#### Answer

$1-i$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To divide the given expression, $
\dfrac{2}{1+i}
,$ multiply both the numerator and the denominator by the complex conjugate of the denominator.
$\bf{\text{Solution Details:}}$
Multiplying both the numerator and the denominator by the complex conjugate of the denominator, the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{2}{1+i}\cdot\dfrac{1-i}{1-i}
\\\\=
\dfrac{2(1-i)}{(1+i)(1-i)}
.\end{array}
Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent
\begin{array}{l}\require{cancel}
\dfrac{2(1-i)}{(1)^2-(i)^2}
\\\\=
\dfrac{2(1-i)}{1-i^2}
.\end{array}
Since $i^2=-1,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{2(1-i)}{1-(-1)}
\\\\=
\dfrac{2(1-i)}{1+1}
\\\\=
\dfrac{2(1-i)}{2}
\\\\=
\dfrac{\cancel{2}(1-i)}{\cancel{2}}
\\\\=
1-i
.\end{array}