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

We begin by rewriting each complex fraction in polar (exponential) form so that taking powers becomes straightforward.

The numerator $$1+i$$ has modulus $$\sqrt{2}$$ and argument $$\dfrac{\pi}{4}$$, so we write it as $$\sqrt{2}\,e^{\,i\pi/4}.$$ The denominator $$1-i$$ also has modulus $$\sqrt{2}$$ but argument $$-\dfrac{\pi}{4}$$, so it is $$\sqrt{2}\,e^{\,-i\pi/4}.$$

Hence

$$\frac{1+i}{1-i}= \frac{\sqrt{2}\,e^{\,i\pi/4}}{\sqrt{2}\,e^{\,-i\pi/4}}=e^{\,i\pi/4+i\pi/4}=e^{\,i\pi/2}=i.$$

In the second fraction the denominator is $$i-1 = -(1-i).$$ Multiplying by $$-1$$ multiplies the modulus by $$1$$ (unchanged) but adds $$\pi$$ to the argument. Therefore

$$\frac{1+i}{i-1}= -\frac{1+i}{1-i}= -\,e^{\,i\pi/2}=e^{\,-i\pi/2}= -i.$$

Now the equation of the question can be rewritten as

$$\left(i\right)^{\,m/2}=1\qquad\text{and}\qquad\left(-i\right)^{\,n/3}=1,$$

with $$m,n\in\mathbb N.$$ We handle the two conditions separately.

First condition. Using the identity $$e^{i\theta}=1\Longleftrightarrow\theta=2\pi k,\ k\in\mathbb Z,$$ we write

$$i=e^{\,i\pi/2},\qquad\text{so}\qquad i^{\,m/2}=e^{\,i(\pi/2)\,m/2}=e^{\,i\pi m/4}.$$

Setting this equal to 1 gives

$$e^{\,i\pi m/4}=1\Longrightarrow \frac{\pi m}{4}=2\pi k\Longrightarrow m=8k.$$

The least positive natural value is obtained when $$k=1,$$ so $$m_{\min}=8.$$

Second condition. The number $$-i$$ has modulus $$1$$ and, because the argument of a complex number is defined up to addition of multiples of $$2\pi$$, we may choose any of its valid arguments: $$-i=e^{\,i\left(-\pi/2+2\pi k\right)},\quad k\in\mathbb Z.$$

Raising to the power $$n/3$$ we get

$$(-i)^{\,n/3}=e^{\,i\left(-\pi/2+2\pi k\right)n/3}=1.$$

Again using $$e^{i\theta}=1\Longleftrightarrow\theta=2\pi r,\ r\in\mathbb Z,$$ we equate arguments:

$$\left(-\frac{\pi}{2}+2\pi k\right)\frac{n}{3}=2\pi r.$$

Dividing by $$\pi$$ and simplifying,

$$\left(-\frac{1}{2}+2k\right)\frac{n}{3}=2r\Longrightarrow n\,(4k-1)=12r.$$

To find the least positive $$n$$ we can choose integers $$k,r$$ to make $$n$$ as small as possible.

• If $$k=0$$, the factor $$4k-1=-1$$, giving $$n=12(-r),$$ whose least positive value is $$12.$$
• If $$k=1$$, the factor $$4k-1=3$$, giving $$3n=12r\Longrightarrow n=4r.$$ Taking $$r=1$$ yields $$n=4,$$ which is smaller than 12.
• If $$k\ge 2,$$ the factor $$4k-1\ge 7,$$ which makes $$n$$ even larger.

Thus the least positive value satisfying the second condition is $$n_{\min}=4.$$

Finally we need the greatest common divisor of these least values:

$$\gcd(m_{\min},n_{\min})=\gcd(8,4)=4.$$

So, the answer is $$4.$$