Given Equation
$z(\bar z+2+i)+k(2+3i)=0$
where
$k\in\mathbb{R}$
and $z\in\mathbb{C}$.
We need values of $k$ for which at least one solution exists.
Step 1 : Let
$z=x+iy$
Then
$\bar z=x-iy$
Now,
$z(\bar z+2+i)=(x+iy)(x-iy+2+i)$
Write
$\bar z+2+i=(x+2)+i(1-y)$
Hence,
$z(\bar z+2+i)=(x+iy)\big((x+2)+i(1-y)\big)$
Multiplying,
$=(x(x+2)-y(1-y))+i\big(x(1-y)+y(x+2)\big)$
Real part:
$x^2+2x-y+y^2$
Imaginary part:
$x-yx+yx+2y=x+2y$
Thus,
$z(\bar z+2+i)=(x^2+y^2+2x-y)+i(x+2y)$
Equation becomes
$(x^2+y^2+2x-y)+i(x+2y)+k(2+3i)=0$
Equating real and imaginary parts:
$x^2+y^2+2x-y+2k=0$
$x+2y+3k=0$
Step 2 : Express $k$
From second equation,
$k=-\dfrac{x+2y}{3}$
Substitute into first equation:
$x^2+y^2+2x-y-\dfrac{2(x+2y)}{3}=0$
Multiply by $3$:
$3x^2+3y^2+6x-3y-2x-4y=0$
$3x^2+3y^2+4x-7y=0$
Complete squares:
$3\left(x^2+\dfrac43x\right)+3\left(y^2-\dfrac73y\right)=0$
$3\left[\left(x+\dfrac23\right)^2-\dfrac49\right]+3\left[\left(y-\dfrac76\right)^2-\dfrac{49}{36}\right]=0$
$3\left(x+\dfrac23\right)^2+3\left(y-\dfrac76\right)^2=\dfrac43+\dfrac{49}{12}$
$3\left(x+\dfrac23\right)^2+3\left(y-\dfrac76\right)^2=\dfrac{65}{12}$
$\left(x+\dfrac23\right)^2+\left(y-\dfrac76\right)^2=\dfrac{65}{36}$
So the circle has center
$\left(-\dfrac23,\dfrac76\right)$
and radius
$\dfrac{\sqrt{65}}{6}$
Step 3 : Find range of $k$
Since
$k=-\dfrac{x+2y}{3}$
we find range of $x+2y$.
For a circle, maximum and minimum of $x+2y$ are
$\left(-\dfrac23\right)+2\left(\dfrac76\right)\pm\dfrac{\sqrt{1^2+2^2}\sqrt{65}}{6}$
$=\dfrac53\pm\dfrac{\sqrt5\sqrt{65}}{6}$
$=\dfrac53\pm\dfrac{5\sqrt{13}}{6}$
Thus,
$x+2y\in\left[\dfrac53-\dfrac{5\sqrt{13}}{6},\dfrac53+\dfrac{5\sqrt{13}}{6}\right]$
Therefore,
$k\in\left[-\dfrac13\left(\dfrac53+\dfrac{5\sqrt{13}}{6}\right),-\dfrac13\left(\dfrac53-\dfrac{5\sqrt{13}}{6}\right)\right]$
Hence,
$\alpha+\beta=-\dfrac{10}{9}$
Therefore,
$9(\alpha+\beta)=-10$
Final Answer
$\boxed{-10}$
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