Answer
$\dfrac{19-4i}{29}$
Work Step by Step
Multiplying by the conjugate of the denominator, the given expression, $
\dfrac{2+3i}{2+5i}
,$ is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{2+3i}{2+5i}\cdot\dfrac{2-5i}{2-5i}
\\\\=
\dfrac{2(2)+2(-5i)+3i(2)+3i(-5i)}{(2)^2-(5i)^2}
\\\\=
\dfrac{4-10i+6i-15i^2}{4-25i^2}
\\\\=
\dfrac{4-10i+6i-15(-1)}{4-25(-1)}
\\\\=
\dfrac{4-10i+6i+15}{4+25}
\\\\=
\dfrac{(4+15)+(-10i+6i)}{4+25}
\\\\=
\dfrac{19-4i}{29}
\end{array}
