Is this analytic function linked to Barnes multiple gamma function ? Is it entire?

Question

I came across this series of following analytic functions $$ f_n\ :\ z\mapsto\frac{1}{\Big(\Pi_{k=0}^{n-1}\Gamma(1+ e^{2i\frac{k}{n}\pi}z)\Big)^{\frac{1}{n}}} $$
One can get easily a local (around zero) expansion (as an exponential, hence my question) with convergence radius $R=1$ and I wonder whether it can be continued as an entire function. Is it linked to some multiple gamma function defined by Barnes ?

Answer 1

Using a product representation of the gamma function your term is holomorphic in $\mathbb{C}$ without $-z\in\mathbb{N}$.

For $n\ge 2$ and on the right side $\,|z|<1\,$ (and $z=1)$ your term is equivalent to:

$$\left(\prod\limits_{k=1}^\infty\left(1+(-1)^{n-1}\left(\frac{z}{k}\right)^n\right)\right)^{\frac{1}{n}}=\exp\left(-\sum\limits_{k=1}^\infty\frac{(-z)^{nk}\zeta(nk)}{nk}\right)$$

Answer 2

This is the problem of n-th root of an entire function within the space of entire functions. For $n\geq 2$, there is no solution as soon as the original function has (at least) a simple zero. Considering $$ g_n\ :\ z\mapsto\frac{1}{\Big(\Pi_{k=0}^{n-1}\Gamma(1+ e^{2i\frac{k}{n}\pi}z)\Big)} $$ we have $g_n=f_n^n$ for $|z|<1$. The function $g_n$ is entire and has zeroes at the points such that $$ 1+ e^{2i\frac{k}{n}\pi}z\in \mathbb{Z}_{\leq 0} $$ for some $0\leq k<n$, this is the following disjoint union (each term of the union being the set of - simple - zeroes of the corresponding factor with the same $k$) $$ Z_n=\cup_{0\leq k\leq n-1} \{ne^{2i\pi\frac{(n-k)}{n}}\}_{n\leq -1} $$ hence the zeroes are simple whence the impossibility that $f_n$ be entire.