Answer
$\dfrac{2-3i}{2+i}=\dfrac{1}{5}-\dfrac{8}{5}i$
Work Step by Step
$\dfrac{2-3i}{2+i}$
Multiply the numerator and the denominator of this expression by the complex conjugate of the denominator:
$\dfrac{2-3i}{2+i}=\dfrac{2-3i}{2+i}\cdot\dfrac{2-i}{2-i}=\dfrac{(2-3i)(2-i)}{2^{2}-i^{2}}=...$
$...=\dfrac{(2-3i)(2-i)}{4-i^{2}}=\dfrac{4-2i-6i+3i^{2}}{4-i^{2}}=\dfrac{4-8i+3i^{2}}{4-i^{2}}=...$
Substitute $i^{2}$ with $-1$ and simplify:
$...=\dfrac{4-8i+3(-1)}{4-(-1)}=\dfrac{4-8i-3}{4+1}=\dfrac{1-8i}{5}=\dfrac{1}{5}-\dfrac{8}{5}i$