Closed form of a function $f(z)$ whose $\it{only}$ zeros are $n\pi$,$n\pi\omega$ and $n\pi\omega^2$

Question

I need a closed form of a function $f(z)$ whose $\it{only}$ zeros are $n\pi$,$n\pi \omega$ and $n\pi\omega^2 $ where $\omega$ is a cube root of unity and $n\in\mathbb{N}$.

So we need a function of type $$f(z)=\prod_{n=1}^{\infty}\left(1-\frac{z}{n\pi}\right) \left(1-\frac{z}{n\pi\omega}\right) \left(1-\frac{z}{n\pi\omega^2}\right) $$

To find a closed form for above function we take logarithmic derivatives assuming $z$ which are not the points of zeros to get $$\frac{f'(z)}{f(z)}=\sum_{n=1}^{\infty}\left(\frac{1}{z-n\pi}\right)+\sum_{n=1}^{\infty}\left(\frac{1}{z-n\pi\omega}\right)+ \sum_{n=1}^{\infty}\left(\frac{1}{z-n\pi\omega^2}\right) $$

Any help would be highly appreciated.

Answer 1

Gamma function has no zeros and only simple poles in $0, -1,-2,-3, \ldots$ so $\frac{1}{\Gamma\left(1-\frac{z}{\pi}\right)}$ has only simple zeros at $\pi,2\pi,3\pi,\ldots$. Now this means that:

$$\frac{1}{\Gamma\left(1-\frac{z}{\pi}\right)\Gamma\left(1-\frac{z}{\omega\pi}\right)\Gamma\left(1-\frac{z}{\omega^2\pi}\right)}$$

is the function you are looking for.