Let $\omega$ be the complex number $\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}$. Then, the number of distinct complex numbers $z$ satisfying $$\begin{vmatrix} z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega \end{vmatrix} = 0$$ is equal to ____
Step-by-Step Solution
Key Concept: Recognize that ω is a primitive cube root of unity (1 + ω + ω² = 0 and ω³ = 1), and use row operations to factor the determinant as a product involving (z - ω² - ω) and (z + ω - ω²), which both simplify using the relation 1 + ω + ω² = 0 to yield z² = 0.
With $\omega = \cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}$, row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$ are applied to the given determinant. This simplifies to $(z - \omega^2 - \omega)(z + \omega - \omega^2) - (1-\omega)(1-\omega^2) = 0$, which reduces to $z^2 = 0$, yielding only $z = 0$ as the solution.
Correct Answer: 1