Let for some real numbers $\alpha$ and $\beta$, $a=\alpha-i \beta$. If the system of equations $4 i x+(1+i) y=0$ 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 :

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Determinants

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 + \overset{a}{ˉ} y = 0

has more than one solution then

\frac{α}{β}

is equal to :

Answer: (B)

For the system to have more than one solution, the determinant must be zero:

4 i 8 e^{i 2 π /3} 1 + i \overset{a}{ˉ} = 0

4 i \overset{a}{ˉ} - (1 + i) 8 e^{i 2 π /3} = 0

Since

e^{i 2 π /3} = cos \frac{2 π}{3} + i sin \frac{2 π}{3} = \frac{- 1 + i 3}{2}

and

\overset{a}{ˉ} = α + i β

4 i (α + i β) - 8 (1 + i) (\frac{- 1 + i 3}{2}) = 0

i α - β + 1 + 3 + i (1 - 3) = 0

Equating real and imaginary parts:

β = 3 + 1

and

α = 3 - 1

.

Therefore,

\frac{α}{β} = \frac{3 - 1}{3 + 1} = \frac{( 3 - 1 ) ^{2}}{2} = \frac{4 - 2 3}{2} = 2 - 3

.

Answer: (B)

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Single correct option