Answer
$\dfrac{(1+2i)(3-i)}{2+i}=3+i$
Work Step by Step
$\dfrac{(1+2i)(3-i)}{2+i}$
Evaluate the product in the numerator:
$\dfrac{(1+2i)(3-i)}{2+i}=\dfrac{3-i+6i-2i^{2}}{2+i}=\dfrac{3+5i-2i^{2}}{2+i}=...$
Substitute $i^{2}$ with $-1$:
$...=\dfrac{3+5i-2(-1)}{2+i}=\dfrac{5+5i}{2+i}=...$
Multiply the fraction by $\dfrac{2-i}{2-i}$:
$...=\Big(\dfrac{5+5i}{2+i}\Big)\Big(\dfrac{2-i}{2-i}\Big)=\dfrac{10-5i+10i-5i^{2}}{4-i^{2}}=\dfrac{10+5i-5i^{2}}{4-i^{2}}$
Once again, substitute $i^{2}$ with $-1$:
$...=\dfrac{10+5i-5(-1)}{4-(-1)}=\dfrac{15+5i}{5}=3+i$
