Let for some real numbers $$\alpha$$ and $$\beta$$, $$a = \alpha - i\beta$$. If the system of equations $$4ix + (1+i)y = 0$$ and $$8\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)x + \bar{a}y = 0$$ has more than one solution then $$\frac{\alpha}{\beta}$$ is equal to

We are given $$a = \alpha - i\beta$$ and the system of equations:

$$4ix + (1+i)y = 0 \quad \cdots(1)$$

$$8\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)x + \bar{a}y = 0 \quad \cdots(2)$$

Since $$\cos\frac{2\pi}{3} = -\frac{1}{2}$$ and $$\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}$$, it follows that $$8\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -4 + 4\sqrt{3}\,i$$. Also, $$\bar{a} = \alpha + i\beta$$.

For the homogeneous system to have more than one solution, the determinant of the coefficient matrix must be zero:

$$\begin{vmatrix} 4i & 1+i \\ -4 + 4\sqrt{3}\,i & \alpha + i\beta \end{vmatrix} = 0$$

Expanding this determinant gives

$$4i(\alpha + i\beta) - (1+i)(-4 + 4\sqrt{3}\,i) = 0$$

In the first term, $$4i\alpha + 4i^2\beta = 4\alpha i - 4\beta$$. In the second term,

$$ (1+i)(-4 + 4\sqrt{3}\,i) = -4 + 4\sqrt{3}\,i - 4i + 4\sqrt{3}\,i^2 = -4 + 4\sqrt{3}\,i - 4i - 4\sqrt{3} = -(4 + 4\sqrt{3}) + (4\sqrt{3} - 4)i $$.

Putting these together leads to

$$(-4\beta + 4\alpha i) - (-(4 + 4\sqrt{3}) + (4\sqrt{3} - 4)i) = 0$$

which simplifies to

$$(-4\beta + 4 + 4\sqrt{3}) + (4\alpha - 4\sqrt{3} + 4)i = 0$$

Equating the real and imaginary parts to zero yields

Real part: $$-4\beta + 4 + 4\sqrt{3} = 0 \quad\Longrightarrow\quad \beta = 1 + \sqrt{3}$$

Imaginary part: $$4\alpha - 4\sqrt{3} + 4 = 0 \quad\Longrightarrow\quad \alpha = \sqrt{3} - 1$$

Finally,

$$\frac{\alpha}{\beta} = \frac{\sqrt{3} - 1}{1 + \sqrt{3}} = \frac{(\sqrt{3} - 1)}{(\sqrt{3} + 1)} \times \frac{(\sqrt{3} - 1)}{(\sqrt{3} - 1)} = \frac{(\sqrt{3} - 1)^2}{3 - 1} = \frac{3 - 2\sqrt{3} + 1}{2} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$$

Therefore, $$\frac{\alpha}{\beta} = 2 - \sqrt{3}$$.

The correct answer is Option A: $$2 - \sqrt{3}$$.