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Let $z_{1}$ and $z_{2}$ be two imaginary roots of $z^{2}+p z+q=0$, where $p$ and $q$ are real. The points $z_{1}, z_{2}$ and origin form an equilateral triangle if
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Verified Answer
The correct answer is:
$p^{2}=3 q$
Hint:
$\begin{aligned} & \mathrm{O}^{2}+\mathrm{z}_{1}^{2}+\mathrm{z}_{2}^{2}=\mathrm{z}_{1} \mathrm{z}_{2} \\ & \Rightarrow \mathrm{z}_{1}^{2}+\mathrm{z}_{2}^{2}=\mathrm{z}_{1} \mathrm{z}_{2} \\ & \Rightarrow\left(\mathrm{z}_{1}+\mathrm{z}_{2}\right)^{2}=3 \mathrm{z}_{1} \mathrm{z}_{2} \\ & \Rightarrow \mathrm{p}^{2}=3 \mathrm{q} \end{aligned}$
$\begin{aligned} & \mathrm{O}^{2}+\mathrm{z}_{1}^{2}+\mathrm{z}_{2}^{2}=\mathrm{z}_{1} \mathrm{z}_{2} \\ & \Rightarrow \mathrm{z}_{1}^{2}+\mathrm{z}_{2}^{2}=\mathrm{z}_{1} \mathrm{z}_{2} \\ & \Rightarrow\left(\mathrm{z}_{1}+\mathrm{z}_{2}\right)^{2}=3 \mathrm{z}_{1} \mathrm{z}_{2} \\ & \Rightarrow \mathrm{p}^{2}=3 \mathrm{q} \end{aligned}$
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