Question

If \( \omega \) is a complex cube root of unity and \( x = \omega^2 - \omega + 2 \) then

• A. \( x^2 - 4x + 7 = 0 \)
• B. \( x^2 + 4x + 7 = 0 \)
• C. \( x^2 - 2x + 4 = 0 \)
• D. \( x^2 + 2x + 4 = 0 \)

Hint:

If \( x = a + ib \), then \( (x-a)^2 = (ib)^2 = -b^2 \). This rearranges to \( x^2 - 2ax + a^2 + b^2 = 0 \), which is a quick way to find the quadratic equation.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:

We use the property \( 1 + \omega + \omega^2 = 0 \) to simplify \( x \), find its value in terms of standard complex numbers, and then determine the quadratic equation it satisfies.
Step 2: Key Formula or Approach:

1. \( \omega^2 = -1 - \omega \) 2. \( \omega = \frac{-1 + i\sqrt{3}}{2} \)
Step 3: Detailed Explanation:

Given \( x = \omega^2 - \omega + 2 \). Substitute \( \omega^2 = -1 - \omega \): \[ x = (-1 - \omega) - \omega + 2 = 1 - 2\omega \] Substitute \( \omega = \frac{-1 + i\sqrt{3}}{2} \): \[ x = 1 - 2\left(\frac{-1 + i\sqrt{3}}{2}\right) = 1 - (-1 + i\sqrt{3}) = 1 + 1 - i\sqrt{3} = 2 - i\sqrt{3} \] So, \( x = 2 - i\sqrt{3} \). The conjugate root must be \( 2 + i\sqrt{3} \) (since coefficients of the required quadratic are real). Sum of roots \( S = (2 - i\sqrt{3}) + (2 + i\sqrt{3}) = 4 \). Product of roots \( P = (2 - i\sqrt{3})(2 + i\sqrt{3}) = 4 - 3i^2 = 4 + 3 = 7 \). The quadratic equation is \( X^2 - SX + P = 0 \): \[ x^2 - 4x + 7 = 0 \]
Step 4: Final Answer:

The equation is \( x^2 - 4x + 7 = 0 \).