If 1 + i 1 - i m 2 = 1 + i i - 1 n 3 = 1 ,     m ,   n ∈ N then the greatest common…

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

Rationalize both the terms

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

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

We know $i^2=-1$

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

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

For ${\left(i\right)}^{\frac{m}{2}}=1 \Rightarrow \frac{m}{2}=4k \Rightarrow m=8k, k\in I$

Since $m$ is natural number so the least value of $m$ is $8$.

For ${\left(-i\right)}^{\frac{n}{3}}=1 \Rightarrow \frac{n}{3}=4l \Rightarrow n=12l, l\in I$

Since $n$ is natural number so the least value of $n$ is $12$.

So greatest common factor of $8$ and $12$ is $4$.