I ask about this claim:
let $f$ be an entire function satisfying $f(s)=u(s)f(a-s)$. Assume that $s$ and $a-s$ are not zeroes of $f$ and $f (bar)(a-s)=f(s)$ in a region $D$ ($f(bar)$ is the conjugate of $f$). Then the module of $f(s)/f(bar)(a-s)$ is equal to $1$, implying that the module of $u(s)$ is also $1$. The question is: Does this result implies that in fact the function $u(s)$ is constant.
Thank you in advance.



Functional equation and constant functions
Technically, the answer is yes $u(s)=1$. 


However, for a quite boring reason (as hinted at in the comment of Xogn Ambandl):


If $\overline{f(a-s)}=f(s)$, then $2 \ \Re{f}(s)$ would be holomorphic in that region, as it is equal to $f(s)+\overline{f(s)}= f(s) + f(a-s)$ a sum of holomorphic functions. Yet, then 
as a *real-valued* holomorphic function $\Re{f}(s)$ is constant. And, so $f(s)$ is constant. 


Thus, the only functions $f$ fulfilling your assumptions are constant functions, for which what you ask about is clear.


