Does analytic continuation preserve $\zeta(\overline{s})=\overline{\zeta(s)}$?

Question

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.

Answer 1

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 $\Omega$.

Answer 2

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 $\operatorname{Re}z>0.$