Let $S=\{z \in \mathbb{C}: z^2+4z+16=0\}$. Then $\sum_{z \in S}|z+\sqrt{3}i|^2$ is equal to:

Given the equation $z^2 + 4z + 16 = 0$, we find its roots using the quadratic formula:

$z = \dfrac{-4 \pm \sqrt{16 - 64}}{2} = \dfrac{-4 \pm \sqrt{-48}}{2} = -2 \pm 2\sqrt{3}i$

The set $S$ contains two complex numbers: $z_1 = -2 + 2\sqrt{3}i$ and $z_2 = -2 - 2\sqrt{3}i$.

We need to evaluate the sum $\sum_{z \in S} |z + \sqrt{3}i|^2$, which is $|z_1 + \sqrt{3}i|^2 + |z_2 + \sqrt{3}i|^2$.

For $z_1 = -2 + 2\sqrt{3}i$:

$|z_1 + \sqrt{3}i|^2 = |-2 + 2\sqrt{3}i + \sqrt{3}i|^2 = |-2 + 3\sqrt{3}i|^2$

$|-2 + 3\sqrt{3}i|^2 = (-2)^2 + (3\sqrt{3})^2 = 4 + 27 = 31$

For $z_2 = -2 - 2\sqrt{3}i$:

$|z_2 + \sqrt{3}i|^2 = |-2 - 2\sqrt{3}i + \sqrt{3}i|^2 = |-2 - \sqrt{3}i|^2$

$|-2 - \sqrt{3}i|^2 = (-2)^2 + (-\sqrt{3})^2 = 4 + 3 = 7$

Adding these values together:

$\sum_{z \in S} |z + \sqrt{3}i|^2 = 31 + 7 = 38$

Answer: $38$