Let for some real numbers α and β , a = α - i β . If the system of equations 4 i x + 1 +…

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

So, $\frac{4i}{8\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)}=\frac{1+i}{\alpha +i\beta }$    $\left(\because \begin{array}{rl}a& =\alpha -i\beta \\ \Rightarrow \overline{a}& =\alpha +i\beta \end{array}\right)$

$\Rightarrow \frac{4i}{8\left(\frac{-1}{2}+i\frac{\sqrt{3}}{2}\right)}=\frac{1+i}{\alpha +i\beta }$

$\Rightarrow \alpha i-\beta =-1-i+\sqrt{3}i-\sqrt{3}$

$\Rightarrow -\beta +\alpha i=\left(-1-\sqrt{3}\right)+\left(-1+\sqrt{3}\right)i$

$\Rightarrow \alpha =-1+\sqrt{3}$ & $-\beta =-1-\sqrt{3}$

$\therefore \frac{\alpha }{\beta }=\frac{-1+\sqrt{3}}{1+\sqrt{3}}$$=\frac{-1+\sqrt{3}}{1+\sqrt{3}}\times \frac{1-\sqrt{3}}{1-\sqrt{3}}$

$=\frac{-{\left(1-\sqrt{3}\right)}^2}{1-3}=\frac{1+3-2\sqrt{3}}{2}=2-\sqrt{3}$