Calculate Sum of Eighth Powers of Roots for Complex Quadratic

To find the value of $\sum_{z \in S} z^8$, where $S = \{z \in \mathbb{C} : z^2 + \sqrt{6} iz - 3 = 0\}$, we first solve the given quadratic equation for $z$.

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

Using the method of completing the square, we rewrite the equation as: $\left(z + \frac{\sqrt{6}i}{2}\right)^2 - \left(\frac{\sqrt{6}i}{2}\right)^2 - 3 = 0$

Since $\left(\frac{\sqrt{6}i}{2}\right)^2 = \frac{6i^2}{4} = -\frac{3}{2}$, the equation becomes: $\left(z + \frac{\sqrt{6}i}{2}\right)^2 - \left(-\frac{3}{2}\right) - 3 = 0$ $\left(z + \frac{\sqrt{6}i}{2}\right)^2 + \frac{3}{2} - 3 = 0$ $\left(z + \frac{\sqrt{6}i}{2}\right)^2 = \frac{3}{2}$

Taking the square root on both sides: $z + \frac{\sqrt{6}i}{2} = \pm \sqrt{\frac{3}{2}}$

Thus, the two roots $z_1$ and $z_2$ are: $z_1 = \sqrt{\frac{3}{2}} - \frac{\sqrt{6}}{2}i = \sqrt{\frac{3}{2}}(1 - i)$ $z_2 = -\sqrt{\frac{3}{2}} - \frac{\sqrt{6}}{2}i = \sqrt{\frac{3}{2}}(-1 - i)$

Now, we compute $z_1^8$ and $z_2^8$:

For $z_1$: $z_1^8 = \left(\sqrt{\frac{3}{2}}\right)^8 (1 - i)^8$

Note that: $(1 - i)^2 = 1 + i^2 - 2i = -2i$ $(1 - i)^4 = (-2i)^2 = -4$ $(1 - i)^8 = (-4)^2 = 16$

Thus: $z_1^8 = \left(\frac{3}{2}\right)^4 \cdot 16 = \frac{81}{16} \cdot 16 = 81$

For $z_2$: $z_2^8 = \left(\sqrt{\frac{3}{2}}\right)^8 (-1 - i)^8$

Note that: $(-1 - i)^2 = 1 + i^2 + 2i = 2i$ $(-1 - i)^4 = (2i)^2 = -4$ $(-1 - i)^8 = (-4)^2 = 16$

Thus: $z_2^8 = \left(\frac{3}{2}\right)^4 \cdot 16 = \frac{81}{16} \cdot 16 = 81$

Finally, we find the sum of the eighth powers of all roots in $S$: $\sum_{z \in S} z^8 = z_1^8 + z_2^8 = 81 + 81 = 162$