Let for some real numbers α and β, a = α - iβ . If the system of...

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

So, $\dfrac{4i}{8\left(cos\dfrac{2\pi}{3} + isin\dfrac{2\pi}{3}\right)} = \dfrac{1 + i}{\alpha + i\beta}$ \left(∵ $\begin{cases}a & = \alpha - i\beta \\ ⇒ a & = \alpha + i\beta\end{cases}$\right)$</p><p>$⇒ \dfrac{4i}{8\left(\dfrac{- 1}{2} + i\dfrac{\sqrt{3}}{2}\right)} = \dfrac{1 + i}{\alpha + i\beta}$</p><p>$⇒ \alphai - \beta = - 1 - i + \sqrt{3}i - \sqrt{3}$</p><p>$⇒ - \beta + \alphai = \left(- 1 - \sqrt{3}\right) + \left(- 1 + \sqrt{3}\right)i$</p><p>$⇒ \alpha = - 1 + \sqrt{3}&- \beta = - 1 - \sqrt{3}$</p><p>$∴ \dfrac{\alpha}{\beta} = \dfrac{- 1 + \sqrt{3}}{1 + \sqrt{3}}$$= \dfrac{- 1 + \sqrt{3}}{1 + \sqrt{3}} \times \dfrac{1 - \sqrt{3}}{1 - \sqrt{3}}$

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