I ask about this claim: let $f$ be an entire function satisfying $f(s)=u(s)f(as)$. Assume that $s$ and $as$ are not zeroes of $f$ and $f (bar)(as)=f(s)$ in a region $D$ ($f(bar)$ is the conjugate of $f$). Then the module of $f(s)/f(bar)(as)$ 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.

Technically, the answer is yes $u(s)=1$. However, for a quite boring reason (as hinted at in the comment of Xogn Ambandl): If $\overline{f(as)}=f(s)$, then $2 \ \Re{f}(s)$ would be holomorphic in that region, as it is equal to $f(s)+\overline{f(s)}= f(s) + f(as)$ a sum of holomorphic functions. Yet, then as a realvalued 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. 

