Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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. )

share|improve this question
1  
A minor notational nit. $z!$ is not defined, except for natural numbers, but $\Gamma(z+1)$ is. –  Thomas Andrews Aug 5 at 23:25
2  
Yes, the factorial is to be interpreted as the value assigned to it by the gamma function. –  Assaultous2 Aug 5 at 23:26
1  
For people who want to see a few sample values, take a look at Wolfram|Alpha. (Not an answer, but may help some people.) –  anorton Aug 5 at 23:33

1 Answer 1

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.

enter image description here

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 $151$:

enter image description here

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):

enter image description here

share|improve this answer
    
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 at 23:32
    
How did you obtain this graph? ( so that I can do it myself in the future) –  Assaultous2 Aug 5 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 at 23:36
    
Thanks. The values do seem to lie on a straight line... –  Assaultous2 Aug 5 at 23:38
1  
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 at 0:32

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.