Answer
$\dfrac{16}{17}-\dfrac{4}{17}i$
Work Step by Step
Multiplying the numerator and the denominator by the conjugate of the denominator, the given expression, $
\dfrac{4}{4+i}
,$ is equivalent to
\begin{align*}\require {cancel}
&
\dfrac{4}{4+i}\cdot\dfrac{4-i}{4-i}
\\\\&=
\dfrac{4(4-i)}{(4)^2-(i)^2}
&\left( \text{use }(a+b)(a-b)=a^2-b^2 \right)
\\\\&=
\dfrac{4(4-i)}{16-i^2}
\\\\&=
\dfrac{16-4i}{16-i^2}
&\left( \text{use Distributive Property } \right)
\\\\&=
\dfrac{16-4i}{16-(-1)}
&\left( \text{use }i^2=-1 \right)
\\\\&=
\dfrac{16-4i}{16+1}
\\\\&=
\dfrac{16-4i}{17}
\\\\&=
\dfrac{16}{17}-\dfrac{4}{17}i
.\end{align*}
Hence, in the form $a\pm bi,$ the given expression is equivalent to $
\dfrac{16}{17}-\dfrac{4}{17}i
$.
