Value of Expression from Range of Parameter in Complex Equation

To find the range of values for $k \in \mathbb{R}$ such that the equation $z(\bar{z} + 2 + i) + k(2 + 3i) = 0$ has at least one complex solution $z \in \mathbb{C}$, we represent $z$ in Cartesian form.

Let $z = x + iy$, where $x, y \in \mathbb{R}$.

Using the property $z \bar{z} = |z|^2 = x^2 + y^2$, the given equation can be rewritten as: $(x^2 + y^2) + (x + iy)(2 + i) + k(2 + 3i) = 0$

Expanding and regrouping terms into real and imaginary parts: $(x^2 + y^2 + 2x - y + 2k) + i(x + 2y + 3k) = 0$

For a complex number to equal zero, both its real and imaginary parts must simultaneously equal zero:

1. Imaginary Part: $x + 2y + 3k = 0 \implies x = -2y - 3k \quad \text{--- (1)}$

2. Real Part: $x^2 + y^2 + 2x - y + 2k = 0 \quad \text{--- (2)}$

Substituting Equation (1) into Equation (2): $(-2y - 3k)^2 + y^2 + 2(-2y - 3k) - y + 2k = 0$

Expanding this expression: $(4y^2 + 12ky + 9k^2) + y^2 - 4y - 6k - y + 2k = 0$

Combining like terms to form a quadratic equation in $y$: $5y^2 + (12k - 5)y + (9k^2 - 4k) = 0$

For a solution $z \in \mathbb{C}$ to exist, $y$ must be a real number. Therefore, the discriminant $D$ of this quadratic equation in $y$ must be greater than or equal to zero ($D \ge 0$): $D = (12k - 5)^2 - 4(5)(9k^2 - 4k) \ge 0$

Expanding and simplifying the inequality: $144k^2 - 120k + 25 - 20(9k^2 - 4k) \ge 0$ $144k^2 - 120k + 25 - 180k^2 + 80k \ge 0$ $-36k^2 - 40k + 25 \ge 0$

Multiplying by $-1$ (and reversing the inequality sign): $36k^2 + 40k - 25 \le 0$

Let $\alpha$ and $\beta$ be the roots of the quadratic equation $36k^2 + 40k - 25 = 0$. Since the coefficient of $k^2$ is positive ($36 > 0$), the solution to $36k^2 + 40k - 25 \le 0$ is $k \in [\alpha, \beta]$.

Using Vieta's formulas for $36k^2 + 40k - 25 = 0$, the sum of the roots is: $\alpha + \beta = -\frac{40}{36} = -\frac{10}{9}$

We are required to find the value of $9(\alpha + \beta)$: $9(\alpha + \beta) = 9 \left(-\frac{10}{9}\right) = -10$

Hence, the correct option is A.