TITLE: Are all zeros of $\Gamma(s) \pm \Gamma(1-s)$ on a line with real part = $\frac12$ ?
QUESTION [52 upvotes]: 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!

REPLY [7 votes]: Just to point out there are very good approximation to complex zeros off your line of
$$ \Gamma(s)-\Gamma(1-s) \qquad(1)$$
At $\rho \approx -1.69711183621729718874218687438 - 0.305228379993226071272967719419 i$ (1) appears to vanish while $\Gamma(\rho) \approx 1.4648039 + 0.3642699441 i$
Root finding with better precision converges to $\rho$ while (1) still appear to vanish in both sage and gp/pari (modulo bugs).
Checked to precision $5000$ digits and (1) still appears to vanish.
Here is $\rho$ with $100$ digits of precision:
-1.697111836217297188742186874382163077146364585981726518217373889827452772242797069678994954785699956 - 0.3052283799932260712729677194188512919331197338088909477524842921187943642970297308885952936796125572*I


... for $ \Gamma(s)+\Gamma(1-s)$ approximation of zero appears $\rho \approx -0.60940537628997711023 - 0.82913081575572747216 i$ checked to $5000$ digits of precision.
With 100 digits:
 -0.6094053762899771102337308158313839002012166649163876907688596366808893391382113824494098816671945331 - 0.8291308157557274721587141536678087800797120641344787653174391388417832472543392187032283839972409848*I

Edit In comments juan suggested using x-ray to investigate the zeros.
The primary reference for x-ray I found is X-Ray of Riemann zeta-function, J. Arias-de-Reyna 
AFAICT x-ray are the plots of Re(f(s))=0 and Im(f(s))=0. The zeros are the intersection.
The x-ray and juan's comments suggest the above quadruples of zeros are indeed zeros off $\frac12$ and possibly there are no more complex zeros zeros off the line.
Here is the x-ray of $ \Gamma(s)-\Gamma(1-s)$. Blue is $\Re(\Gamma(s)-\Gamma(1-s))=0$ and red is $\Im(\Gamma(s)-\Gamma(1-s))=0$.

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