Question

Let \( S = \{z \in \mathbb{C} : z^2 + \sqrt{6}\,iz - 3 = 0 \}. \) Then \( \displaystyle \sum_{z \in S} z^8 \) is equal to:

• A. \(162\)
• B. \(184\)
• C. \(262\)
• D. \(324\)

Answer 1

The Correct Option is A

Step 1: Understanding the Question
We are given a quadratic equation in the complex variable \(z\). The set \(S\) contains the roots of this equation. We need to find the sum of the 8th powers of these roots.
Step 2: Key Formula or Approach
1. Solve the quadratic equation using the quadratic formula: \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
2. Convert the complex roots into polar form \( z = r(\cos\theta + i\sin\theta) \) or \( r\,\text{cis}(\theta) \).
3. Use De Moivre's Theorem to compute high powers of the complex numbers: \( [r(\cos\theta + i\sin\theta)]^n = r^n(\cos(n\theta) + i\sin(n\theta)) \).
Step 3: Detailed Explanation
The given equation is \( z^2 + (\sqrt{6}i)z - 3 = 0 \).
Using the quadratic formula with \( a=1, b=\sqrt{6}i, c=-3 \):
\[ z = \frac{-\sqrt{6}i \pm \sqrt{(\sqrt{6}i)^2 - 4(1)(-3)}}{2(1)} \] \[ z = \frac{-\sqrt{6}i \pm \sqrt{-6 + 12}}{2} \] \[ z = \frac{-\sqrt{6}i \pm \sqrt{6}}{2} \] The two roots, let's call them \(z_1\) and \(z_2\), are:
\[ z_1 = \frac{\sqrt{6} - \sqrt{6}i}{2} = \frac{\sqrt{6}}{2}(1 - i) \] \[ z_2 = \frac{-\sqrt{6} - \sqrt{6}i}{2} = \frac{\sqrt{6}}{2}(-1 - i) \] Now, let's convert these roots to polar form.
For \(z_1\): Modulus \( |z_1| = \left|\frac{\sqrt{6}}{2}(1 - i)\right| = \frac{\sqrt{6}}{2}|1 - i| = \frac{\sqrt{6}}{2}\sqrt{1^2 + (-1)^2} = \frac{\sqrt{6}}{2}\sqrt{2} = \sqrt{3} \).
Argument: \( \arg(z_1) = \arg(1-i) = -\frac{\pi}{4} \). So, \( z_1 = \sqrt{3}\,\text{cis}\left(-\frac{\pi}{4}\right) \).
For \(z_2\): Modulus \( |z_2| = \left|\frac{\sqrt{6}}{2}(-1 - i)\right| = \frac{\sqrt{6}}{2}|-1 - i| = \frac{\sqrt{6}}{2}\sqrt{(-1)^2 + (-1)^2} = \frac{\sqrt{6}}{2}\sqrt{2} = \sqrt{3} \).
Argument: \( \arg(z_2) = \arg(-1-i) = -\frac{3\pi}{4} \). So, \( z_2 = \sqrt{3}\,\text{cis}\left(-\frac{3\pi}{4}\right) \).
Now we compute \( z_1^8 \) and \( z_2^8 \) using De Moivre's Theorem.
\[ z_1^8 = \left(\sqrt{3}\,\text{cis}\left(-\frac{\pi}{4}\right)\right)^8 = (\sqrt{3})^8\,\text{cis}\left(8 \times -\frac{\pi}{4}\right) = 3^4\,\text{cis}(-2\pi) \] Since \( \text{cis}(-2\pi) = \cos(-2\pi) + i\sin(-2\pi) = 1 + 0i = 1 \), we have \( z_1^8 = 81 \times 1 = 81 \).
\[ z_2^8 = \left(\sqrt{3}\,\text{cis}\left(-\frac{3\pi}{4}\right)\right)^8 = (\sqrt{3})^8\,\text{cis}\left(8 \times -\frac{3\pi}{4}\right) = 3^4\,\text{cis}(-6\pi) \] Since \( \text{cis}(-6\pi) = \cos(-6\pi) + i\sin(-6\pi) = 1 + 0i = 1 \), we have \( z_2^8 = 81 \times 1 = 81 \).
The required sum is \( \sum_{z \in S} z^8 = z_1^8 + z_2^8 \).
\[ \text{Sum} = 81 + 81 = 162 \] Step 4: Final Answer
The value of the sum is 162.