Is $\Gamma(z,1)\not=0$ for all $z$ with $\Re(z)<0$?

Question

I found this paper online which appears to present zeros of the incomplete gamma function within the right half plane. It makes me think that there are no zeros in the left half plane. Not sure how to prove this. Looking at Wikipedia I found the following continued fraction by Gauss,

$$e\cdot\Gamma(z,1)=\frac{1}{1+\frac{1-z}{1+\frac{1}{1+\frac{2-z}{1+\frac{2}{1+\frac{3-z}{\ddots}}}}}}$$

But this representation neither confirms that there are zeros in the right half plane nor that there are no zeros in the left half plane. Any clarification would be helpful.

Answer 1

Denote $z=-a-bi$, $a>0$. We have $$\Gamma(z,1)=\int_1^\infty t^{z-1}e^{-t}dt=\int_0^\infty e^{(-a-1-bi)s}e^{-e^s}d(e^s)=\int_0^\infty e^{-ibs}h(s)ds$$ for a positive decreasing function $h(s)=e^{-as-e^s}$. If $b=0$, the integral is certainly positive. If $b\ne 0$, we have $$\Im\, \Gamma(z,1)=- \int_0^\infty \sin bs\, h(s)ds=\int_0^\infty h(s) d\left(\frac{\cos bs-1}b\right)=\int_0^\infty\frac{1-\cos bs}b h'(s)ds\ne 0.$$