Does analytic continuation preserve $\zeta(\overline{s})=\overline{\zeta(s)}$? It is easy to show that the for the series representing the Riemann zeta function for $\Re(s)>1$
$$\zeta(s)=\sum\frac{1}{n^s}$$
the symmetry $\zeta(\overline{s})=\overline{\zeta(s)}$ holds.
Question: Does this symmetry hold for any analytic continuation of this series? If so, why?

For example, by doing the laborious algebra, we can show the symmetry also holds for the following continuation valid for $\Re(s)>0$
$$\zeta(s)=\frac{1}{1-2^{1-s}}\sum\frac{(-1)^{n+1}}{n^{s}}$$
I would like to state that the symmetry holds by referring to a general principle of complex functions - which I don't know.
 A: Yes. Consider the function
$$ f(s) = \zeta(s) - \overline{\zeta(\overline{s})}. $$
Since $\zeta(s)$ is analytic on the domain $\Omega = \mathbb{C}\setminus\{1\}$, $f(s)$ is also analytic on $\Omega$ and satisfies $f(s) = 0$ over $(1, \infty)$. So by the principle of analytic continuation, $f$ is identically zero on all of $\Omega$.
A: More generally: If $f(z)$ is analytic on $U\subseteq\mathbb C,$ then $\overline{f(\overline z)}$ is analytic on $\overline U=\{\overline u\mid u\in U\}.$
If also $f(z)$ is real for some open interval on the real line, then $g(z)=f(z)-\overline {f(\overline z)}$ is analytic on $U\cap \overline U,$ including that open interval on the real line.
So $g(z)$ is analytic and $0$ on an open interval of the real line, hence $g$ is zero at least in the connected component containing that interval.
But $g(z)=0$ means $f(z)=\overline{f(\overline z)}$ means: $$\overline{f(z)}=f(\overline z).$$
In the case of $\zeta,$ $U=\mathbb C\setminus \{1\}$ has $\overline U=U.$

An example of where they are not always equal everywhere is if we define $f(z)=\sqrt z$ on $\mathbb C\setminus i\mathbb R^{\geq 0}.$ Then $f(z)$ and $\overline {f(\overline z))}$ are only equal when $\operatorname{Re}z>0.$
