Answer
$1-\dfrac{3}{2}i$
Work Step by Step
Using $(a+b)^2=a^2+2ab+b^2$ or the square of a binomial, the given expression, $
\dfrac{3+2i}{(1+i)^2}
,$ is equivalent to
\begin{align*}
&
\dfrac{3+2i}{(1)^2+2(1)(i)+(i)^2}
\\\\&=
\dfrac{3+2i}{1+2i+i^2}
\\\\&=
\dfrac{3+2i}{1+2i+(-1)}
&\text{ (use $i^2=-1$)}
\\\\&=
\dfrac{3+2i}{2i}
.\end{align*}
Multiplying both the numerator and the denominator of the expression above by the denominator, then
\begin{align*}\require{cancel}
&
\dfrac{3+2i}{2i}\cdot\dfrac{2i}{2i}
\\\\&=
\dfrac{3(2i)+2i(2i)}{2i(2i)}
&\text{ (use Distributive Property)}
\\\\&=
\dfrac{6i+4i^2}{4i^2}
\\\\&=
\dfrac{6i+4(-1)}{4(-1)}
&\text{ (use $i^2=-1$)}
\\\\&=
\dfrac{6i-4}{-4}
\\\\&=
\dfrac{-\cancel{4}^2+\cancel6^3i}{-\cancel4^2}
&\text{ (divide by $2$)}
\\\\&=
\dfrac{-2+3i}{-2}
\\\\&=
\dfrac{-2}{-2}+\dfrac{3}{-2}i
\\\\&=
1-\dfrac{3}{2}i
.\end{align*}
Hence, the simplified form of the given expression is $
1-\dfrac{3}{2}i
$.