Answer
$\dfrac{1+i}{1-i}=i$
Work Step by Step
$\dfrac{1+i}{1-i}$
Multiply the numerator and the denominator of the fraction by the complex conjugate of the denominator:
$\dfrac{1+i}{1-i}=\dfrac{1+i}{1-i}\cdot\dfrac{1+i}{1+i}=\dfrac{(1+i)^{2}}{1^{2}-i^{2}}=\dfrac{1+2i+i^{2}}{1-i^{2}}=...$
Substitute $i^{2}$ by $-1$ and simplify:
$...=\dfrac{1+2i-1}{1-(-1)}=\dfrac{2i}{1+1}=\dfrac{2i}{2}=i$