Let $\Gamma(z,1) $ be the well-known incomplete gamma function : $\int_1^{\infty} e^{-t} t^{z-1} dt $
Now I am curious to find solutions to
$$\Gamma(z,1)^2 + \Gamma(z,1) = 0 $$
We can ofcourse reduce this too all complex solutions of either
$$\Gamma(z,1) = 0 $$
or
$$\Gamma(z,1) = -1 $$
I heard that this function is an entire function so the equation $\Gamma(z,1)^2 + \Gamma(z,1) = 0 $ must have solutions.
I think that $\Gamma(z,1) = 0 $ has no (finite) solutions.
I assume most solutions to $\Gamma(z,1) = -1 $ must have $-3 < Re(z) < 3 $ due to the functional equation this function satisfies. ( $\Gamma(z+1,1) = z \Gamma(z,1) + e^{-1} $ )
I know a bit about complex analysis and contour integrals. But I have no efficient method to find the zero's. I would like to have the values and some insight into them. How far are the zero's apart from eachother ?
Im aware of another identity for strictly real $z < 1$ :
$$ \Gamma(z,1)=\frac{e^{z-1}}{\Gamma(1-z)} \int_0^\infty \frac{e^{-t} t^{-z}}{1+t} dt$$
Not sure if that could help.
I was only able to find $t=4.86853..+5.66062..i$ and its complex conjugate as solutions to $\Gamma(t,1) = -1$
edit :
Perhaps the following limits might be helpful :
$$ \lim_{x \to +\infty} Re(x^2 \Gamma(1+e+x i,1) ) = -1$$
$$ \lim_{x \to +\infty} Im(x \Gamma(1+e+ x i,1) ) = e^{-1} i $$