If ((1 + i)/(1 - i))(m)/(2) = ((1 + i)/(i - 1))(n)/(3) = 1, (m, n ∈...

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

Rationalize both the terms

\left(\dfrac{1 + i}{1 - i} \times \dfrac{1 + i}{1 + i}\right)\dfrac{m}{2} = \left(\dfrac{1 + i}{i - 1} \times \dfrac{- 1 - i}{- 1 - i}\right)\dfrac{n}{3} = 1$</p><p>\left(\dfrac{1 + {i}^{2} + 2i}{1 - {i}^{2}}\right)\dfrac{m}{2} = \left(\dfrac{- 1 - {i}^{2} - 2i}{{\left(- 1\right)}^{2} - {i}^{2}}\right)\dfrac{n}{3} = 1$

We know ${i}^{2} = - 1$

\left(\dfrac{2i}{2}\right)\dfrac{m}{2} = \left(\dfrac{- 2i}{2}\right)\dfrac{n}{3} = 1$</p><p>\left(i\right)\dfrac{m}{2} = \left(- i\right)\dfrac{n}{3} = 1$

For \left(i\right)\dfrac{m}{2} = 1 ⇒ \dfrac{m}{2} = 4k ⇒ m = 8k, k \in I$</p><p>Since$m$is natural number so the least value of$m$is$8$.</p><p>For \left(- i\right)\dfrac{n}{3} = 1 ⇒ \dfrac{n}{3} = 4l ⇒ 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 $4$.