Answer
$\dfrac{12}{17} - \dfrac{3}{17}i$
Work Step by Step
Rationalize the denominator by multiplying the conjugate of the denominator, which is $4-i$, to both the numerator and the denominator:
$=\dfrac{3(4-i)}{(4+i)(4-i)}
\\=\dfrac{12-3i}{(4+i)(4-i)}$
Simplify using the rule $(a+b)(a-b) = a^2-b^2$ to obtain:
$=\dfrac{12-3i}{4^2-i^2}
\\=\dfrac{12-3i}{16-i^2}$
Use the rule $i^2=-1$ to obtain:
$=\dfrac{12-3i}{16-(-1)}
\\=\dfrac{12-3i}{16+1}
\\=\dfrac{12-3i}{17}
\\=\dfrac{12}{17} - \dfrac{3}{17}i$
