Solving the Complex Number Equation

The problem asks for the sum of the 8th powers of the roots of the complex quadratic equation:

$z^2 + \sqrt{6} iz - 3 = 0$

Let the set of roots be $S = \{z_1, z_2\}$. We need to calculate $\sum_{z \in S} z^8 = z_1^8 + z_2^8$.

Finding the Roots

We apply the quadratic formula $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1$, $b=\sqrt{6}i$, and $c=-3$.

$z = \frac{-\sqrt{6}i \pm \sqrt{(\sqrt{6}i)^2 - 4(1)(-3)}}{2(1)}$

$z = \frac{-\sqrt{6}i \pm \sqrt{6i^2 + 12}}{2}$

$z = \frac{-\sqrt{6}i \pm \sqrt{-6 + 12}}{2}$

$z = \frac{-\sqrt{6}i \pm \sqrt{6}}{2}$

The two roots are:

• $z_1 = \frac{\sqrt{6}}{2} - \frac{\sqrt{6}}{2}i$
• $z_2 = -\frac{\sqrt{6}}{2} - \frac{\sqrt{6}}{2}i$

Calculating Powers of the Roots

We convert the roots to polar form ($re^{i\theta}$) to simplify the power calculation using De Moivre's theorem.

For $z_1 = \frac{\sqrt{6}}{2} - \frac{\sqrt{6}}{2}i$:

• Magnitude: $|z_1| = \sqrt{(\frac{\sqrt{6}}{2})^2 + (-\frac{\sqrt{6}}{2})^2} = \sqrt{\frac{6}{4} + \frac{6}{4}} = \sqrt{3}$
• Argument: $\arg(z_1) = \arctan\left(\frac{-\sqrt{6}/2}{\sqrt{6}/2}\right) = -\frac{\pi}{4}$ (as $z_1$ is in Quadrant IV)
• Polar form: $z_1 = \sqrt{3} e^{-i\pi/4}$

For $z_2 = -\frac{\sqrt{6}}{2} - \frac{\sqrt{6}}{2}i$:

• Magnitude: $|z_2| = \sqrt{(-\frac{\sqrt{6}}{2})^2 + (-\frac{\sqrt{6}}{2})^2} = \sqrt{\frac{6}{4} + \frac{6}{4}} = \sqrt{3}$
• Argument: $\arg(z_2) = \arctan\left(\frac{-\sqrt{6}/2}{-\sqrt{6}/2}\right) = \arctan(1) = -\frac{3\pi}{4}$ (as $z_2$ is in Quadrant III)
• Polar form: $z_2 = \sqrt{3} e^{-i3\pi/4}$

Summing the 8th Powers

Using De Moivre's theorem $(re^{i\theta})^n = r^n e^{in\theta}$:

Calculate $z_1^8$:

$z_1^8 = (\sqrt{3} e^{-i\pi/4})^8 = (\sqrt{3})^8 (e^{-i\pi/4})^8 = 3^4 e^{-i2\pi} = 81 \times 1 = 81$

Calculate $z_2^8$:

$z_2^8 = (\sqrt{3} e^{-i3\pi/4})^8 = (\sqrt{3})^8 (e^{-i3\pi/4})^8 = 3^4 e^{-i6\pi} = 81 \times 1 = 81$

The sum $\sum_{z \in S} z^8$ is:

$z_1^8 + z_2^8 = 81 + 81 = 162$

Thus, the sum is 162.