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Wilson's theorem states $n \in \mathbb N$ is prime iff$(n-1)! \equiv -1\pmod n$. $\Gamma$-function extends the usual factorial to complex numbers.

What are the complex numbers such that $\Gamma(z)+1 = nz$ , $n \in \mathbb Z$?

Eisenstein or Gaussian primes don't necessarily satisfy the requirement, take for example $2+\omega$ and $5+12i$ respectively.

What I've tried:

Let $z=a+ib$. From the definition of the $\Gamma$-function, we have

$\Gamma(z)=(z-1)\Gamma(z-1)$.

$\Gamma(a+ib)=(a-1+ib)\Gamma(a-1+ib)$

$=(a-1+ib)(a-2+ib)\Gamma(a-2+ib)$

$=(a-1+ib)(a-2+ib)...(a-k+ib)\Gamma(a-k+ib)$

$=\Gamma(ib)\prod_{k=0}^{a-1}{(k+ib)}$

Now, turning to the imaginary-$\Gamma$, a brick wall I ran into...

$\Large\Gamma(ib)=\Large\int_0^{\infty}\huge\frac{t^{-1+ib}}{e^t}\large\mathrm{d}t$

... and cannot evaluate.

Questions

  1. How do we evaluate $\Gamma(a+ib)$?
  2. How should we go about solving for $z$ once (1) is done?

Computational 'evidence'

Wolfram|Alpha thinks these $z$ exist, infact they seem plentiful. I'm not sure if approximation is muddling the results, but I doubt it.

I'm using $\mathrm{solve\space Gamma(a+ib)+1=n(a+ib)}$ and plugging in values of a,b,n.

share|improve this question
    
I assume you mean complex numbers which are not natural numbers? –  Alex Becker Aug 26 '12 at 7:00
1  
If you want to generalize Wilson's theorem to rings of integers in number fields this is not the way to go about doing it. For a Gaussian integer $a + bi$, for example, it is much more natural to look at the product of the nonzero elements in $\mathbb{Z}[i]/(a + bi)$. –  Qiaochu Yuan Aug 26 '12 at 7:01
    
@Qiaochu Yuan My motive was not Wilson related so much as exploring the Gamma function, being new to it and all. Perhaps 'primality' is an unsuitable term? To Alex, I'll try to be careful =P $a+bi$ where $b$ is non-zero. –  Furlox Aug 26 '12 at 7:10

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