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I'm reading on the extension of $\Gamma$ to the complex plane and there is written:

Corollary

$$\Gamma(z) \not = 0 \qquad \forall z \in \mathbb{C}\setminus\{0,-1,-2, \dots\}$$

Proof

$(\forall z \in \mathbb{C}\setminus\mathbb{Z}) \quad \Gamma(1-z)\Gamma(z) = \dfrac{\pi}{\sin \pi z}$ implies this. For $n\in \mathbb{N}$ is $\Gamma(n+1) = n! \not = 0$.

Why?

Can someone explain why that is true? Why does $\Gamma(1-z)\Gamma(z) = \dfrac{\pi}{\sin \pi z}$ imply $\Gamma(z) \not = 0$?

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dietervdf
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  • is it because $\sin (\pi z)$ is bounded for bounded $z?$ – abel Dec 29 '14 at 03:58
  • @abel Hmm, suddenly it seems super-obvious. I have no idea why I was overlooking David's answer :). Math can be a strange creature :). (I still find the last statement about $n\in \mathbb{N}$ strange, what does that contribute?) – dietervdf Dec 29 '14 at 04:05
  • Got it (answer below) – dietervdf Dec 29 '14 at 04:09
  • may be because $\sin (n\pi) = 0$ will force $\Gamma(x)$ to have a vertical asymptote at $1 - n, n = 1, 2, \cdots$ – abel Dec 29 '14 at 04:09

2 Answers2

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If $\Gamma(z_0)=0$ then

$$0 = \Gamma(1-z_0)\Gamma(z_0) = \dfrac{\pi}{\sin(\pi z_0)}$$

And $\dfrac{1}{z}\neq 0$ for any $z\in\mathbb{C}$

David P
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  • :) Why did i miss this. Do you have any idea what the last statement about $\Gamma(n+1)$ contributes? – dietervdf Dec 29 '14 at 04:06
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    Ah, okay got it. It's there to handle the deleted elements from $\mathbb{N}$ which are omitted in $\mathbb{C}\setminus\mathbb{Z}$ – dietervdf Dec 29 '14 at 04:08
  • Not quite. For instance, we might have $\Gamma(1-z)\to\pm\infty$, in which case the product may not necessarily be $0$. In other words, it would probably be better to add a few words covering the special case when $z_0\in\mathbb N^*$. But that situation is easily ruled out, since $0\not\in\Gamma(\mathbb N)$. – Lucian Dec 29 '14 at 09:10
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The RHS is nonzero, so no factor of the LHS can be zero.

MPW
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