Question

$4z^{2} + \bar{z} = 0$

let $z = x + iy$

$${4(x + iy)^{2} + x - iy = 0 }{4x^{2} - 4y^{2} + 8xyi + x - iy = 0 }{4x^{2} - 4y^{2} + x = 0\& y(8x - 1) = 0 }$$$\Rightarrow y = 0$ or $x = \frac{1}{8}$

If $y = 0,4x^{2} + x = 0$

$${x = 0,\frac{- 1}{4} }{\therefore z_{1} = 0 + 0.i\ \left| z_{1} \right|^{2} = 0 }{z_{2} = 0 - \frac{1}{4}i\ \left| z_{2} \right|^{2} = \frac{1}{16} }$$If $x = \frac{1}{8}$,

$$\begin{matrix} & 4 \times \frac{1}{64} - 4y^{2} + \frac{1}{8} = 0 \\ & \ \Rightarrow 4y^{2} = \frac{3}{16} \Rightarrow y = \pm \frac{\sqrt{3}}{8} \\ & \ \therefore z_{3} = \frac{1}{8} + \frac{\sqrt{3}}{8}i\ \left| z_{3} \right|^{2} = \frac{1}{64} + \frac{3}{64} = \frac{1}{16} \\ & z_{4} = \frac{1}{8} - \frac{\sqrt{3}}{8}i\ \left| z_{4} \right|^{2} = \frac{1}{64} + \frac{3}{64} = \frac{1}{16} \\ & \ \therefore\sum_{i = 1}^{n}\mspace{2mu}\mspace{2mu}\left| z_{i} \right|^{2} = 0 + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} = \frac{3}{16} \end{matrix}$$