Let $z_1$ and $z_2$ be the non-real roots of the equation $3z^2+3z+b=0$. If the origin together with the points represented by $z_1$ and $z_2$ form an equilateral triangle, then the value of $b$ is :
Step-by-Step Solution
Key Concept: For an equilateral triangle with vertex at origin, the non-real roots must be complex conjugates with equal modulus satisfying the geometric constraint $|z_1| = |z_2| = |z_1 - z_2|$.
For the quadratic $3z^2 + 3z + b = 0$, by Vieta's formulas: $z_1 + z_2 = -1$ and $z_1z_2 = b/3$. Since the origin $O$, $z_1$, and $z_2$ form an equilateral triangle, we need $|z_1| = |z_2| = |z_1 - z_2|$. From $|z_1| = |z_2|$ and $z_1z_2 = b/3$ being real, we get $z_2 = \overline{z_1}$. Using $z_1 + z_2 = -1$, we have $2\text{Re}(z_1) = -1$, so $\text{Re}(z_1) = -1/2$. For an equilateral triangle: $|z_1 - z_2|^2 = |z_1|^2 + |z_2|^2$, which gives $4|\text{Im}(z_1)|^2 = 2|z_1|^2$. With $z_1 = -1/2 + iy$, we get $4y^2 = 2(1/4 + y^2)$, yielding $y^2 = 1/4$. Therefore $b/3 = |z_1|^2 = 1/4 + 1/4 = 1/2$, giving $b = 3/2$... Wait, checking: $|z_1|^2 = 1$, so $b/3 = 1$ and $b = 3$.
Correct Answer: 1