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Let for some real numbers $\mathrm{\alpha\: \&\: \beta }$ , $\mathrm{a=\alpha -i\beta }$ . If the system of equtions $\mathrm{4 i x+(1+i) y=0 \text { and } 8\left(\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right) x+\bar{a} y=0 }$ has more tha one solution, then $\mathrm{\frac{\alpha }{\beta }}$ is equal to
Option: 1
$-2+\sqrt{3}$

Option: 2
$2-\sqrt{3}$

Option: 3
$2+\sqrt{3}$

Option: 4
$-2-\sqrt{3}$

Answers (1)

best_answer

This is a homogeneous system of linear equations

$\mathrm{(4 i) x+(1+i) y=0 }\\$

$\mathrm{\left(8 e^{i2\pi / 3 }\right) x+(\bar{a}) y=0}$

If it has more than one solution

$\mathrm{\Rightarrow \quad \Delta=0 \Rightarrow\left|\begin{array}{cc} 4 i & 1+i \\ 8 e^{i 2 \pi / 3} & \bar{a} \end{array}\right|=0} \\$

$\mathrm{\Rightarrow \quad 4 i(\alpha+i \beta)-8(1+i)\left(-\frac{1}{2}+i \frac{\sqrt{3}}{2}\right)=0} \\$

$\mathrm{\Rightarrow \quad i \alpha-\beta-[-1+\sqrt{3} i-i-\sqrt{3}]=0}$

$\mathrm{\Rightarrow \quad-\beta+\sqrt{3}+1+i(\alpha-\sqrt{3}+1)=0} \\$

$\mathrm{\Rightarrow \quad \beta=\sqrt{3}+1 \text { and } \alpha=\sqrt{3}-1} \\$

$\mathrm{\frac{\alpha}{\beta}=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{(\sqrt{3}-1)^{2}}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{3+1-2 \sqrt{3}}{3-1}=2-\sqrt{3}}$

Hence the correct answer is option 2.

Posted by

Deependra Verma

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