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

Solution 1:

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 $$i! = \Gamma(i+1) = \int_0^{\infty} e^{-t}t^{i}\, dt \approx 0.498015668−0.154949828i$$

See also here and here.

Solution 2:

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

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