Suppose $f(x+1)=f(x)$ is real analytic for $x\in \mathbb{R}\setminus \mathbb{Z}$. Moreover, it is once-differentiable at $x\in \mathbb{Z}$.
In addition, suppose that on the interval $(0,1)$, we know the analytic continuation into the complex plane. Let us call this function $g(z)$, where $0<\mathrm{Re}(z)<1$. Finally, we also know that $g(z+n)=g(z)$ for some integer $n>1$. So really, we know the analytic continuation of $f(x)$ for $\mathrm{Re}(z)\in (mn,mn+1)$ for $m\in\mathbb{Z}$.
I want to find the analytic continuation valid for any $\mathrm{Re}(z)\in \mathbb{R}\setminus \mathbb{Z}$. An obvious continuation is that one takes $g([z])$, where the fractional part is defined as: $$[z]=z+k \quad, k\in\mathbb{Z}\quad \text{such that}\quad 0<z+k<1.$$ This function agrees with $f(x)$ on $\mathbb{R}\setminus \mathbb{Z}$. For the specific function at hand, it turns out that $g(z)$ at $z=0,1$ is once differentiable, even though the fractional part is not defined there. This matches with the behaviour of $f(x)$.
My question: is this continuation unique?