Answer
$-\dfrac{1}{2}-\dfrac{1}{2}i$
Work Step by Step
Multiplying the numerator and the denominator by the conjugate of the denominator, the given expression, $
\dfrac{2-3i}{1+5i}
,$ is equivalent to
\begin{align*}\require{cancel}
&
\dfrac{2-3i}{1+5i}\cdot\dfrac{1-5i}{1-5i}
\\\\&=
\dfrac{2(1)+2(-5i)-3i(1)-3i(-5i)}{(1+5i)(1-5i)}
&\text{ (use FOIL Method)}
\\\\&=
\dfrac{2(1)+2(-5i)-3i(1)-3i(-5i)}{(1)^2-(5i)^2}
&\text{ (use $(a+b)(a-b)=a^2-b^2)$}
\\\\&=
\dfrac{2-10i-3i+15i^2}{1-25i^2}
\\\\&=
\dfrac{2-10i-3i+15(-1)}{1-25(-1)}
&\text{ (use $i^2=-1$)}
\\\\&=
\dfrac{2-10i-3i-15}{1+25}
\\\\&=
\dfrac{(2-15)+(-10i-3i)}{1+25}
&\text{ (combine like terms)}
\\\\&=
\dfrac{-13-13i}{26}
\\\\&=
\dfrac{-\cancel{13}^1-\cancel{13}^1i}{\cancel{26}^2}
\\\\&=
\dfrac{-1-i}{2}
\\\\&=
-\dfrac{1}{2}-\dfrac{1}{2}i
.\end{align*}
Hence, the given expression simplifies to $
-\dfrac{1}{2}-\dfrac{1}{2}i
$.
