MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It has been proven that:

1) if $s$ is a non trivial zero $\rho$ of $\zeta(s)$ then so is $1−s$.

2) $\zeta(s) = 2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \Gamma(1-s) \zeta(1-s)$

3) $ 0 < \Re(\rho) <1$

From this it follows that when $s \to \rho$:

$\displaystyle \lim_{s \to \rho} |\dfrac{\zeta(s)}{\zeta(1-s)}| = |2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \Gamma(1-s)|=1$

It is easy to see that the outcome will be $1$ for all $y$ in $s=\frac12 + y i$.

But if a $\rho$ would lie off this critical line, it also must reside in 'spots' where $\displaystyle \lim_{s \to \rho} |\dfrac{\zeta(s)}{\zeta(1-s)}|=1$.

On which points off the critical line could this occur? I found a surprisingly small domain (no proof).

The blue line shows the only values where:

$\displaystyle |2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \Gamma(1-s)|=1$, $s=x + y i$, $ 0 \le x \le 1$.

Note that $y \to 2\pi$ for both $x=0$ and $x=1$. The $y$ rises only a little in the middle.

This doesn't say anything about whether or not off-line $\rho$'s are actually hiding on this curve. There still is an infinite number to check. However, I wondered if anything more is known about this curve?


Please click here for the picture

share|cite|improve this question
up vote 5 down vote accepted

I believe you're mistaken that $$ \lim_{s\to\rho}\left|\frac{\zeta(s)}{\zeta(1-s)}\right|=1. $$ Write $\zeta(1-s)=\zeta(s)f(s)$ with $f(s)$ as implied by your equation (2). The series expansion for $\zeta(s)$ at $s=\rho$ is $$ \zeta(s)=\zeta^\prime(\rho)(s-\rho)+O(s-\rho)^2. $$ The series expansion for $\zeta(1-s)$ at $s=\rho$ is $$ \zeta(1-s)=\zeta(s)f(s)=\zeta^\prime(\rho)f(\rho)(s-\rho)+O(s-\rho)^2. $$ By standard manipulation of series, $$ \frac{\zeta(s)}{\zeta(1-s)}=\frac{1}{f(\rho)}+O(s-\rho), $$ so the limit should equal $1/f(\rho)$.

share|cite|improve this answer

There is some mild confusion here. Yes, for $s$ with real part $1/2$ the function $f(s)=2^{s}\pi^{s-1} \sin(\pi s/2)\Gamma(1-s)$ has magnitude $1$, which is an easy consequence of $|\Gamma(1/2+it)|=\sqrt{\frac{\pi}{\cosh {\pi t}}}$, but $|f(s)| \neq 1$ in general since a meromorphic function whose magnitude is constant on any open set is necessarily a constant. The limit $\lim_{s \to \rho} \frac{\zeta(s)}{\zeta(1-s)}$ is $f(\rho)$, but this doesn't put any constraint on $\rho$...

share|cite|improve this answer
Many thanks David and Stopple. The mistake is clear and I got a good learning out of it. I am also convinced now that the reflection formula is not going to reveal any information about the $\rho$'s. They seem to originate from somewhere deep down inside $\zeta(s)$ itself. – Agno Jan 10 '12 at 21:43

Let $\chi(s)=2^s \pi^{s-1}\sin(\pi s/2) \Gamma(1-s)$ so that $\zeta(s)=\chi(s)\zeta(1-s)$. You are asking about the curve $|\chi(s)|=1$.

As you have observed, $|\chi(1/2+it)|=1$ for real $t$. There is a partial converse to this statement, namely that there is a positive absolute constant $C_0$ such that if $|\chi(\sigma+it)| = 1$ with $0 \le \sigma \le 1$ and $|t| \ge C_0$, then $\sigma=1/2$.

A simple proof can be found in Lemma 6.1 of S. M. Gonek "Finite Euler products and the Riemann hypothesis" Trans. Amer. Math. Soc. 364 (2012), 2157-2191. This paper is also on the arXiv. Gonek states that $C_0<6.3$ so it seems that phenomena in your pictures stops shortly after the ranges you plotted.

share|cite|improve this answer
Thanks Micah. That is exactly what I have observed. To be even more precise: the phenomenon where $\chi(s)=1$ for $\sigma \ne \frac12$ and $0 \le \sigma \le 1$ only occurs when $2 \pi \le |t| \le 6.2898359888...$. All other values for $|t|$ only generate a single $1$ at $\sigma = \frac12$. – Agno Jan 10 '12 at 22:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.