# Why is $i! = 0.498015668 - 0.154949828i$?

While moving my laptop the other day, I ended up mashing the keyboard a little, and by pure chance managed to do a google search for i!.

Curiously, Google's calculator dutifully informed me that $i!$ was, in fact, $0.498015668 - 0.154949828i$.

Why is this?

I know what a factorial is, so what does it actually mean to take the factorial of a complex number? Also, are those parts of the complex answer rational or irrational? Do complex factorials give rise to any interesting geometric shapes/curves on the complex plane?

– Pedro
Sep 25, 2012 at 12:39
• This should give you a jumping off point: wolframalpha.com/input/?i=i%21 Sep 25, 2012 at 12:40
• Strictly speaking, under the usual definition the factorial is only defined for natural arguments, so you will have to use a generalized definition for $i!$. Sep 25, 2012 at 12:43
• $\Gamma(i+1)$ of course. Sep 25, 2012 at 12:44
• I changed the "number theory" tag to "complex analysis". Sep 25, 2012 at 15:14

It is sort of an abuse of what is meant by factorial. The usual definition of $$n! = \prod_{k=1}^n k$$ obviously cannot apply because you can sit and count integers until the end of time and beyond and you'll never find $i$.

However, we can generalise what we mean by factorial by using a property of the gamma function, which is defined to be $$\Gamma(z) = \int_0^{\infty} e^{-t}t^{z-1}\, dt$$ This has the useful property that, for any $n \in \mathbb{N}$, $$\Gamma(n) = (n-1)!$$ which has an easy proof by induction on $n$. It also has lots of nice analytical properties which make it a good choice for an extension of the factorial function.

Anyway, since the gamma function can be defined (after analytic continuation; see LVK's comment) on the entire complex plane, minus the non-positive integers, for a general $z \in \mathbb{C} - \{ -1, -2, \cdots \}$ we can put $$z! \overset{\text{def}}{=} \Gamma(z+1)$$ For this reason we get $$i! = \Gamma(i+1) = \int_0^{\infty} e^{-t}t^{i}\, dt \approx 0.498015668−0.154949828i$$

• The result comes just from numerical computation? Can it be expressed in terms of elementary functions or well known constants? Sep 25, 2012 at 13:41
• @leonbloy: Yes it's computed numerically. If there is a nice expression for it then neither I nor Google nor Wolfram Alpha knows what it is ;) Sep 25, 2012 at 13:44
• This is interesting stuff, but I have a follow-up: is factorial the only case where a function which only works on a certain set of numbers has been 'abused' and turned into a different function which also works across a bigger set? Or is factorial a bit unique here? Sep 25, 2012 at 15:19
• @growse Riemann zeta-function is another example where original definition works only in some range of numbers, but can be profitably extended. "Abused" really means "extended" here.
– user31373
Sep 25, 2012 at 15:21
• No need to exclude $0$ from $z!=\Gamma(z+1)$, by the way.
– user31373
Sep 25, 2012 at 15:21

Do complex factorials give rise to any interesting geometric shapes/curves on the complex plane?

There are a couple of Gamma fractals shown on Wolfram's reference article for Gamma under "Neat Examples":

• DensityPlot[ Arg[Nest[Gamma, x + I y, 3]], {x, -1.25, -0.6}, {y, -0.25, 0.25}] // Quiet
• ArrayPlot[ Table[c = N[cr + I ci]; Length @NestWhileList[ If[Abs[#] > 20., Indeterminate, Gamma[#/c]] &, c, (# =!= Indeterminate) &, 1, 20], {ci, -2.5, 2.5, 5/100}, {cr, -2, 2, 4/100}]] // Quiet

See also Christopher Olah's blog post, Gamma Fractals, and from there, Bidimensional zoom in on the Z=Gamma(Z) iteration with display of the arguments, both of which have some nice images.

$$i!=\Gamma(i+1)=\int_0^{\infty}e^{-x} x^{i}dx$$ where $\Gamma(n)$ represents the Gamma Function

Note $$x^i=e^{i\ln x}=\cos(\ln x)+i\sin(\ln x)$$

• @LarsH Are you sure? It only see an equation with an integral, but no explanation of what that really means. It is not even mentioned that $\Gamma$ is the Gamma function, for example.
• @LarsH I'm implying that if someone asks "How/why is $i!=\text{bleh}$"? then it is probably because they know nothing about the Gamma function and thus answering $$i! = \Gamma \left( {i + 1} \right) = \int\limits_0^\infty {{x^i}{e^{ - x}}dx}$$ is pretty useless. Let alone that the OP is not supposed to know $\Gamma$ is the Greek letter gamma. Lately we seem to think leaving short "smart" answers that a few can understand (and will probably upvote) will help the OP, and that is not the case. Compare to Clive N.'s answer.