Solution
let $\mathrm{z}=\mathrm{x}+\mathrm{iy}, \overline{\mathrm{z}}=\mathrm{x}-\mathrm{iy} \Rightarrow \overline{\mathrm{z}}^{2}=\mathrm{x}^{2}-\mathrm{y}^{2}-2 \mathrm{ixy} \Rightarrow(\overline{\mathrm{z}})^{2}+|\mathrm{z}|=0$
\[
\Rightarrow x^{2}-y^{2}-2 i x y+\sqrt{x^{2}+y^{2}}=0 \Rightarrow x^{2}-y^{2}+\sqrt{x^{2}+y^{2}}-2 x y i=0 \Rightarrow x^{2}-y^{2}+\sqrt{x^{2}+y^{2}}=0 \text { and } x y=0
\]
Case $\mathrm{I}: \mathrm{x}=0 \quad$ Case II : $\mathrm{y}=0$
\[
\begin{array}{ll}
-\mathrm{y}^{2}+|\mathrm{y}|=0 \Rightarrow|\mathrm{y}|=|\mathrm{y}|^{2} & \mathrm{x}^{2}+|\mathrm{x}|=0 \\
\Rightarrow|\mathrm{y}|(|\mathrm{y}|-1)=0 & \Rightarrow \mathrm{x}=0 \\
\Rightarrow \mathrm{y}=0, \pm 1 & \Rightarrow \mathrm{z}=0
\end{array}
\]
$\mathrm{z}=0, \mathrm{i},-\mathrm{i}$ are roots $\alpha=\mathrm{i}-\mathrm{i}=0 ; \beta=\mathrm{i}(-\mathrm{i})=1 \Rightarrow 4\left(\alpha^{2}+\beta^{2}\right)=4(0+1)=4$
.
