Complex Numbers Part 2

Concept:The problem involves solving a complex equation by separating real and imaginary parts. The sum of squares of moduli is then computed.Explanation:Let $z = x + iy$, then $\overline{z} = x - iy$.Substitute into $4z^2 + \overline{z} = 0$:$4(x+iy)^2 + (x-iy) = 0$Expand: $4(x^2 + 2ixy - y^2) + x - iy = 0$Group real and imaginary parts:$(4x^2 - 4y^2 + x) + i(8xy - y) = 0$Real part zero: $4x^2 - 4y^2 + x = 0$Imaginary part zero: $y(8x - 1) = 0$Case 1: $y = 0$Then $4x^2 + x = 0 \Rightarrow x(4x+1)=0$ So $x = 0$ or $x = -\frac{1}{4}$.$z_1 = 0$, $|z_1|^2 = 0$$z_2 = -\frac{1}{4}$, $|z_2|^2 = \frac{1}{16}$Case 2: $x = \frac{1}{8}$Substitute into real part: $4\left(\frac{1}{64}\right) - 4y^2 + \frac{1}{8} = 0$Simplify: $\frac{1}{16} - 4y^2 + \frac{1}{8} = 0 \Rightarrow \frac{3}{16} - 4y^2 = 0$Thus $y^2 = \frac{3}{64}$, so $y = \pm \frac{\sqrt{3}}{8}$.$z_3 = \frac{1}{8} + i\frac{\sqrt{3}}{8}$, $|z_3|^2 = \left(\frac{1}{8}\right)^2 + \left(\frac{\sqrt{3}}{8}\right)^2 = \frac{1}{64} + \frac{3}{64} = \frac{1}{16}$$z_4 = \frac{1}{8} - i\frac{\sqrt{3}}{8}$, $|z_4|^2 = \frac{1}{16}$Sum of squares: $0 + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} = \frac{3}{16}$Answer:$\frac{3}{16}$ (Option D)