For $s=\sigma+it$, the Riemann Zeta series
$$\zeta(s)=\sum \frac{1}{n^s}$$
valid for $\sigma>1$, and the extended version
$$\zeta(s)=\frac{1}{1-2^{1-s}}\sum\frac{(-1)^{n+1}}{n^{s}}$$
valid for $\sigma>0$, and both symmetric about the real axis.
That is, $\zeta(\sigma+it)=\zeta(\sigma-it)$.
Question: How can I prove the functions are symmetric about the real-axis, that is, $\zeta(\sigma+it)=\zeta(\sigma-it)$
Note - the suggested question doesn't answer this question sufficiently clearly: Is Riemann Zeta Function symmetrical about the real axis?