#### Answer

$\dfrac{-5-12i}{13}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To divide the given expression, $
\dfrac{2-3i}{2+3i}
,$ multiply both the numerator and the denominator by the complex conjugate of the denominator. Use special products to multiply the resulting expression and 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{2-3i}{2+3i}\cdot\dfrac{2-3i}{2-3i}
\\\\=
\dfrac{(2-3i)(2-3i)}{(2+3i)(2-3i)}
\\\\=
\dfrac{(2-3i)^2}{(2+3i)(2-3i)}
.\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{(2-3i)^2}{(2)^2-(3i)^2}
\\\\=
\dfrac{(2-3i)^2}{4-9i^2}
.\end{array}
Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{(2)^2-2(2)(3i)+(3i)^2}{4-9i^2}
\\\\=
\dfrac{4-12i+9i^2}{4-9i^2}
.\end{array}
Since $i^2=-1,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{4-12i+9(-1)}{4-9(-1)}
\\\\=
\dfrac{4-12i-9}{4+9}
\\\\=
\dfrac{-5-12i}{13}
.\end{array}