# Gamma function, zeros

• May 4th 2010, 11:25 AM
pascal4542
Gamma function, zeros
Show that $\Gamma(z)$ has no zeros.

I am not sure how to prove this. I would appreciate a few hints or suggestions. $\Gamma(z)$ is the gamma function. The gamma function is meromorphic. It is defined in the right half-plane by $\Gamma(z)= \int_0^{\infty} e^{-t}t^{z-1}dt$ for $\text{Re}(z)>0$. There is also another representation of $\Gamma(z)=\frac{\Gamma(z+m)}{(z+m-1) \cdots (z+1)z}$ where the right-hand side is defined and meromorphic for $\text{Re}(z)>-m$ with simple poles at $z=0, -1, \ldots, -m+1$. Thank you.
• May 4th 2010, 11:35 AM
Bruno J.
What about $\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin{\pi z}}$?
• May 4th 2010, 11:39 AM
chisigma
The most simple way to demonstrate that $\Gamma (z)$ has no zeroes is to expand its inverse as 'infinite product'...

$\frac{1}{\Gamma (z)} = z\cdot e^{\gamma z}\cdot \prod_{n=1}^{\infty} (1+\frac{z}{n})\cdot e^{-\frac{z}{n}}$ (1)

The (1) shows that $\frac{1}{\Gamma (z)}$ is an entire function, so that it has no poles... and therefore $\Gamma (z)$ has no zeroes...

Kind regards

$\chi$ $\sigma$