Let $$z$$ and $$w$$ be two complex numbers such that $$w = z\bar{z} - 2z + 2$$, $$\left|\frac{z+i}{z-3i}\right| = 1$$ and $$\text{Re}(w)$$ has minimum value. Then, the minimum value of $$n \in N$$ for which $$w^n$$ is real, is equal to ________.

Given $$\left|\frac{z+i}{z-3i}\right| = 1$$, this means the point $$z$$ is equidistant from $$-i$$ and $$3i$$ in the complex plane. Setting $$z = x + iy$$, we get $$|z + i|^2 = |z - 3i|^2$$, which gives $$x^2 + (y+1)^2 = x^2 + (y-3)^2$$. Expanding: $$y^2 + 2y + 1 = y^2 - 6y + 9$$, so $$8y = 8$$, hence $$y = 1$$. Therefore $$z = x + i$$ for real $$x$$.

Now $$w = z\bar{z} - 2z + 2$$. With $$z = x + i$$, we have $$z\bar{z} = |z|^2 = x^2 + 1$$. So $$w = x^2 + 1 - 2(x + i) + 2 = x^2 - 2x + 3 - 2i$$.

The real part of $$w$$ is $$\text{Re}(w) = x^2 - 2x + 3 = (x-1)^2 + 2$$. This is minimized when $$x = 1$$, giving $$\text{Re}(w) = 2$$.

At $$x = 1$$: $$w = 2 - 2i$$. We can write $$w = 2 - 2i = 2\sqrt{2}\,e^{-i\pi/4}$$.

Then $$w^n = (2\sqrt{2})^n \, e^{-in\pi/4}$$. For $$w^n$$ to be real, we need $$e^{-in\pi/4}$$ to be real, which requires $$\frac{n\pi}{4} = k\pi$$ for some integer $$k$$, i.e., $$n = 4k$$.

The minimum natural number $$n$$ is $$n = 4$$.