Another question about proof that $\zeta(s) \neq 0$ for $\Re(s) = 1$ 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 $\Re(s) > 0$.

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 $v\geq 0$.
