How to find complex solutions for the following equation:
$\Gamma(x)=-x$
We can find the real solutions to be only negative by plotting both on 2D plane.
But how to find complex solution? If possible can we get an upper bound on real part of complex solution?
For me, this is a purely numerical work. Grphing shows where are, more or less the solutions and Newton method converges like a charm. For example, some iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.100000\pm1.000000\,i \\ 1 & 0.090610\pm0.765230 \, i \\ 2 & 0.088850\pm0.794651 \,i \\ 3 & 0.088566\pm0.795376 \,i \\ 4 & 0.088566\pm0.795377 \,i \end{array} \right)$$
$$\left( \begin{array}{cc} n & x_n \\ 0 & -1.50000\pm0.50000 \,i \\ 1 & -1.73488\pm0.22858 \,i \\ 2 & -1.66916\pm0.25561 \,i \\ 3 & -1.66190\pm0.26103 \,i \\ 4 & -1.66187\pm0.26111 \,i \end{array} \right)$$
Edit
Beside numerical methods, for the real roots, you can have quite good approximations using series expansion. For example, around $x=-3^-$
$$\Gamma(x)+x=-\frac{1}{6 (x+3)}+\left(\frac{\gamma }{6}-\frac{119}{36}\right)+O\left((x+3)\right)$$ gives $$x \sim -3+\frac{6}{6 \gamma -119}=-3.05193$$ while the "exact" solution is $-3.05141$.
Similarly, around $x=-4^+$
$$\Gamma(x)+x=\frac{1}{24 (x+4)}+\left(-\frac{1127}{288}-\frac{\gamma }{24}\right)+O\left((x+4)\right)$$ gives $$x \sim -4+\frac{12}{1127+12 \gamma }=-3.98942$$ while the "exact" solution is $-3.98939$.