Let for some real numbers α and β,a=α−iβ. If the system of equations 4ix+(1+i)y=0 and 8(cos⁡2π3+isin⁡2π3)x+a¯y=0 has more than one solution then αβ is equal to :

a=α-iβ;α∈R;β∈ℝ,4ix+1+iy=0 8(cos2π3+isin2π3)x+a¯y=04i1+i8ei2π3a=0⇒4ia¯−(1+i)8ei2π/3=0⇒4i(α+iβ)−8(1+i)−1+i32=0⇒iα−β+1+3+i(1−3)=0⇒β=3+1,α=3−1∴αβ=3−13+1=2−3

Let for some real numbers α and β,a=α−iβ. If the system of equations 4ix+(1+i)y=0 and 8(cos⁡2π3+isin⁡2π3)x+a¯y=0 has more than one solution then αβ is equal to :

a=α-iβ;α∈R;β∈ℝ,4ix+1+iy=0 8(cos2π3+isin2π3)x+a¯y=04i1+i8ei2π3a=0⇒4ia¯−(1+i)8ei2π/3=0⇒4i(α+iβ)−8(1+i)−1+i32=0⇒iα−β+1+3+i(1−3)=0⇒β=3+1,α=3−1∴αβ=3−13+1=2−3

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Let for some real numbers $α$ and $β, a = α - i β$ . If the system of equations $4 i x + (1 + i) y = 0$ and $8 (\cos \frac{2 π}{3} + i \sin \frac{2 π}{3}) x + \bar{a} y = 0$ has more than one solution then $\frac{α}{β}$ is equal to :

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-2+ $\sqrt{3}$

2- $\sqrt{3}$

2+ $\sqrt{3}$

-2- $\sqrt{3}$

answer is B.

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Detailed Solution

$\begin{array}{l} a = α - i β; α \in R; β \in ℝ, 4 i x + (1 + i) y = 0 \end{array}$ $8 (cos \frac{2 π}{3} + i sin \frac{2 π}{3}) x + \bar{a} y = 0$

$|\begin{matrix} 4 i & 1 + i \\ 8 e^{\frac{i 2 π}{3}} & a \end{matrix}| = 0$

$\begin{array}{l} \Rightarrow 4 i \bar{a} - (1 + i) 8 e^{i 2 π / 3} = 0 \\ \Rightarrow 4 i (α + i β) - 8 (1 + i) (\frac{- 1 + i \sqrt{3}}{2}) = 0 \\ \Rightarrow i α - β + 1 + \sqrt{3} + i (1 - \sqrt{3}) = 0 \\ \Rightarrow β = \sqrt{3} + 1, α = \sqrt{3} - 1 \\ ∴ \frac{α}{β} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = 2 - \sqrt{3} \end{array}$