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

The given equation is $z\bar{z} + z(2+i) + k(2+3i) = 0$.
Let $z = x + iy$ where $x, y \in \mathbb{R}$.
Substitute $z$ and $z\bar{z} = x^2 + y^2$ into the equation:
$(x^2 + y^2) + (x + iy)(2 + i) + 2k + 3ki = 0$
$(x^2 + y^2) + (2x + ix + 2iy - y) + 2k + 3ki = 0$

Group the real and imaginary parts:
Real part: $x^2 + y^2 + 2x - y + 2k = 0$ ... (1)
Imaginary part: $x + 2y + 3k = 0 \implies x = -2y - 3k$ ... (2)

Now substitute $x$ from (2) into (1):
$(-2y - 3k)^2 + y^2 + 2(-2y - 3k) - y + 2k = 0$
$(4y^2 + 12ky + 9k^2) + y^2 - 4y - 6k - y + 2k = 0$
$5y^2 + (12k - 5)y + (9k^2 - 4k) = 0$

For the complex number $z$ to exist, $y$ must be a real number. Therefore, the discriminant ($D$) of this quadratic equation in $y$ must be non-negative ($D \ge 0$):
$D = B^2 - 4AC = (12k - 5)^2 - 4(5)(9k^2 - 4k) \ge 0$
$144k^2 - 120k + 25 - 20(9k^2 - 4k) \ge 0$
$144k^2 - 120k + 25 - 180k^2 + 80k \ge 0$
$-36k^2 - 40k + 25 \ge 0$
$36k^2 + 40k - 25 \le 0$

The roots of $36k^2 + 40k - 25 = 0$ are the boundaries $\alpha$ and $\beta$ of the interval for $k$.
By the sum of roots formula for $ak^2 + bk + c = 0$, $\alpha + \beta = -b/a = -40/36 = -10/9$.

The question asks for $9(\alpha + \beta)$:
$9 \times (-10/9) = -10$.