Find the smallest positive integer value of m for which

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

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

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

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

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

$=-2 i\left(i^{m}\right)$

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

For this to be real, the smallest positive value of $m$ will be 1 .

Thus, $i^{1+1}=i^{2}=-1$, which is real.