Question

Let the set of all values of $k \in \mathbb{R}$ such that the equation
$z(\bar{z} + 2 + i) + k(2 + 3i) = 0, z \in \mathbb{C}$, has at least one solution, be the interval $[\alpha, \beta]$. Then $9(\alpha + \beta)$ is equal to:

• A. -10
• B. -8
• C. $10\sqrt{13}$
• D. $8\sqrt{13}$

Hint:

Substitute $z = x + iy$ and separate the equation into real and imaginary parts. Eliminate $x$ to get a quadratic in $y$, then apply the condition for real roots ($D \ge 0$).

Answer 1

The Correct Option is A

Given equation: $|z|^2 + z(2+i) + k(2+3i) = 0$.
Let $z = x+iy$. Separating real and imaginary components of the equation:
Real: $x^2 + y^2 + 2x - y + 2k = 0$
Imaginary: $x + 2y + 3k = 0$

From the imaginary part, we isolate $x$: $x = -2y - 3k$.
Plugging this into the real part equation:
$(-2y - 3k)^2 + y^2 + 2(-2y - 3k) - y + 2k = 0$
$4y^2 + 9k^2 + 12ky + y^2 - 4y - 6k - y + 2k = 0$
$5y^2 + (12k - 5)y + 9k^2 - 4k = 0$

For $y$ (and hence $z$) to have at least one real solution, the discriminant of this quadratic in $y$ must be greater than or equal to zero.
$\Delta = (12k - 5)^2 - 4(5)(9k^2 - 4k) \ge 0$
$144k^2 - 120k + 25 - 180k^2 + 80k \ge 0$
$-36k^2 - 40k + 25 \ge 0$
$36k^2 + 40k - 25 \le 0$

Let the roots of $36k^2 + 40k - 25 = 0$ be $\alpha$ and $\beta$. Then the solution for $k$ is the interval $[\alpha, \beta]$.
The sum of roots $\alpha + \beta$ is given by $-\frac{\text{coefficient of } k}{\text{coefficient of } k^2} = -\frac{40}{36} = -\frac{10}{9}$.
Finally, $9(\alpha + \beta) = 9 \times (-10/9) = -10$.