# Complex Factorial Equaling One

For what complex values of $z$ is $$z! =1?$$ Are they even all known? Are there finitely many or infinitely many?

(Yes, the trivial $z$ are 0 and 1. )

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A minor notational nit. $z!$ is not defined, except for natural numbers, but $\Gamma(z+1)$ is. –  Thomas Andrews Aug 5 '14 at 23:25
Yes, the factorial is to be interpreted as the value assigned to it by the gamma function. –  Assaultous2 Aug 5 '14 at 23:26
For people who want to see a few sample values, take a look at Wolfram|Alpha. (Not an answer, but may help some people.) –  apnorton Aug 5 '14 at 23:33

I assume what you mean is $\Gamma(z+1) = 1$.

Here's a plot of the curves $\text{Re}(\Gamma(z+1)) = 1$ (blue) and $\text{Im}(\Gamma(z+1)) = 0$ (red). Each intersection of a red and blue curve corresponds to a solution. Assuming the pattern continues, it certainly appears that there are infinitely many.

The first $10$ solutions in the first quadrant are $$\begin {array}{c} 1\\ 3.213486150+ 4.253693352\,i\\ 4.447352283+ 6.904660210\,i\\ 5.449043370+ 9.238727110\,i \\ 6.328673500+ 11.39926303\,i\\ 7.129370000+ 13.44405135\,i\\ 7.873424830+ 15.40369196\,i\\ 8.574168470+ 17.29686175\,i \\ 9.240338285+ 19.13602021\,i\\ 9.878036600+ 20.93000503\,i\end {array}$$

Here's a plot of the first $151$:

It certainly looks like they lie on a curve.

EDIT: OK, something analytic can be said. The asymptotic

$$\Gamma(z) \sim \sqrt{2\pi} e^{-z} z^{z-1/2} = \sqrt{2\pi} \exp(-z + (z - 1/2) \log(z)) \ \text{as}\ |z| \to \infty$$

holds for $|\arg z| < \pi$ with the principal branch of the log. If $z = t e^{i\theta}$ with $\theta \in (0, \pi/2)$, $$\text{Re}(-z + (z-1/2) \log(z)) = t \ln(t) \cos(\theta) - (\theta \sin(\theta) + \cos( \theta)) t - \ln(t)/2$$ If this is $\log(1/\sqrt{2\pi})$, indicating that $|\Gamma(z) \approx 1$, then $\ln(t) \approx 1 + \theta \tan(\theta)$. Note that the right side goes to $\infty$ as $\theta \to (\pi/2)-$. The roots should be approximately on this curve. And indeed, here is the previous plot together with the curve $\ln(t) = 1 + \theta \tan(\theta)$ (in red):

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Do they fit on some approximative line? Or is the pattern more ugly when you look at it on a wider scale? –  Patrick Da Silva Aug 5 '14 at 23:32
How did you obtain this graph? ( so that I can do it myself in the future) –  Assaultous2 Aug 5 '14 at 23:33
In Maple: plots:-implicitplot([Re(GAMMA(x+Iy+1))-1,Im(GAMMA(x+Iy+1))],x=-5..15,y=0..15,c‌​olour=[blue,red],gridrefine=3); –  Robert Israel Aug 5 '14 at 23:36
Thanks. The values do seem to lie on a straight line... –  Assaultous2 Aug 5 '14 at 23:38
It may be worth looking up the properties of $\log\Gamma(x)$ rather than $\Gamma(x)-1$ directly, since they both have the same zeroes but the former seems a bit more 'natural.' @Assaultous2 –  Semiclassical Aug 6 '14 at 0:32