Answer
$\dfrac{23+41i}{65}$
Work Step by Step
Multiplying by the conjugate of the denominator, the given expression is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{5+3i}{7-4i}
\\\\=
\dfrac{5+3i}{7-4i}\cdot\dfrac{7+4i}{7+4i}
\\\\=
\dfrac{(5+3i)(7+4i)}{(7-4i)(7+4i)}
.\end{array}
Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel}
\dfrac{(5+3i)(7+4i)}{(7-4i)(7+4i)}
\\\\=
\dfrac{5(7)+5(4i)+3i(7)+3i(4i)}{(7-4i)(7+4i)}
\\\\=
\dfrac{35+20i+21i+12i^2}{(7-4i)(7+4i)}
\\\\=
\dfrac{35+41i+12i^2}{(7-4i)(7+4i)}
.\end{array}
Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent
\begin{array}{l}\require{cancel}
\dfrac{35+41i+12i^2}{(7)^2-(4i)^2}
\\\\=
\dfrac{35+41i+12i^2}{49-16i^2}
.\end{array}
Using $i^2=-1$ and then combining like terms, the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{35+41i+12i^2}{49-16i^2}
\\\\=
\dfrac{35+41i+12(-1)}{49-16(-1)}
\\\\=
\dfrac{35+41i-12}{49+16}
\\\\=
\dfrac{23+41i}{65}
.\end{array}
