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The factorial function is defined as $$n!=\prod_{k=1}^{n}k$$ For $n>0$, $n$ being an integer. This definition can be extended to the complex numbers with positive real part by the Gamma function: $$\Gamma(n)=(n-1)!$$ This is the plot of the factorial function, in which the values of the factorial of non-integral numbers are evaluated by the gamma function: enter image description here It can be seen that for the positive numbers, the factorial is always nonzero. Why is it so?
According to mathguy's comment, the factorial function is positive for all the complex numbers. Is this true? If yes, then why is it so?

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  • $\begingroup$ In fact,(according to Wolfram alpha) no solutions of $n!=0$ exist for all the complex numbers. $\endgroup$ Commented Oct 30, 2020 at 3:12
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    $\begingroup$ Use the (say integral) definition of $\Gamma$—$\Gamma(n):=\int_0^\infty e^{-x}x^{n-1}dx$; the integrand is positive over $(0,\infty)$ for the positive numbers desired. $\endgroup$ Commented Oct 30, 2020 at 3:14
  • $\begingroup$ The factorial domain is $\displaystyle\mathbb{C}\setminus\left\{-1,-2,-3,\ldots\right\}$. So, you can have some factorial with $\underline{\mbox{negative real part}}$: For example, $\displaystyle\left(-\,{1 \over 2}\right){\LARGE !} = \sqrt{\pi}$. $\endgroup$ Commented Oct 30, 2020 at 3:26

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The complex variable function $z! = \Gamma(z+1)$ has no zeros anywhere in the complex plane. Why? One way to see this is to use Euler's reflection formula, $$ z!(-z)! = \frac{\pi z}{\sin \pi z}, \quad z \neq 0 $$

See ProofWiki's "Euler's Reflection Formula" entry for a proof.

If $z!$ were ever zero, then the above formula would also be zero, but that obviously is not the case. In fact $1/(z!)$ is an entire function, that is it is analytic for all $z$, and in particular has no poles, implying $z!$ has no zeros.

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This does not necessarily preclude the case where $z!$ has a pole (is infinite) and $(-z)!$ is zero. However, the only poles of $z!$ are at the negative integers, $z=-1, -2, \cdots$, which can be seen from the product formula, $$ z! = \Gamma(z+1) = \Pi_{n=1}^\infty \frac{n}{n+z}\Big( 1+\frac{1}{n}\Big)^z $$ convergent for all $z \neq -1, -2, \cdots$. At these values $-z$ is a positive integers and then $(-z)!$ is known to be non zero.

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  • $\begingroup$ Can we rule out the case that the factorial function has a pole at $z$ and a zero at $-z$? $\endgroup$ Commented Oct 30, 2020 at 7:01
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    $\begingroup$ Fair point: however, all poles are at the negative integers and in those cases, $z!$ is non-zero. $\endgroup$ Commented Oct 30, 2020 at 7:05
  • $\begingroup$ I agree it's obvious for all poles that actually do exist, but I worry that this passes the burden of proof onto figuring out whether there are any non-obvious poles, rather than non-obvious zeros. Is there a good way to see that only the negative integers are poles? $\endgroup$ Commented Oct 30, 2020 at 7:42
  • $\begingroup$ Well, that the product formula converges for all $z$ except the negative integers answers the question of whether there are other poles. Proof of its convergence is somewhat longer than can be contained here but is widely available. $\endgroup$ Commented Oct 30, 2020 at 11:52

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