The function $\Gamma(s)$ does not have zeros, but $\Gamma(s)\pm \Gamma(1-s)$ does.


Ignoring the real solutions for now and assuming $s \in \mathbb{C}$ then:


$\Gamma(s)-\Gamma(1-s)$ yields zeros at:


$\frac12 \pm 2.70269111740240387016556585336 i$
 $\frac12 \pm 5.05334476784919736779735104686 i$
 $\frac12 \pm 6.82188969510663531320292827393 i$
 $\frac12 \pm 8.37303293891455628139008877004 i$
 $\frac12 \pm 9.79770751746885191388078483695 i$
 $\frac12 \pm 11.1361746342106720656243966380 i$
 $\frac12 \pm 12.4106273718343980402685363665 i$


$\dots$


and 


$\Gamma(s)+\Gamma(1-s)$ gives zeros at: 


$\frac12 \pm 4.01094805906156869492043027819 i$
 $\frac12 \pm 5.97476992595365858561703252235 i$
 $\frac12 \pm 7.61704024553573658642606787126 i$
 $\frac12 \pm 9.09805003388841581320246381948 i$
 $\frac12 \pm 10.4760650707765536619292369200 i$
 $\frac12 \pm 11.7804020877663106830617193188 i$
 $\frac12 \pm 13.0283749883477570386353012761 i$


$\dots$


By multiplication, both functions can be combined into: $\Gamma(s)^2 - \Gamma(1-s)^2$


After playing with the domain of $s$ and inspecting the associated 3D output charts, I now dare to conjecture that all 'complex' zeros of this function must have a real part of $\frac12$. 


Has this been proven? If not, appreciate any thoughts on possible approaches.


Thanks!



Are all zeros of $\Gamma(s) \pm \Gamma(1-s)$ on a line with real part = $\frac12$ ?
 **In the first part, we show that there are no zeros for $z = s + i t$
with $|t| \ge 4$** .


Let $\psi(z):= \Gamma'(z)/\Gamma(z)$ be the digamma function.
If $z = s + i t$, then
$$\frac{d}{ds} |\Gamma(z)|^2 = \frac{d}{ds} \Gamma(z) \Gamma(\overline{z})
= |\Gamma(z)|^2 \left(\psi(z) + \psi(\overline{z})\right).$$
(Both $\Gamma(z)$ and $\psi(z)$ are real for real $z$, and so satisfy
the Schwartz reflection principle.)
The product formula for the Gamma function implies that there is an identity
$$\psi(z) = - \ \gamma + \sum\_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{z + n} \right)
= 1 - \gamma + \sum\_{n=1}^{\infty} \left(\frac{1}{n + 1} - \frac{1}{z + n} \right),$$
and hence
$$\psi(z) + \psi(\overline{z}) = 2(1 - \gamma) + 
\sum\_{n=1}^{\infty} \left(\frac{2}{n + 1} - \frac{1}{z + n} - \frac{1}{\overline{z} + n} \right).$$
Suppose that $z = s + i t$, and that $s \in [0,1]$.
Then
$$ \frac{2}{n + 1} - \frac{1}{s + i t + n} - \frac{1}{s - i t + n} 
= \frac{2(s^2 + t^2 + n s - s - n)}{(1+n)(n^2 + 2 n s + s^2 + t^2)} \ge
\frac{-2}{(n^2 + t^2)}.$$
(The last inequality comes from ignoring
all the positive terms in the numerator, and then
setting $s = 0$ in the denominator.)
It follows that
$$\psi(z) + \psi(\overline{z}) \ge 2(1 - \gamma) - 
\sum\_{n=1}^{\infty} \frac{2}{n^2 + t^2},$$
which is positive for $t$ big enough, e.g. $|t| \ge 4$.
On the other hand,
$$\psi(z + 1) + \psi(\overline{z} + 1) = \psi(z) + \psi(\overline{z}) + \frac{1}{z} + \frac{1}{\overline{z}} = 
\psi(z) + \psi(\overline{z}) + \frac{2s}{|z|^2}.$$
In particular, if $\psi(z) + \psi(\overline{z})$ is positive for $s \in [0,1]$ for some particular $t$, it is positive for
all $s$ and that particular $t$.
It follows that, if $|t| > 4$, that $|\Gamma(s + it)|^2$ is increasing as a function of $s$.
In particular, if $|t| > 4$, then any equality
$$|\Gamma(s + i t)| = |\Gamma(1 - (s + i t))| = |\Gamma(1 - s + i t)|$$
implies that $s = 1/2$.


