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Let $a, b \in \mathbf{R}$ and the roots $\alpha, \beta$ of the equation $z^2+a z+b=0$ be complex. If the origin, $\alpha$ and $\beta$ represent the vertices of an equilateral triangle on the Argand plane, then
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The correct answer is:
$a^2=3 b$
We have, $\alpha$ and $\beta$ are roots of equation
$z^2+a z+b=0$
$\therefore \alpha+\beta=-a$ and $\alpha \beta=b$
Now, $0 < \alpha, \beta$ forms a equilateral triangle
$\begin{array}{ll}\therefore & \beta=\alpha\left(\cos 60^{\circ}+i \sin 60\right) \\ \Rightarrow & \beta=\alpha\left(\frac{1}{2}+\frac{i \sqrt{3}}{2}\right) \\ \Rightarrow & 2 \beta=\alpha+\sqrt{3} \alpha i \\ \Rightarrow & 2 \beta-\alpha=\sqrt{3} \alpha i \Rightarrow(2 \beta-\alpha)^2=3 \alpha^2 i^2 \\ \Rightarrow & 4 \beta^2+\alpha^2-4 \alpha \beta=-3 \alpha^2 \\ \Rightarrow & 4 \alpha^2+4 \beta^2=4 \alpha \beta \Rightarrow \alpha^2+\beta^2=\alpha \beta \\ \Rightarrow & (\alpha+\beta)^2-2 \alpha \beta=\alpha \beta \Rightarrow(-a)^2=3(b) \\ \Rightarrow & a^2=3 b\end{array}$
$z^2+a z+b=0$
$\therefore \alpha+\beta=-a$ and $\alpha \beta=b$
Now, $0 < \alpha, \beta$ forms a equilateral triangle
$\begin{array}{ll}\therefore & \beta=\alpha\left(\cos 60^{\circ}+i \sin 60\right) \\ \Rightarrow & \beta=\alpha\left(\frac{1}{2}+\frac{i \sqrt{3}}{2}\right) \\ \Rightarrow & 2 \beta=\alpha+\sqrt{3} \alpha i \\ \Rightarrow & 2 \beta-\alpha=\sqrt{3} \alpha i \Rightarrow(2 \beta-\alpha)^2=3 \alpha^2 i^2 \\ \Rightarrow & 4 \beta^2+\alpha^2-4 \alpha \beta=-3 \alpha^2 \\ \Rightarrow & 4 \alpha^2+4 \beta^2=4 \alpha \beta \Rightarrow \alpha^2+\beta^2=\alpha \beta \\ \Rightarrow & (\alpha+\beta)^2-2 \alpha \beta=\alpha \beta \Rightarrow(-a)^2=3(b) \\ \Rightarrow & a^2=3 b\end{array}$
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