$\left(\frac{1+i}{1-i}\right)^{m / 2}=\left(\frac{1+i}{i-1}\right)^{n / 3}=1$

$\Rightarrow\left(\frac{(1+i)^{2}}{2}\right)^{m / 2}=\left(\frac{(1+i)^{2}}{-2}\right)^{n / 3}=1$

$\Rightarrow \quad( i )^{ m / 2}=(- i )^{ n / 3}=1$

$\Rightarrow \frac{ m }{2}=4 k _{1}$ and $\frac{ n }{3}=4 k _{2}$

$\Rightarrow m =8 k _{1}$ and $n =12 k _{2}$

Least value of $m =8$ and $n =12$

$\therefore \quad GCD =4$