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

Concept: If \(z_1, z_2\) are roots of the quadratic equation \[ z^2 + az + b = 0 \] then: \[ z_1 + z_2 = -a, \qquad z_1 z_2 = b \] Higher powers of roots can be simplified using the original equation to reduce powers step-by-step. Step 1: {Find the roots of the equation.} Given equation: \[ z^2 + \sqrt{6}\,iz - 3 = 0 \] Using quadratic formula: \[ z = \frac{-\sqrt{6}i \pm \sqrt{(\sqrt{6}i)^2 - 4(1)(-3)}}{2} \] \[ (\sqrt{6}i)^2 = -6 \] \[ \sqrt{-6 + 12} = \sqrt{6} \] Thus, \[ z = \frac{-\sqrt{6}i \pm \sqrt{6}}{2} \] \[ z = \frac{\sqrt{6}}{2}(\pm1 - i) \] Step 2: {Write the roots in polar form.} \[ z_1 = \frac{\sqrt{6}}{2}(1-i), \qquad z_2 = \frac{\sqrt{6}}{2}(-1-i) \] Magnitude: \[ |z_1| = |z_2| = \sqrt{3} \] Arguments: \[ 1-i = \sqrt{2}\,\text{cis}\left(-\frac{\pi}{4}\right), \qquad -1-i = \sqrt{2}\,\text{cis}\left(-\frac{3\pi}{4}\right) \] Thus \[ z_1 = \sqrt{3}\,\text{cis}\left(-\frac{\pi}{4}\right), \qquad z_2 = \sqrt{3}\,\text{cis}\left(-\frac{3\pi}{4}\right) \] Step 3: {Compute \(z^8\) using De Moivre’s theorem.} \[ z_1^8 = (\sqrt{3})^8 \, \text{cis}\left(-2\pi\right) = 81 \] \[ z_2^8 = (\sqrt{3})^8 \, \text{cis}\left(-6\pi\right) = 81 \] Step 4: {Find the required sum.} \[ \sum_{z \in S} z^8 = z_1^8 + z_2^8 \] \[ = 81 + 81 \] \[ = 162 \]