Question

Sol. [Key Concept: Split the complex equation into real and imaginary parts, eliminate $x$ to get a quadratic in $y$; a real solution requires discriminant $\geq 0$, giving the interval $\lbrack\alpha,\beta\rbrack$.]

Put $z = x + iy,\bar{z} = x - iy$

$${\Rightarrow \overline{z} + 2 + i = x - iy + 2 + i }{\Rightarrow z(\overline{z} + 2 + i) = (x + iy)(x + 2 + i(1 - y)) }$$⇒ Real part $= x(x + 2) - y(1 - y)$

Imaginary part $= x(1 - y) + y(x + 2)$

$$\begin{array}{r} \Rightarrow x^{2} + y^{2} + 2x - y + 2k = 0\#(1) \end{array}$$

and $x + 2y + 3k = 0$

eliminating $x$ from (1) and (2)

$$5y^{2} + (12k - 5)y + 9k^{2} - 4k = 0 $$since $y \in R$

use $D \geq 0$

$${(12k - 5)^{2} - 4.5\left( 9k^{2} - 4k \right) \geq 0 }{\Rightarrow 36k^{2} + 40k - 25 \leq 0 }{\Rightarrow \alpha + \beta = \frac{- 10}{9} }{\Rightarrow 9(\alpha + \beta) = - 10 }$$