Find the least positive integer n for which

We have, $\left(\frac{1+i}{1-i}\right)^{n}=1$

Now, $\frac{1+\mathrm{i}}{1-\mathrm{i}}=\frac{1+\mathrm{i}}{1-\mathrm{i}} \times \frac{1+\mathrm{i}}{1+\mathrm{i}}$

$=\frac{(1+i)^{2}}{1^{2}-i^{2}}$

$=\frac{1^{2}+2 i+i^{2}}{1-(-1)}$

$=\frac{1+2 i-1}{2}$

$=i$

$\therefore\left(\frac{1+\mathrm{i}}{1-\mathrm{i}}\right)^{\mathrm{n}}=(\mathrm{i})^{\mathrm{n}}=1 \Rightarrow \mathrm{n}$ is multiple of 4

$\therefore$ The least positive integer $\mathrm{n}$ is 4