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

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