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(B) Given $a = \alpha - i \beta$,where $\alpha, \beta \in \mathbb{R}$.The system of equations is:$4ix + (1 + i)y = 0$$8(\cos \frac{2\pi}{3} + i \sin \

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$2 - \sqrt{3}$

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(B) Given $a = \alpha - i \beta$,where $\alpha, \beta \in \mathbb{R}$.The system of equations is:$4ix + (1 + i)y = 0$$8(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3})x + \bar{a}y = 0$Since the system has more than one solution,the determinant of the coefficient matrix must be zero:$\begin{vmatrix} 4i & 1 + i \ 8e^{i2\pi/3} & \bar{a} \end{vmatrix} = 0$$4i\bar{a} - (1 + i)8e^{i2\pi/3} = 0$$4i(\alpha + i\beta) - 8(1 + i)(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}) = 0$$4i\alpha - 4\beta - 8(1 + i)(-\frac{1}{2} + i\frac{\sqrt{3}}{2}) = 0$$4i\alpha - 4\beta - 4(1 + i)(-1 + i\sqrt{3}) = 0$$4i\alpha - 4\beta - 4(-1 + i\sqrt{3} - i - \sqrt{3}) = 0$$i\alpha - \beta - (-1 - \sqrt{3} + i(\sqrt{3} - 1)) = 0$$i\alpha - \beta + 1 + \sqrt{3} - i(\sqrt{3} - 1) = 0$Equating real and imaginary parts to zero:Real part: $-\beta + 1 + \sqrt{3} = 0 \Rightarrow \beta = \sqrt{3} + 1$Imaginary part: $\alpha - (\sqrt{3} - 1) = 0 \Rightarrow \alpha = \sqrt{3} - 1$Therefore,$\frac{\alpha}{\beta} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} - 1)^2}{3 - 1} = \frac{3 + 1 - 2\sqrt{3}}{2} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$.

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

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