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# Are all zeros of $\Gamma(s) \pm \Gamma(1-s)$ on a line with real part = $\frac12$ ?

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!