JEE Challenger
More from Determinants
MathMedAdv 2010 • Paper 1

Number of Distinct Complex Solutions of Determinant Equation

Let ω\omega be the complex number cos⁡2π3+isin⁡2π3\cos \frac{2\pi}{3} + i\sin \frac{2\pi}{3}. Then the number of distinct complex numbers z satisfying

∣z+1ωω2ωz+ω21ω21z+ω∣=0\begin{vmatrix} z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega \end{vmatrix} = 0

is equal to

Official Numerical Answer1

Topics & Concepts

DeterminantsDeterminantsComplex NumbersDe Moivre's Theorem

Step-by-Step Solution

To find the number of distinct complex numbers zz satisfying the given determinant equation, we begin by simplifying the determinant.

Given:

Δ=∣z+1ωω2ωz+ω21ω21z+ω∣=0\Delta = \begin{vmatrix} z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega \end{vmatrix} = 0

where ω=cos⁡2π3+isin⁡2π3\omega = \cos \frac{2\pi}{3} + i\sin \frac{2\pi}{3} is a non-real cube root of unity, which satisfies:

1+ω+ω2=0andω3=11 + \omega + \omega^2 = 0 \quad \text{and} \quad \omega^3 = 1

Applying the row operation R1→R1+R2+R3R_1 \to R_1 + R_2 + R_3:

Δ=∣z+(1+ω+ω2)z+(1+ω+ω2)z+(1+ω+ω2)ωz+ω21ω21z+ω∣\Delta = \begin{vmatrix} z + (1+\omega+\omega^2) & z + (1+\omega+\omega^2) & z + (1+\omega+\omega^2) \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega \end{vmatrix}

Since 1+ω+ω2=01 + \omega + \omega^2 = 0, the first row simplifies to:

Δ=∣zzzωz+ω21ω21z+ω∣=z∣111ωz+ω21ω21z+ω∣\Delta = \begin{vmatrix} z & z & z \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega \end{vmatrix} = z \begin{vmatrix} 1 & 1 & 1 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega \end{vmatrix}

Now, applying column operations C2→C2−C1C_2 \to C_2 - C_1 and C3→C3−C1C_3 \to C_3 - C_1:

Δ=z∣100ωz+ω2−ω1−ωω21−ω2z+ω−ω2∣\Delta = z \begin{vmatrix} 1 & 0 & 0 \\ \omega & z+\omega^2-\omega & 1-\omega \\ \omega^2 & 1-\omega^2 & z+\omega-\omega^2 \end{vmatrix}

Expanding along the first row:

Δ=z[(z−(ω−ω2))(z+(ω−ω2))−(1−ω)(1−ω2)]\Delta = z \left[ (z - (\omega - \omega^2))(z + (\omega - \omega^2)) - (1-\omega)(1-\omega^2) \right] Δ=z[z2−(ω−ω2)2−(1−ω−ω2+ω3)]\Delta = z \left[ z^2 - (\omega - \omega^2)^2 - (1 - \omega - \omega^2 + \omega^3) \right]

Using the algebraic properties of ω\omega:

  1. ω−ω2=(−12+i32)−(−12−i32)=i3\omega - \omega^2 = \left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) - \left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = i\sqrt{3}, so:
    (ω−ω2)2=(i3)2=−3(\omega - \omega^2)^2 = (i\sqrt{3})^2 = -3
  2. 1−(ω+ω2)+ω3=1−(−1)+1=31 - (\omega + \omega^2) + \omega^3 = 1 - (-1) + 1 = 3

Substituting these values back into the expression:

Δ=z[z2−(−3)−3]=z(z2+3−3)=z3\Delta = z \left[ z^2 - (-3) - 3 \right] = z(z^2 + 3 - 3) = z^3

Setting the determinant to zero:

z3=0  ⟹  z=0z^3 = 0 \implies z = 0

Thus, the only solution is z=0z = 0, which means there is only 11 distinct complex number satisfying the equation.

PreviousNext