Solving the Complex Equation

We need to find the set $S = \{z \in \mathbb{C} : 4z^2 + \bar{z} = 0\}$ and then compute the sum $\sum_{z \in S} |z|^2$. Let $z = x + iy$, where $x, y \in \mathbb{R}$. Then $\bar{z} = x - iy$. Substituting into the equation:

$4(x + iy)^2 + (x - iy) = 0$

Expand the equation:

$4(x^2 + 2ixy - y^2) + x - iy = 0$

$4x^2 + 8ixy - 4y^2 + x - iy = 0$

Group the real and imaginary parts:

$(4x^2 - 4y^2 + x) + i(8xy - y) = 0$

For this complex number to be zero, both the real and imaginary parts must be zero:

• Equation 1 (Real part): $4x^2 - 4y^2 + x = 0$
• Equation 2 (Imaginary part): $8xy - y = 0$

Finding the Roots $z \in S$

First, solve Equation 2:

$y(8x - 1) = 0$

This yields two cases:

• Case A: $y = 0$
• Case B: $8x - 1 = 0 \implies x = \frac{1}{8}$

Case A: $y = 0$

Substitute $y = 0$ into Equation 1:

$4x^2 - 4(0)^2 + x = 0$

$4x^2 + x = 0 \implies x(4x + 1) = 0$

This gives $x = 0$ or $x = -\frac{1}{4}$. The solutions are $z_1 = 0 + i(0) = 0$ and $z_2 = -\frac{1}{4} + i(0) = -\frac{1}{4}$.

Case B: $x = \frac{1}{8}$

Substitute $x = \frac{1}{8}$ into Equation 1:

$4(\frac{1}{8})^2 - 4y^2 + \frac{1}{8} = 0$

$4(\frac{1}{64}) - 4y^2 + \frac{1}{8} = 0$

$\frac{1}{16} - 4y^2 + \frac{2}{16} = 0$

$\frac{3}{16} - 4y^2 = 0 \implies 4y^2 = \frac{3}{16} \implies y^2 = \frac{3}{64}$

This gives $y = \pm \sqrt{\frac{3}{64}} = \pm \frac{\sqrt{3}}{8}$. The solutions are $z_3 = \frac{1}{8} + i\frac{\sqrt{3}}{8}$ and $z_4 = \frac{1}{8} - i\frac{\sqrt{3}}{8}$.

The set of solutions is $S = \{0, -\frac{1}{4}, \frac{1}{8} + i\frac{\sqrt{3}}{8}, \frac{1}{8} - i\frac{\sqrt{3}}{8}\}$.

Calculating the Sum of Magnitudes Squared

Now, calculate $|z|^2$ for each element in $S$:

• For $z_1 = 0$: $|z_1|^2 = 0^2 = 0$.
• For $z_2 = -\frac{1}{4}$: $|z_2|^2 = (-\frac{1}{4})^2 = \frac{1}{16}$.
• For $z_3 = \frac{1}{8} + i\frac{\sqrt{3}}{8}$: $|z_3|^2 = (\frac{1}{8})^2 + (\frac{\sqrt{3}}{8})^2 = \frac{1}{64} + \frac{3}{64} = \frac{4}{64} = \frac{1}{16}$.
• For $z_4 = \frac{1}{8} - i\frac{\sqrt{3}}{8}$: $|z_4|^2 = (\frac{1}{8})^2 + (-\frac{\sqrt{3}}{8})^2 = \frac{1}{64} + \frac{3}{64} = \frac{4}{64} = \frac{1}{16}$.

Final Summation

The sum is:

$\sum_{z \in S} |z|^2 = |z_1|^2 + |z_2|^2 + |z_3|^2 + |z_4|^2 = 0 + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} = \frac{3}{16}$.