$$\begin{aligned} & (\bar{z})^2+|z|=0 \quad \text{... (1)}\\ & z^2+|\bar{z}|=0 \quad \text{... (2)} \end{aligned}$$

From equation (1) and (2)

$$\begin{aligned} & \text { as }|z|=|\bar{z}| \\ & \Rightarrow \quad(\bar{z})^2=z^2 \\ & \Rightarrow \quad z=\bar{z} \text { or } z=-\bar{z} \\ & \Rightarrow \operatorname{Im}(z)=0 \text { or } \operatorname{Re}(z)=0 \end{aligned}$$

Case I : If $$\operatorname{Im}(z)=0$$

$$\Rightarrow z=x$$

Putting value of $$z$$ in equation (1)

$$\begin{aligned} & x^2+|x|=0 \\ & \Rightarrow x=0 \quad \text{[Rejected] } \end{aligned}$$

Case II : If $$\operatorname{Re}(z)=0$$

$$\Rightarrow z=i y$$

Putting value of $$z$$ in equation (1)

$$\begin{aligned} & -y^2+|y|=0 \\ & y= \pm 1 \text { as } y \neq 0 \end{aligned}$$

Hence, $$z= \pm i$$ are the solution of the given equation

$$\begin{aligned} & \Rightarrow \alpha=i-i=0 \\ & \text { and } \beta=i(-i)=1 \\ & \Rightarrow \quad 4\left(\alpha^2+\beta^2\right)=4(0+1) \\ & \quad=4 \end{aligned}$$

$$\therefore$$ Option (3) is correct