 **The second part is a continuation of the argument above, which completes the argument. (merged from a different answer.)**


Let $C\_n$ denote the square with vertices $[n \pm 1/2, \pm 4 I]$ for a positive integer $n$.
We have the following inequalities for $z \in C\_n$ and $n \ge 15$:
$$|\sin(\pi z)| \ge 1, \quad z \in C\_n.$$
$$|\Gamma(z)| \ge \frac{1}{2} \Gamma(n - 1/2),$$
$$|\Gamma(1-z)| \le \frac{\pi}{\Gamma(n - 1/2)} \le 1,$$
$$|\psi(1-z)|, |\psi(z)| \le 2 \log(n), $$


The first is easy, the second follows from Stirling's formula (this requires $n$ to be big enough, and also
requires $z$ to have imaginary part at most $4$), the third follows from the previous
two by the reflection formula for $\Gamma(z)$, the last follows by induction and by the formula
$\psi(z+1) = \psi(z) + 1/z$.
It follows that
$$\left| \frac{1}{2 \pi i} \oint\_{C\_n} \frac{\Gamma'(z)}{\Gamma(z)} - \frac{d/dz (\Gamma(z) + \theta 
\cdot \Gamma(1-z))}{\Gamma(z) + \theta\cdot \Gamma(1-z)} \right|$$
$$= \left| \frac{1}{2 \pi i} \oint\_{C\_n} \frac{\theta \Gamma(1-z) (\psi(1-z) + \psi(z))}
{\Gamma(z) + \theta \cdot \Gamma(1-z)} \right|$$
$$ \le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)} 
 \oint\_{C\_n} \frac{1}
{|\Gamma(z) + \theta \cdot \Gamma(1-z)|}$$
$$ \le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)} \cdot \frac{1}{1/2 \Gamma(n - 1/2) + 1} \ll 1,$$
where $\theta = \pm 1$ (or anything small) and $n \ge 15$, where the final inequality holds by a huuuge margin.
It follows that
$\Gamma(z) + \theta \cdot\Gamma(1-z)$ and $\Gamma(z)$ have the same number of zeros minus the
number of poles in $C\_n$. Since $\Gamma(z)$ has no zeros and poles in $C\_n$, it follows that $\Gamma(z) + \theta\cdot\Gamma(1-z)$
has the same number of zeros and poles. It has exactly one pole, and thus exactly one zero.
If $\theta = \pm 1$ (and so in particular is real), by the Schwarz
reflection principle, this zero is forced to be real.
By symmetry, the same argument applies in the region $z = s + i t$ with $|t| \le 4$ and
$s \le -15$.
Combined with the above argument, this reduces the claim to $z = s + i t$ with $|s| \le 15$ and $|t| \le 4$ where the
claim can be checked directly. 


 **Hence all the zeros outside the box $z = s + it$ with $|t| \le 4$ and $|s| \le 15$ are either in $\mathbf{R}$, or lie on the line $1/2 + i \mathbf{R}$.** 


 **EDIT**  To clarify, I didn't actually *check* that there were no ``exceptional'' zeros in the box $\pm 15 \pm 4 I$, since I presumed that the original poster had done so. 
If $F(z) = \Gamma(z) - \Gamma(1-z)$, then computing the integral
$$\frac{1}{2 \pi i} \oint \frac{F'(z)}{F(z)} dz$$
around that box, one obtains (numerically, and thus exactly) $1$. There are (assuming
the OP at least computed the critical line zeros correctly) $2$ zeros in that range on the critical line. Along the real line in that range, there are $30$ poles and $25$ zeros. This means that there must be $1 + 30 - 25 = 6$ unaccounted for zeros. For such a zero
$\rho$ off the line, by symmetry one also has $\overline{\rho}$, $1 - \rho$ and
$1 - \overline{\rho}$ as zeros. Hence there must be either $1$ or $3$ pairs of zeros on the critical line, and either $1$ or $0$ quadruples of roots off the line. Varying the parameters of the integral, one can confirm there is a zero with $\rho \sim 2.7 + 0.3 i$, which is one of the four
conjugates of the root found by joro. A similar argument applies 
for $\Gamma(z)+\Gamma(1-z)$. Hence:


 **Any zero of $\Gamma(z) - \Gamma(1-z)$ is either in $\mathbf{R}$, on the line $1/2 + i \mathbf{R}$, or is one of the four exceptional zeros $\{\rho,1-\rho,\overline{\rho},1-\overline{\rho}\}$. A similar calculation implies the same
for $\Gamma(z) + \Gamma(1-z)$, except now with an exceptional set
$\{\mu,1-\mu,\overline{\mu},1-\overline{\mu}\}$.**


