Solution
let $\mathrm{z}=\mathrm{x}+\mathrm{iy}, 4 \mathrm{z}^{2}+\overline{\mathrm{z}}=0 \Rightarrow 4(\mathrm{x}+\mathrm{iy})^{2}+(\mathrm{x}-\mathrm{iy})=0$
\[
\begin{aligned}
& \Rightarrow 4\left(x^{2}-y^{2}+2 x y i\right)+x-i y=0 \Rightarrow 4 x^{2}-4 y^{2}+x+i y(8 x-1)=0+i 0 \Rightarrow 4 x^{2}-4 y^{2}+x=0 \text { and } y(8 x-1)=0 \\
& \text { Case I: } y=0 ; 4 x^{2}-0+x=0 \Rightarrow x=0,-\frac{1}{4} \Rightarrow \Rightarrow Z_{1}=0+i 0, Z_{2}=-\frac{1}{4}+i 0 \Rightarrow\left|z_{1}\right|^{2}=0 ;\left|z_{2}\right|^{2}=\frac{1}{16} \\
& \text { Case II: } 8 x-1=0 \Rightarrow x=\frac{1}{8} \Rightarrow 4 x^{2}-4 y^{2}+x=0 \Rightarrow \frac{4}{64}-y^{2}+\frac{1}{8}=0 \Rightarrow 4 y^{2}=\frac{3}{16} \Rightarrow y= \pm \frac{\sqrt{3}}{8} \\
& z_{3}=\frac{1}{8}+\frac{\sqrt{3}}{8} i, z_{4}=\frac{1}{8}-\frac{\sqrt{3}}{8} i \Rightarrow\left|z_{3}\right|^{2}=\frac{1}{64}+\frac{3}{64}=\frac{1}{16} ;\left|z_{4}\right|^{2}=\frac{1}{64}+\frac{3}{64}=\frac{1}{16} \therefore \sum_{i=1}^{n}\left|z_{i}\right|^{2}=0+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{3}{16}
\end{aligned}
\]
.
