Given $a = \alpha - i\beta$ and $4ix + (1 + i)y = 0$ ...... (i) $8\left( {\cos {{2\pi } \over 3} + i\sin {{2\pi } \over 3}} \right)x + \overline a y = 0$ .... (ii) By (i) ${x \over y} = {{ - (1 + i)} \over {4i}}$ ...... (iii) By (ii) ${x \over y} = {{ - \overline a } \over {8\left( {{{ - 1} \over 2} + {{\sqrt 3 i} \over 2}} \right)}}$ ..... (iv) Now by (iii) and (iv) ${{1 + i} \over {4i}} = {{\overline a } \over {4\left( { - 1 + \sqrt 3 i} \right)}}$ $\Rightarrow \overline a = \left( {\sqrt 3 - 1} \right) + \left( {\sqrt 3 + 1} \right)i$ $\Rightarrow \alpha + i\beta = \left( {\sqrt 3 - 1} \right) + \left( {\sqrt 3 + 1} \right)i$ $\therefore$ ${\alpha \over \beta } = {{\sqrt 3 - 1} \over {\sqrt 3 + 1}} = 2 - \sqrt 3$