Answer
$\dfrac{18-i}{25}$
Work Step by Step
Multiplying by the conjugate of the denominator, the given expression, $
\dfrac{3+2i}{4+3i}
,$ is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{3+2i}{4+3i}\cdot\dfrac{4-3i}{4-3i}
\\\\=
\dfrac{3(4)+3(-3i)+2i(4)+2i(-3i)}{(4)^2-(3i)^2}
\\\\=
\dfrac{12-9i+8i-6i^2}{16-9i^2}
\\\\=
\dfrac{12-9i+8i-6(-1)}{16-9(-1)}
\\\\=
\dfrac{12-9i+8i+6}{16+9}
\\\\=
\dfrac{(12+6)+(-9i+8i)}{16+9}
\\\\=
\dfrac{18-i}{25}
\end{array}
