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Wilson's theorem states $n \in \mathbb N$ is prime iff $(n-1)! \equiv -1\pmod n$. The $\Gamma$-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






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

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

... and cannot evaluate.


  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 solve Gamma(a+ib) + 1= n(a+ib) and plugging in values of $a,b,n$.

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I assume you mean complex numbers which are not natural numbers? – Alex Becker Aug 26 '12 at 7:00
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
Your $z$ almost certainly are plentiful - but they're (likely to be) all transcendental. I don't even believe an explicit formula for $\Gamma(i)$ is known (Alpha has it as $-0.1549498\ldots+0.4980\ldots i$), much less $\Gamma(a+bi)$ for even integral $a,b$. – Steven Stadnicki Dec 15 '15 at 20:19
(Once you use the gamma function and complex arguments, alpha has no constraints about the integrality of your parameters, so it treats your $a$ and $b$ as arbitrary real numbers rather than integers. Under those circumstances it's no surprise that there are numerous solutions, just as there's surely a real solution (but not an integral one) to the equation $\Gamma(x)+1=4x$. – Steven Stadnicki Dec 15 '15 at 20:23

The roadblock that you are hitting is, as Wikipedia indicates, attributable to the fact that the integral definition of the Gamma function as you presented it only holds when the Re(z) > 0.

For $z=i*b$, $Re(z)$=0

You would be better off using a product definition, insofar as calculations are concerned.

So to answer your first question, we may use your integral iff Re(z)>0 Otherwise we must use one of these alternate definitions:

Product Definitions of the Gamma Function (stolen from Wikipedia)

We may also employ approximations, or if Im(z)=0 and 0

Now as to your second question. If we have evaluated $Γ(a+ib)=Γ(z)$ then I am afraid that the only way we could go about recovering z would be by using inverse approximations, proper coverage of which you may find here:

Hope that helps; please let me know how I can improve the answer.

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