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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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