Answer
$4+i$
Work Step by Step
Multiplying by the conjugate of the denominator of $
\dfrac{3+5i}{1+i}
$ results to
\begin{array}{l}
\dfrac{3+5i}{1+i}
\cdot
\dfrac{1-i}{1-i}
\\\\=
\dfrac{3(1)+3(-i)+5i(1)+5i(-i)}{(1)^2-(i)^2}
\\\\=
\dfrac{3-3i+5i-5i^2}{1-i^2}
\\\\=
\dfrac{3-3i+5i-5(-1)}{1-(-1)}
\\\\=
\dfrac{3-3i+5i+5}{1+1}
\\\\=
\dfrac{8+2i}{2}
\\\\=
4+i
.\end{array}
