A common strategy taken by mathematicians wrestling with the Riemann Hypothesis is to explore the properties of its zeros. Here we'll take our first steps along this path.
The video for this topic is here [youtube], and the slides are here [pdf].
Property $\zeta(\overline{s})=\overline{\zeta(s)}$
We previously showed the property
$$\zeta(\overline{s})=\overline{\zeta(s)}$$
holds for the series $\zeta(s)=\sum1/n^{s}$, valid for $\sigma>1$.
We can show the new series extending $\zeta(s)$ to $\sigma>0$ maintains this property.
Instead of doing this with slightly laborious algebra, we'll take this opportunity to introduce the more elegant and rather powerful principle of analytic continuation, which we'll define properly in a separate blog post.
Using Analytic Continuation
Let's construct a function
$$f(s)=\zeta(s)-\overline{\zeta(\overline{s})}$$
Wherever $\zeta(s)$ is analytic, so is $f(s)$.
We know $f(s)=0$ along the real line where $\sigma>1$. Using analytic continuation, $f(z)$ must also be zero in any domain that $f(s)$ is analytic, as long as that domain includes the real line $\sigma>1$.
So $f(s)=0$ in the complex half-plane $\sigma>1$, but also $\sigma>0$ because we extended $\zeta(s)$ to this larger domain, where it remains analytic except at $s=1$.
But $f(s)=0$ means $\zeta(\overline{s})=\overline{\zeta(s)}$, which means this property holds in $\sigma>0$. If later we are able to extend $\zeta(s)$ into $\sigma<0$, this property will continue to hold there too.
Symmetric Zeros
If $\zeta(s)=0$ then the property $\zeta(\overline{s})=\overline{\zeta(s)}$ tells us $\zeta(\overline{s})=0$.
$$\zeta(s)=0\implies\zeta(\overline{s})=0$$
This means the zeros exist in symmetric pairs $\sigma+it$ and $\sigma-it$, or exist on the real line where $t=0$.
Thoughts
For a first attempt, this is quite an enlightening insight into the zeros of the Riemann Zeta function $\zeta(s)$.
Zeros existing in symmetric pairs $\sigma\pm it$ is compatible with the Riemann Hypothesis, but sadly it doesn't mean they all lie on a single line $\sigma=a$, never mind the holy grail $\sigma=1/2$.
