Question

If \(\alpha\) and \(\beta\) are the roots of the equation \(z^2 - \sqrt{6}i z - 3 = 0\), then find \(\alpha^8 + \beta^8\).

Hint:

When asked for high powers of complex roots, always convert to polar form. It turns a difficult algebraic expansion into a simple multiplication of the angle.

Answer 1

Correct Answer: 162

Step 1: Understanding the Concept:
We find the roots of the quadratic equation using the quadratic formula. Then, we express the roots in polar form to easily calculate their 8th powers using De Moivre's Theorem.

Step 2: Key Formula or Approach:
Quadratic Formula: \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). De Moivre's Theorem: \((r e^{i\theta})^n = r^n e^{in\theta}\).

Step 3: Detailed Explanation:
1. Find roots: \[ z = \frac{\sqrt{6}i \pm \sqrt{(\sqrt{6}i)^2 - 4(1)(-3)}}{2} = \frac{\sqrt{6}i \pm \sqrt{-6 + 12}}{2} = \frac{\sqrt{6}i \pm \sqrt{6}}{2} \] \[ \alpha = \sqrt{\frac{6}{4}} (1 + i) = \sqrt{\frac{3}{2}}(1 + i), \quad \beta = \sqrt{\frac{3}{2}}(-1 + i) \] 2. Convert to polar form: \[ \alpha = \sqrt{\frac{3}{2}} \cdot \sqrt{2} e^{i\pi/4} = \sqrt{3} e^{i\pi/4} \] \[ \beta = \sqrt{\frac{3}{2}} \cdot \sqrt{2} e^{i3\pi/4} = \sqrt{3} e^{i3\pi/4} \] 3. Calculate 8th powers: \[ \alpha^8 = (\sqrt{3})^8 (e^{i\pi/4})^8 = 3^4 e^{i2\pi} = 81(1) = 81 \] \[ \beta^8 = (\sqrt{3})^8 (e^{i3\pi/4})^8 = 3^4 e^{i6\pi} = 81(1) = 81 \] 4. Sum: \[ \alpha^8 + \beta^8 = 81 + 81 = 162 \]

Step 4: Final Answer:
The value of \(\alpha^8 + \beta^8\) is 162.