Question

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

Answer 1

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

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

$\Rightarrow (i{)}^{m/2}=(-i{)}^{n/3}=1$
$\Rightarrow \frac{m}{2}=4k_1$ and $\frac{n}{3}=4k_2$
$\Rightarrow m=8k_1$ and $n=12k_2$
Least value of $m=8$ and $n=12$
Greatest common divisor of $m$ and $n$ is $4.$