If $\lambda \in \mathbb{R}$ such that the origin and the non-real roots of the equation $2z^2 + 2z + \lambda = 0$ form the vertices of an equilateral triangle in the argand plane, then $\lambda$ is equal to
Step-by-Step Solution
Key Concept: Convert complex number equations to real coordinates by setting $z = x + iy$ and separating imaginary and real parts
Let $z = x + iy$. We have $\text{Im}\left(\frac{(x+iy)^2}{x+iy-1}\right) = -1$. Simplifying, $\text{Im}\left(\frac{(x+iy)^2}{(x-1)+iy}\right) = -1$. Expanding and separating real and imaginary parts: $2x^2 + 2y^2 - y - 1 = 0$ and $x^2 + y^2 + y - 2 = -x^2 - y^2 + 2y - 1$. Solving these equations gives the circle $x^2 + y^2 = 1$. The radius is $\sqrt{\left(\frac{1}{2}\right)^2 + \frac{1}{4}} = \frac{1}{4}$, giving answer $0.75$.
Correct Answer: 0.75