Let w be the complex number cos \frac{2 \pi}{3} + isin \frac{2 \pi}{3} . Then the number of distinct complex number z satisfying \left|\right. \textrm{ } z + 1 & \omega & \left(\omega\right)^{2} \\ \omega & z + \left(\omega\right)^{2} & 1 \\ \left(\omega\right)^{2} & 1 & z + \omega \textrm{ } \left|\right. = 0 is equal to]

Engineering

Mathematics

Cube Roots of Unity

Question

Let w be the complex number $\cos \frac{2 π}{3} + isin \frac{2 π}{3}$ . Then the number of distinct complex number z satisfying $| \begin{matrix} z + 1 & ω & ω^{2} \\ ω & z + ω^{2} & 1 \\ ω^{2} & 1 & z + ω \end{matrix} |$ = 0 is equal to]

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We have $| \begin{matrix} z + 1 & w & w^{2} \\ w & z + w^{2} & 1 \\ w^{2} & 1 & z + w \end{matrix} | = 0$

Apply C₁ $\to_{}^{}$ C₁ + C₂ + C₃ and taking out (z + 1 + w + w²) as common from C₁, we get

(z + 1 + w + w²) $| \begin{matrix} 1 & w & w^{2} \\ 1 & z + w^{2} & 1 \\ 1 & 1 & z + w \end{matrix} | = 0$

Apply R₁ $\to_{}^{}$ R₁ + R₂ + R₃, we get

$z | \begin{matrix} 3 & z & z \\ 1 & z + w^{2} & 1 \\ 1 & 1 & z + w \end{matrix} | = 0$

Expanding along R₁, we get

z [3 {(z + w) (z + w²)–1}– z {z + w –1} + z (1–z–w²) ] = 0

$\Rightarrow$ z (z²) = 0

$\Rightarrow$ z³ = 0

$∴$ z = 0

Hence only one value of z will satisfy above equation.