If, then find the least positive integral value of m.

$\left(\frac{1+i}{1-i}\right)^{m}=1$

$\Rightarrow\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^{m}=1$

$\Rightarrow\left(\frac{(1+i)^{2}}{1^{2}+1^{2}}\right)^{m}=1$

$\Rightarrow\left(\frac{1^{2}+i^{2}+2 i}{2}\right)^{m}=1$

$\Rightarrow\left(\frac{1-1+2 i}{2}\right)^{m}=1$

$\Rightarrow\left(\frac{2 i}{2}\right)^{w}=1$

$\Rightarrow i^{m}=1$

$\therefore m=4 k$, where $k$ is some integer.

Therefore, the least positive integer is 1 .

Thus, the least positive integral value of $m$ is $4(=4 \times 1)$.