## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$i$
Multiplying by the conjugate of the denominator, then \begin{array}{l}\require{cancel} \dfrac{1+i}{1-i} \\\\= \dfrac{1+i}{1-i}\cdot\dfrac{1+i}{1+i} \\\\= \dfrac{(1+i)(1+i)}{(1-i)(1+i)} \\\\= \dfrac{(1+i)^2}{(1-i)(1+i)} .\end{array} Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{(1)^2+2(1)(i)+(i)^2}{(1-i)(1+i)} \\\\= \dfrac{1+2i+i^2}{(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{1+2i+i^2}{(1)^2-(i)^2} \\\\= \dfrac{1+2i+i^2}{1-i^2} .\end{array} Using $i^2=-1$ and then combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{1+2i+i^2}{1-i^2} \\\\= \dfrac{1+2i+(-1)}{1-(-1)} \\\\= \dfrac{1+2i-1}{1+1} \\\\= \dfrac{2i}{2} \\\\= i .\end{array}