Answer
$i$
Work Step by Step
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}
