#### 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{\cancel2(1+i)}{\cancel2}
\\\\=
1+i
.\end{array}