QJEEM24Mad4ff: JEE Main 2024 Mathematics - Class 11 JEEM24Mad4ff Solution: Let and be the sum and the product of all the non-zero…

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

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