Let $$S= \left\{z \in \mathbb{C}: 4z^{2}+ \overline{z}=0 \right\}$$. Then $$\sum_{z\in S} |z|^{2}$$ is equal to:

We wish to determine all complex numbers $$z$$ satisfying the equation $$4z^2 + \bar{z} = 0$$ and then compute the sum of their squared moduli, $$\sum_{z\in S}|z|^2\,. $$

First, note that substituting $$z=0$$ into the equation gives $$4(0)^2 + \overline{0}=0\,, $$ so $$z=0$$ is indeed a solution, contributing $$|0|^2=0$$ to the desired sum.

For nonzero solutions, the original equation can be rewritten as $$\bar{z}=-4z^2\,. $$ Taking the modulus of both sides yields $$|\bar{z}|=|{-4z^2}|=4|z|^2\,, $$ and since $$|\bar{z}|=|z|\,, $$ one obtains $$|z|=4|z|^2\,. $$ Dividing by the nonzero value $$|z|$$ gives $$1=4|z|\implies |z|=\tfrac14\,. $$

To find the exact values of these nonzero solutions, multiply the relation $$\bar{z}=-4z^2$$ by $$z$$ to obtain $$z\bar{z}=-4z^3\,. $$ Since $$z\bar{z}=|z|^2=\tfrac1{16}\,, $$ it follows that $$\tfrac1{16}=-4z^3\implies z^3=-\tfrac1{64}\,. $$

Writing the right‐hand side in polar form as $$-\tfrac1{64}=\tfrac1{64}e^{i\pi}\,, $$ the three cube roots are given by $$z_k=\tfrac14\,e^{i(\pi+2k\pi)/3}\quad(k=0,1,2)\,. $$ Explicitly, these are $$z_0=\tfrac14e^{i\pi/3}=\tfrac14\bigl(\cos60^\circ+i\sin60^\circ\bigr),\quad z_1=-\tfrac14,\quad z_2=\tfrac14e^{i5\pi/3}=\tfrac14\bigl(\cos300^\circ+i\sin300^\circ\bigr).$$ Each of these satisfies $$|z_k|=\tfrac14\,, $$ so $$|z_k|^2=\tfrac1{16}\,. $$

Altogether, the set of solutions is $$\{0,z_0,z_1,z_2\}$$ and the sum of their squared moduli is $$\sum_{z\in S}|z|^2 =|0|^2+|z_0|^2+|z_1|^2+|z_2|^2 =0+\tfrac1{16}+\tfrac1{16}+\tfrac1{16} =\tfrac3{16}\,. $$

Therefore, the final answer is $$\frac{3}{16}\,. $$