Answer
$\dfrac{8}{13}+\dfrac{12}{13}i$
Work Step by Step
Multiplying both the numerator and the denominator of $
\dfrac{4}{2-3i}
$ by the complex conjugate of the denominator, then
\begin{align*}\require{cancel}
&
\dfrac{4}{2-3i}\cdot\dfrac{2+3i}{2+3i}
\\\\&=
\dfrac{4(2)+4(3i)}{(2-3i)(2+3i)}
&\text{ (use Distributive Property)}
\\\\&=
\dfrac{4(2)+4(3i)}{(2)^2-(3i)^2}
&\text{ (use $(a+b)(a-b)=a^2-b^2$)}
\\\\&=
\dfrac{8+12i}{4-9i^2}
\\\\&=
\dfrac{8+12i}{4-9(-1)}
&\text{ (use $i^2=-1$)}
\\\\&=
\dfrac{8+12i}{4+9}
\\\\&=
\dfrac{8+12i}{13}
\\\\&=
\dfrac{8}{13}+\dfrac{12}{13}i
.\end{align*}
Hence, the simplified form of the given expression is $
\dfrac{8}{13}+\dfrac{12}{13}i
$.
