Answer
$\dfrac{1-8i}{5}$
Work Step by Step
Multiplying by the conjugate of the denominator, the expression $
\dfrac{2-3i}{2+i}
$ simplifies to
\begin{array}{l}
\dfrac{2-3i}{2+i}
\cdot
\dfrac{2-i}{2-i}
\\\\=
\dfrac{2(2)+2(-i)-3i(2)-3i(-i)}{(2)-(i)^2}
\\\\=
\dfrac{4-2i-6i+3i^2}{4-i^2}
\\\\=
\dfrac{4-2i-6i+3(-1)}{4-(-1)}
\\\\=
\dfrac{1-8i}{5}
.\end{array}
