Answer
$\dfrac{3}{4+i}=\dfrac{12}{17}-\dfrac{3}{17}i$
Work Step by Step
$\dfrac{3}{4+i}$
Multiply the numerator and the denominator of the fraction by the complex conjugate of the denominator:
$\dfrac{3}{4+i}=\dfrac{3}{4+i}\cdot\dfrac{4-i}{4-i}=\dfrac{3(4-i)}{4^{2}-i^{2}}=\dfrac{12-3i}{16-i^{2}}=...$
Substitute $i^{2}$ by $-1$ and simplify:
$...=\dfrac{12-3i}{16-(-1)}=\dfrac{12-3i}{16+1}=\dfrac{12-3i}{17}=\dfrac{12}{17}-\dfrac{3}{17}i$
