Let $$z = a + ib$$, $$b \neq 0$$ be complex numbers satisfying $$z^2 = \bar{z} \cdot 2^{1-|z|}$$. Then the least value of $$n \in \mathbb{N}$$, such that $$z^n = (z+1)^n$$, is equal to _____

We need complex numbers $$z = a + ib$$ with $$b \neq 0$$ satisfying $$z^2 = \bar{z} \cdot 2^{1-|z|}$$, and then the least $$n \in \mathbb{N}$$ such that $$z^n = (z+1)^n$$.

Writing $$z = re^{i\theta}$$ where $$r = |z|$$ and $$\theta \neq 0, \pi$$ (since $$b \neq 0$$), and $$\bar{z} = re^{-i\theta}$$, the equation becomes:

$$r^2 e^{2i\theta} = r e^{-i\theta} \cdot 2^{1-r}$$

Comparing magnitudes: $$r^2 = r \cdot 2^{1-r}$$, so $$r = 2^{1-r}$$ (dividing by $$r > 0$$). Testing $$r = 1$$: $$1 = 2^0 = 1$$. This works, and since $$f(r) = r$$ is increasing while $$g(r) = 2^{1-r}$$ is decreasing, $$r = 1$$ is the unique solution.

Comparing arguments: $$2\theta \equiv -\theta \pmod{2\pi}$$, so $$3\theta = 2k\pi$$ for some integer $$k$$, giving $$\theta = \frac{2k\pi}{3}$$.

Since $$b \neq 0$$, we need $$\theta \neq 0, \pi$$. The values $$\theta = \frac{2\pi}{3}$$ (for $$k = 1$$) and $$\theta = \frac{4\pi}{3}$$ (for $$k = 2$$) both satisfy this. These give $$z = e^{2\pi i/3}$$ and its conjugate $$z = e^{-2\pi i/3}$$.

Taking $$z = e^{2\pi i/3}$$, we compute $$z + 1$$:

$$z + 1 = 1 + \cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3} = 1 - \frac{1}{2} + i\frac{\sqrt{3}}{2} = \frac{1}{2} + i\frac{\sqrt{3}}{2} = e^{i\pi/3}$$

Now $$z^n = (z+1)^n$$ requires $$\left(\frac{z}{z+1}\right)^n = 1$$. We compute:

$$\frac{z}{z+1} = \frac{e^{2\pi i/3}}{e^{i\pi/3}} = e^{i(2\pi/3 - \pi/3)} = e^{i\pi/3}$$

For $$e^{in\pi/3} = 1$$, we need $$\frac{n\pi}{3} = 2m\pi$$ for some positive integer $$m$$, which gives $$n = 6m$$. The least such positive integer is $$n = 6$$.

We can verify with $$z = e^{-2\pi i/3}$$: then $$z + 1 = e^{-i\pi/3}$$ and $$z/(z+1) = e^{-i\pi/3}$$, so $$e^{-in\pi/3} = 1$$ also requires $$n = 6$$.

Hence, the correct answer is 6.