Question

If \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( x^2 - 2x + 2 = 0 \), then \( \alpha^{2026} + \beta^{2026} = \)

• A. \( -2^{2026} \)
• B. \( 2^{1014} \)
• C. \( -2^{1014} \)
• D. \( 0 \)

Hint:

For roots like \( 1 \pm i \), always remember the argument is \( \pm \pi/4 \). If the power \( n \) is such that \( n\theta \) is an odd multiple of \( \pi/2 \), the sum of the powers of the roots will always be zero.

Answer 1

The Correct Option is D

Concept: The roots of a quadratic equation can be expressed in polar form to simplify high-power calculations using De Moivre's Theorem.
• Roots of \( x^2 - 2x + 2 = 0 \) are found using the quadratic formula.
• Polar form: \( z = r(\cos \theta + i \sin \theta) = r e^{i\theta} \).
• \( z^n + \overline{z}^n = 2r^n \cos(n\theta) \).

Step 1: Finding the roots \( \alpha \) and \( \beta \).
Using the quadratic formula for \( x^2 - 2x + 2 = 0 \): \[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(2)}}{2(1)} = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm 2i}{2} = 1 \pm i \] So, \( \alpha = 1 + i \) and \( \beta = 1 - i \).

Step 2: Converting to polar form.
For \( \alpha = 1 + i \): \( r = \sqrt{1^2 + 1^2} = \sqrt{2} \) and \( \theta = \tan^{-1}(1/1) = \frac{\pi}{4} \). \[ \alpha = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right), \quad \beta = \sqrt{2} \left( \cos \frac{\pi}{4} - i \sin \frac{\pi}{4} \right) \]

Step 3: Calculating \( \alpha^{2026} + \beta^{2026} \).
Using the property \( z^n + \overline{z}^n = 2r^n \cos(n\theta) \): \[ \alpha^{2026} + \beta^{2026} = 2(\sqrt{2})^{2026} \cos\left( 2026 \times \frac{\pi}{4} \right) \] \[ = 2 \cdot 2^{1013} \cos\left( \frac{1013\pi}{2} \right) = 2^{1014} \cos\left( 506\pi + \frac{\pi}{2} \right) \] Since \( \cos(\text{odd multiple of } \pi/2) = 0 \), the expression equals \( 2^{1014} \times 0 = 0 \).