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

Answer: 4

${\left(\frac{1+i}{1-i}\right)}^{m/2}={\left(\frac{1+i}{i-1}\right)}^{n/3}=1$
$\Rightarrow {\left(\frac{(1+i{)}^2}{2}\right)}^{m/2}={\left(\frac{(1+i{)}^2}{-2}\right)}^{n/3}=1$
$\Rightarrow (i{)}^{m/2}=(-i{)}^{n/3}=1$
$\Rightarrow \frac{m}{2}=4k_1$ and $\frac{n}{3}=4k_2$
$\Rightarrow m=8k_1$ and $n=12k_2$
Least value of $m=8$ and $n=12$.
$\therefore GCD=4$