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

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Answer: B. 2 - \sqrt 3 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}} ight)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}} ight)}} ..... (iv) Now by (iii) and (iv) {{1 + i} \over {4i}} = {{\overline a } \over {4\left( { - 1 + \sqrt 3 i} ight)}} \Rightarrow \overline a = \left( {\sqrt 3 - 1} ight) + \left( {\sqrt 3 + 1} ight)i \Rightarrow \alpha + i\beta = \left( {\sqrt 3 - 1} ight) + \left( {\sqrt 3 + 1} ight)i 	herefore {\alpha \over \beta } = {{\sqrt 3 - 1} \over {\sqrt 3 + 1}} = 2 - \sqrt 3

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

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