## Intermediate Algebra (12th Edition)

$1+i$
$\bf{\text{Solution Outline:}}$ 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}