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

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

Rearranging the terms, we get $z^2 - 3 = -\sqrt{6}iz$.

Squaring both sides, we obtain:

$(z^2 - 3)^2 = (-\sqrt{6}iz)^2$

$z^4 - 6z^2 + 9 = -6z^2$

$z^4 + 9 = 0$

$z^4 = -9$

Squaring again, we get:

$(z^4)^2 = (-9)^2$

$z^8 = 81$

Since the given equation is a quadratic in $z$, it has two roots, say $z_1$ and $z_2$.

The discriminant of the quadratic equation is $\Delta = (\sqrt{6}i)^2 - 4(1)(-3) = -6 + 12 = 6 \neq 0$, which means the roots are distinct.

Thus, the set $S$ contains two distinct elements $z_1$ and $z_2$.

For each root, $z^8 = 81$.

Therefore, $\sum\limits_{z \in S} z^8 = z_1^8 + z_2^8 = 81 + 81 = 162$.

Answer: $162$