Let $\omega$ be the complex number $\cos \frac{2 \pi}{3}+\mathrm{i} \sin \frac{2 \pi}{3}$. Then the number of distinct complex numbers $z$ satisfying $\left|\begin{array}{ccc}z+1 & \omega & \omega^{2} \\ \omega & z+\omega^{2} & 1 \\ \omega^{2} & 1 & z+\omega\end{array}\right|=0$ is equal to

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Complex Numbers

Let

ω

be the complex number

cos \frac{2 π}{3} + i sin \frac{2 π}{3}

. Then the number of distinct complex numbers

z

satisfying

z + 1 ω ω^{2} ω z + ω^{2} 1 ω^{2} 1 z + ω = 0

is equal to

Answer: 1

Given

ω = e^{i 2 π /3}

, we need to solve the determinant equation. Expanding the determinant:

z + 1 ω ω^{2} ω z + ω^{2} 1 ω^{2} 1 z + ω = 0

Factoring out

z

from the first column (after row operations):

z 111 ω z + ω^{2} 1 ω^{2} 1 z + ω = 0

Expanding and using the properties of cube roots of unity (

1 + ω + ω^{2} = 0

and

ω^{3} = 1

), the determinant simplifies to:

z^{3} = 0

Therefore,

z = 0

is the only solution, giving us 1 distinct complex number.

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Integer answer