Let for some real numbers α and β ,a=α -iβ. If the system of equations 4ix+(1+i)y=0 and 8( cos 2π 3+i sin 2π 3)x+ ay=0 has more than one solution…

The given system of equations has more than one solution, then it must have infinitely many solutions.

So, $\frac{4i}{8(\mathrm{cos}\frac{2\pi }{3}+i\mathrm{sin}\frac{2\pi }{3})}=\frac{1+i}{\alpha +i\beta }$8(cos32π​+isin32π​)4i​=α+iβ1+i​ $(\because \begin{matrix}a & =\alpha -i\beta \\ \Rightarrow a & =\alpha +i\beta \end{matrix})$(∵a⇒a​=α−iβ=α+iβ​)

$$\Rightarrow \frac{4i}{8(\frac{-1}{2}+i\frac{\sqrt{3}}{2})}=\frac{1+i}{\alpha +i\beta }$$⇒8(2−1​+i23​​)4i​=α+iβ1+i​

$$\Rightarrow \alpha i-\beta =-1-i+\sqrt{3}i-\sqrt{3}$$⇒αi−β=−1−i+3​i−3​

$$\Rightarrow -\beta +\alpha i=(-1-\sqrt{3})+(-1+\sqrt{3})i$$⇒−β+αi=(−1−3​)+(−1+3​)i

$\Rightarrow \alpha =-1+\sqrt{3}$⇒α=−1+3​ & $-\beta =-1-\sqrt{3}$−β=−1−3​

$\therefore \frac{\alpha }{\beta }=\frac{-1+\sqrt{3}}{1+\sqrt{3}}$∴βα​=1+3​−1+3​​$=\frac{-1+\sqrt{3}}{1+\sqrt{3}}\times \frac{1-\sqrt{3}}{1-\sqrt{3}}$=1+3​−1+3​​×1−3​1−3​​

$$=\frac{-{(1-\sqrt{3})}^{2}}{1-3}=\frac{1+3-2\sqrt{3}}{2}=2-\sqrt{3}$$=1−3−(1−3​)2​=21+3−23​​=2−3​