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I was experimenting with the following simple addition:


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 and it reads:

$$|(-\frac12 + yi)!|^2= \frac{\pi}{\cosh(\pi y)}$$

Is there any way to exactly calculate this outcome (e.g. using $\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}$) ?


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(correction) There are at least $8$ solutions : $\frac {-1\pm\sqrt{3}i}2$, $\frac {3\pm\sqrt{3}i}2$ with four real values near $\ -2.9700120101545954067416063621915\ ,-2.18550554669205987259055000950\ ,3.1855055466920598725905500,\ 3.970012010154595406741606362\ $. – Raymond Manzoni Oct 26 '12 at 19:52
I'll add that the roots are symmetric around $z=\frac 12$ (I don't think that others exist). – Raymond Manzoni Oct 26 '12 at 20:04
Thanks Raymond. I will correct it in the post. – Agno Oct 26 '12 at 21:42
up vote 1 down vote accepted

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).

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Many thanks for the great guidance and approach, Fabian. Nice to see the 4 complex zeros can indeed be directly calculated from $\sin$ and $\cosh$. Maybe the advanced techniques used in this question: $… could help complete the final part of the proof that this is indeed the only complex zero? – Agno Oct 26 '12 at 21:55

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