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

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

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

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

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

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

$$z_1^2 = \left(\frac{\sqrt{6}(1 - i)}{2}\right)^2 = \frac{6(1 - i)^2}{4} = \frac{6(1 - 2i + i^2)}{4}$$

$$z_1^2 = \frac{6(-2i)}{4} = -3i$$

$$z_2^2 = \left(\frac{-\sqrt{6}(1 + i)}{2}\right)^2 = \frac{6(1 + i)^2}{4} = \frac{6(1 + 2i + i^2)}{4} = \frac{6(2i)}{4} = 3i$$

$$z_1^8 = (z_1^2)^4 = (-3i)^4 = (-3)^4 \cdot i^4 = 81 \cdot 1 = 81$$

$$z_2^8 = (z_2^2)^4 = (3i)^4 = (3)^4 \cdot i^4 = 81 \cdot 1 = 81$$

$$\sum_{z \in S} z^8 = z_1^8 + z_2^8 = 81 + 81 = 162$$