Answer
$\dfrac{8}{17}+\dfrac{19}{17}i$
Work Step by Step
Multiplying both the numerator and the denominator of $
\dfrac{4-3i}{-1-4i}
$ by the complex conjugate of the denominator, then
\begin{align*}\require{cancel}
&
\dfrac{4-3i}{-1-4i}\cdot\dfrac{-1+4i}{-1+4i}
\\\\&=
\dfrac{4(-1)+4(4i)-3i(-1)-3i(4i)}{(-1-4i)(-1+4i)}
&\text{ (use FOIL)}
\\\\&=
\dfrac{4(-1)+4(4i)-3i(-1)-3i(4i)}{(-1)^2-(4i)^2}
&\text{ (use $(a+b)(a-b)=a^2-b^2$)}
\\\\&=
\dfrac{-4+16i+3i-12i^2}{1-16i^2}
\\\\&=
\dfrac{-4+16i+3i-12(-1)}{1-16(-1)}
&\text{ (use $i^2=-1$)}
\\\\&=
\dfrac{-4+16i+3i+12}{1+16}
\\\\&=
\dfrac{(-4+12)+(16i+3i)}{1+16}
\\\\&=
\dfrac{8+19i}{17}
\\\\&=
\dfrac{8}{17}+\dfrac{19}{17}i
.\end{align*}
Hence, the simplified form of the given expression is $
\dfrac{8}{17}+\dfrac{19}{17}i
$.
