Let $\alpha$ and $\beta$ be the sum and the product of all the non-zero solutions of the equation $(\bar{z})^2+|z|=0, z \in C$. Then $4(\alpha^2+\beta^2)$ is equal to :

<p>$$\begin{aligned} & (\bar{z})^2+|z|=0 \quad \text{... (1)}\\ & z^2+|\bar{z}|=0 \quad \text{... (2)} \end{aligned}$$</p> <p>From equation (1) and (2)</p> <p>$$\begin{aligned} & \text { as }|z|=|\bar{z}| \\ & \Rightarrow \quad(\bar{z})^2=z^2 \\ & \Rightarrow \quad z=\bar{z} \text { or } z=-\bar{z} \\ & \Rightarrow \operatorname{Im}(z)=0 \text { or } \operatorname{Re}(z)=0 \end{aligned}$$</p> <p>Case I : If $\operatorname{Im}(z)=0$</p> <p>$\Rightarrow z=x$</p> <p>Putting value of $z$ in equation (1)</p> <p>$$\begin{aligned} & x^2+|x|=0 \\ & \Rightarrow x=0 \quad \text{[Rejected] } \end{aligned}$$</p> <p>Case II : If $\operatorname{Re}(z)=0$</p> <p>$\Rightarrow z=i y$</p> <p>Putting value of $z$ in equation (1)</p> <p>$$\begin{aligned} & -y^2+|y|=0 \\ & y= \pm 1 \text { as } y \neq 0 \end{aligned}$$</p> <p>Hence, $z= \pm i$ are the solution of the given equation</p> <p>$$\begin{aligned} & \Rightarrow \alpha=i-i=0 \\ & \text { and } \beta=i(-i)=1 \\ & \Rightarrow \quad 4\left(\alpha^2+\beta^2\right)=4(0+1) \\ & \quad=4 \end{aligned}$$</p> <p>$\therefore$ Option (3) is correct</p>