Question

If $\left(\frac{1+i}{1-i}\right)^{\frac{m}{2}}=\left(\frac{1+i}{i-1}\right)^{\frac{n}{3}}=1,(m, n \in N)$ then the greatest common divisor of the least values of $m$ and $n$ is______.

Answer 1

$\left(\frac{1+i}{1-i}\right)^{m / 2}=\left(\frac{1+i}{i-1}\right)^{n / 3}=1$
$\Rightarrow \left(\frac{(1+i)^{2}}{2}\right)^{m / 2}=\left(\frac{(1+i)^{2}}{-2}\right)^{n / 3}=1$
$\Rightarrow \quad( i )^{ m / 2}=(- i )^{ n / 3}=1$
$\Rightarrow \frac{ m }{2}=4 k _{1}$ and $\frac{ n }{3}=4 k _{2}$
$\Rightarrow m =8 k _{1}$ and $n =12 k _{2}$
Least value of $m =8$ and $n =12$.
$\therefore GCD =4$