Let $\alpha$ and $\beta$ be the sum and the product of all the non-zero solutions of the equation…
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 $4\left(\alpha^2+\beta^2\right)$ is equal to :
6
8
2
4
Solution
$\begin{array}{ll}z=x+i y & \\ \bar{z}=x-i y & \\ \bar{z}^2=x^2-y^2-2 i x y & \\ \Rightarrow x^2-y^2-2 i x y+\sqrt{x^2+y^2}=0 \\ \Rightarrow x=0 \quad \text { or } & y=0 \\ -y^2+|y|=0 & x^2+|x|=0 \\ |y|=|y|^2 & \Rightarrow x=0 \\ y=0, \pm 1 & \Rightarrow \alpha=i-i=0 \\ \Rightarrow i,-i & \beta=i(-i)=1 \\ \text { are roots } & 4(0+1)=4\end{array}$