Question

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

$a=\alpha -i\beta ;\alpha \in R;\beta \in R$
$4ix+(1+i)y=0$ and
$8\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)x+\overline{ay}=0$
$\left|\begin{array}{cc}4i& 1+i\\ 8e^{i2\pi /3}\overline{a}& \end{array}\right|=0$
$\Rightarrow 4i\overline{a}-(1+i)8e^{i2\pi /3}=0$
$\Rightarrow 4i(\alpha +i\beta )-8(1+i)\left(\frac{-1+i\sqrt{3}}{2}\right)=0$
$\Rightarrow i\alpha -\beta +1+\sqrt{3}+i(1-\sqrt{3})=0$
$\Rightarrow \beta =\sqrt{3}+1$
$\alpha =\sqrt{3}-1$
So, $\frac{\alpha }{\beta }=\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3}$