Let S = {z in mathbb{C} : 4z^2 + overline{z} | vedclass

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(A) Given $4z^2 + \overline{z} = 0$. Let $z = x + iy$.Substituting $z$ into the equation: $4(x + iy)^2 + (x - iy) = 0$.$4(x^2 - y^2 + 2xyi) + x - iy =

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$\frac{3}{16}$

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(A) Given $4z^2 + \overline{z} = 0$. Let $z = x + iy$.Substituting $z$ into the equation: $4(x + iy)^2 + (x - iy) = 0$.$4(x^2 - y^2 + 2xyi) + x - iy = 0$.$4x^2 - 4y^2 + x + i(8xy - y) = 0$.Equating real and imaginary parts to zero:$4x^2 - 4y^2 + x = 0$ and $y(8x - 1) = 0$.Case $1$: $y = 0$. Then $4x^2 + x = 0 \Rightarrow x(4x + 1) = 0$. So $x = 0$ or $x = -1/4$.This gives $z_1 = 0$ $(|z_1|^2 = 0)$ and $z_2 = -1/4$ $(|z_2|^2 = 1/16)$.Case $2$: $x = 1/8$. Then $4(1/8)^2 - 4y^2 + 1/8 = 0$.$4/64 - 4y^2 + 1/8 = 0$ $\Rightarrow 1/16 + 1/8 = 4y^2$ $\Rightarrow 3/16 = 4y^2$ $\Rightarrow y^2 = 3/64$.So $y = \pm \sqrt{3}/8$. This gives $z_3 = 1/8 + i\sqrt{3}/8$ and $z_4 = 1/8 - i\sqrt{3}/8$.$|z_3|^2 = (1/8)^2 + (\sqrt{3}/8)^2 = 1/64 + 3/64 = 4/64 = 1/16$.$|z_4|^2 = (1/8)^2 + (-\sqrt{3}/8)^2 = 1/64 + 3/64 = 4/64 = 1/16$.Summing the squares of the moduli: $0 + 1/16 + 1/16 + 1/16 = 3/16$.

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

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