Find the least positive integer $n$ such that $\left(\dfrac{1 + i}{1 - i}\right)^n = 1$
$\begin{aligned}
\dfrac{1 + i}{1 - i} & = \dfrac{\left(1 + i\right)^2}{\left(1 + i\right) \left(1 - i\right)} \\\\
& = \dfrac{1 + 2i + i^2}{1 - i^2} \\\\
& = \dfrac{2i}{2} = i \;\;\;\; \left[\because \; i^2 = -1\right]
\end{aligned}$
$\therefore$ $\;$ $\left(\dfrac{1 + i}{1 - i}\right)^n = i^n$
$\therefore$ $\;$ We have $\;$ $i^n = 1$
Now, $i = \sqrt{-1}$ $\implies$ $i^4 = 1$
$\therefore$ $\;$ The least possible integer value of $n$ is $4$.