.\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}
i(2)+i(-2i)
\\\\=
2i-2i^2
.\end{array}
Since $i^2=-1,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
2i-2(-1)
\\\\=
2i+2
\\\\=
2+2i
.\end{array}