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This is a question distinct from but related to the question I wrote here: Question about proof that $\zeta(s) \neq 0$ for $\Re(s) = 1$, so assume the same things that I wrote there.

The paper then goes to show that the order of this zero of $\zeta$ we called $s_0=1+ia$ has order $\mu \leq 1$ (I have no trouble with this part). We want to show that $\mu=0$ implying that no such zero exists.

To that end, they say "if $\zeta(s)$ has a zero of order $\mu$ at s=1+ia$ (a \in \mathbb{R}, a \neq 0$) and a zero of order $\nu$ at $1+2ia$, then...". I'm okay with the idea of assuming that $1+ia$ is a zero and then proving that its order is zero, but I'm not okay with the additional assumption that $1+2ia$ is a zero.

My question is: why is this legitimate? Surely that $1+2ia$ is a zero follows from assuming $1+ia$ is a zero (if it didn't, then the theorem we are proving would no longer be true). If they are indeed assuming them both simultaneously, then what prevents a possible situation that $1+ia$ is a zero of $\zeta$ while simultaneously $1+2ia$ is NOT a zero of $\zeta$?

I'm currently using the formula $\zeta(s) = \displaystyle\sum_{n=1}^{\infty} \displaystyle\int_n^{n+1} \frac{1}{n^s} - \frac{1}{x^s} dx + \frac{1}{s-1}$ that is defined for $\Re(s) > 0$.

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up vote 3 down vote accepted

The simple answer is that $v$ could be $0$, and there is no zero at all. For this proof, all that matters is that $v$ is not a pole, i.e. that $v\geq 0$.

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That sort of makes sense, but I still have a problem with it. They are defining $1+2ia$ as "a zero" to get to the point they can show $\displaystyle\lim_{\epsilon \searrow 0} \epsilon \Phi(1+\epsilon \pm 2ia) = -\nu$. The argument here should be the same as it was for the zero at $1+ia$, but when I did that that one I used the fact that $1+ia$ was a pole of $\frac{\zeta'(s)}{\zeta(s)}$ to derive $\epsilon \Phi(1+\epsilon+ia) \rightarrow -\mu$. I can't run the same argument for $1+2ai$ if it's not a zero of $\zeta$. – tomcuchta Dec 6 '11 at 22:26
@tomcuchta: If it is not a zero, then $\lim_{\epsilon\rightarrow 0}\epsilon\Phi(1+\epsilon+2ia)$ just equals $0$. This is exactly what we want, because it equals $v$, and $v$ is zero in this case. – Eric Naslund Dec 6 '11 at 22:30

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