We want to show the Riemann zeta function's magnitude is mirrored above and below the real axis, but that it's argument (phase) is inverted:
$$ \zeta(\overline{s}) = \overline{\zeta(s)} $$
We will restrict ourselves to $\sigma>0$ where $s=\sigma+it$.
Question: is the proof correct?
For the simplest case, we'll use $\zeta(s)=\sum 1/n^s$, valid for $\sigma>1$.
We consider each term in the series:
$$n^{-\overline{s}}=e^{-\overline{s}\ln(n)}=\overline{e^{-sln(n)}}=\overline{n^{-s}}$$
The magnitudes at $s$ and $\overline{s}$ are easy to equate.
$$|n^{-\overline{s}}|=|\overline{n^{-s}}|$$
This means $\zeta(\overline{s})$ is the complex conjugate of $\zeta(s)$. So, although the magnitude of $\zeta(s)$ is mirrored above and below the real axis, the sign of the imaginary part is inverted.
For the extended case, we'll use the series valid for $\sigma>0$
$$\zeta(s)=\frac{1}{1-2^{1-s}}\cdot\eta(s)$$
where eta $\eta(s)$ is the alternating zeta function.
Here we can immediately say:
$$\eta(\overline{s})=\overline{\eta(s)}$$
because it is the same as $\sum 1/n^s$ but with alternating signs.
Then we look at the other factor:
$$(1-2^{1-\overline{s}})^{-1} = \overline{\left(1-2^{1-s}\right)^{-1}}$$
because $1/\overline{z}=\overline{1/z}$ which can be shown using polar representations.
Therefore we have:
$$\begin{align} \zeta(\overline{s}) &= ({1-2^{1-\overline{s}})^{-1}}\cdot\eta(\overline{s})\\ \\ &=\overline{(1-2^{1-s})^{-1}}\; \cdot \; \overline{\eta(s)} \\ \\ &= \overline{\zeta(s)} \end{align}$$
Giving us the same conclusions that the magnitude is mirrored in the real axis, but the phase is inverted.
