Single zero for $\Gamma(z)+\Gamma(\overline{1-z})$? I was experimenting with the following simple addition:
$$\Gamma(z)+\Gamma(\overline{1-z})$$
and like to conjecture that for $z=a \pm bi$ and $b>0$, it only becomes zero when:
$$\frac {-1\pm\sqrt{3}i}2,\frac {3\pm\sqrt{3}i}2$$
The only formula I found with $a = -\frac12$ is on http://mathworld.wolfram.com/GammaFunction.html and it reads:
$$|(-\frac12 + yi)!|^2= \frac{\pi}{\cosh(\pi y)}$$
Is there any way to exactly calculate this outcome (e.g. using $\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}$) ?
Thanks!
 A: This is not a complete solution but a few remarks (which are too long for a comment) which maybe somebody finds useful.
Your result is not completely correct. In fact the general set of solutions is
$$ z= \frac{3}{2} + \frac{1}{2}\sqrt{3} i$$ is an additional root.
Maybe some insight into the problem can be obtained using the fact that the $\Gamma$ function is nonzero everywhere in the complex plane:
Multiplying the equation $\Gamma(z) + \Gamma(\overline{1-z}) =\Gamma(z) + \overline{\Gamma(1-z)}$ with $\Gamma(1-z)$, we obtain
$$\frac\pi{\sin(\pi z)} + | \Gamma(1-z)|^2 =0$$
for all zeros of your function. Now $| \Gamma(1-z)|^2 \geq 0$. So we need that $\pi/\sin(\pi z)$ is a negative real number (thus $\sin(\pi z)$ has to be a negative real number). Thus we find that we need to have either that $z \in (1,2) + 2 \mathbb{Z}$ (which are on the real line and thus excluded from your conjecture) or $$\text{Re}(z)=-\frac{1}{2}+2\mathbb{Z}.$$
So let us set $z= -1/2 +2n + i y$ Then (for $n=0$) we have the property
$$|\Gamma(1-z)|^2 = \Gamma(3/2+ i y) \Gamma(3/2- i y) =\frac{\pi(1+4y^2)}{4\cosh(y)} \qquad(1)$$
together with
$$\frac{\pi}{\sin(\pi z)} = - \frac{\pi}{\cosh(\pi y)} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad(2)$$
we find the solution $y=\pm\frac{1}{2}\sqrt{3}$.
Similarly, we find the solutions quoted above for $n=1$. For $n\notin\{0,1\}$, I guess we should bound $|\Gamma(1-z)|^2$. Note that both parts (1) and (2) decay as $e^{-y}$ in general so it is really a matter of prefactors whether a solution can be found (that is also why I did not succeed in proving the extended version of your conjecture).
