
Gamma function, zeros
Show that $\displaystyle \Gamma(z)$ has no zeros.
I am not sure how to prove this. I would appreciate a few hints or suggestions. $\displaystyle \Gamma(z)$ is the gamma function. The gamma function is meromorphic. It is defined in the right halfplane by $\displaystyle \Gamma(z)= \int_0^{\infty} e^{t}t^{z1}dt$ for $\displaystyle \text{Re}(z)>0$. There is also another representation of $\displaystyle \Gamma(z)=\frac{\Gamma(z+m)}{(z+m1) \cdots (z+1)z}$ where the righthand side is defined and meromorphic for $\displaystyle \text{Re}(z)>m$ with simple poles at $\displaystyle z=0, 1, \ldots, m+1$. Thank you.

What about $\displaystyle \Gamma(z)\Gamma(1z) = \frac{\pi}{\sin{\pi z}}$?

The most simple way to demonstrate that $\displaystyle \Gamma (z)$ has no zeroes is to expand its inverse as 'infinite product'...
$\displaystyle \frac{1}{\Gamma (z)} = z\cdot e^{\gamma z}\cdot \prod_{n=1}^{\infty} (1+\frac{z}{n})\cdot e^{\frac{z}{n}}$ (1)
The (1) shows that $\displaystyle \frac{1}{\Gamma (z)}$ is an entire function, so that it has no poles... and therefore $\displaystyle \Gamma (z)$ has no zeroes...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$