Let for some real numbers [] and [], [a = - i ].

Question

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

Accepted Answer

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)

$${{1 + i} \over {4i}} = {{\overline a } \over {4\left( { - 1 + \sqrt 3 i} \right)}}$$

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

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

$$\therefore$$ $${\alpha  \over \beta } = {{\sqrt 3  - 1} \over {\sqrt 3  + 1}} = 2 - \sqrt 3 $$

Solutions · 1