Question

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

• A. $-2+\sqrt{3}$
• B. $2-\sqrt{3}$
• C. $2+\sqrt{3}$
• D. $-2-\sqrt{3}$

Answer 1

Answer: (B)

$a =\alpha- i \beta ; \alpha \in R ; \beta \in R$
$4 ix +(1+ i ) y =0$ and
$8\left(\cos \frac{2 \pi}{3}+ i \sin \frac{2 \pi}{3}\right) x +\overline{ a y}=0$
$\left|\begin{array}{cc}4 i & 1+ i \\ 8 e ^{ i 2 \pi / 3} \overline{ a }\end{array}\right|=0$
$\Rightarrow 4 i \overline{ a }-(1+ i ) 8 e ^{ i 2 \pi / 3}=0$
$\Rightarrow 4 i (\alpha+ i \beta)-8(1+ i )\left(\frac{-1+ i \sqrt{3}}{2}\right)=0$
$\Rightarrow i \alpha-\beta+1+\sqrt{3}+ i (1-\sqrt{3})=0$
$\Rightarrow \beta=\sqrt{3}+1$
$\alpha=\sqrt{3}-1$
So, $\frac{\alpha}{\beta}=\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3}$