Let and be the sum and the product of all the non-zero solutions of the equation (z)^2+|z|…

Question

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

Accepted Answer

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

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

Solutions · 1