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I am currently taking an Analytic Number Theory unit and we're working on the zeros of the zeta function.

In the proof of $\zeta(1+\textit{i}t) \neq 0$ for $t \in \mathbb{R}$, we suppose that $\zeta$(s) had order of vanishing $\textit{m}_t$ at $\sigma + it$ and $\textit{m}_{2t}$ at $\sigma + 2it$.

We then use this to approximate as follows: $- \frac{\zeta'}{\zeta}(\sigma + it) = -\frac{\textit{m}_t}{\sigma-1} + \textit{O}(1)$, $- \frac{\zeta'}{\zeta}(\sigma + 2it) = -\frac{\textit{m}_{2t}}{\sigma-1} + \textit{O}(1)$, and $- \frac{\zeta'}{\zeta}(\sigma) = \frac{1}{\sigma-1} + \textit{O}(1)$.

I don't =understand how to obtain these approximations and I couldn't figure it out by myself, so I'd really appreciate some help!

Thank you.

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In general, when $f$ is a holomorphic function in a disk centered at $z_0$, and $z_0$ is a pole of order $1$ with residue $m$, we have $$f(z) = \frac{m}{z-z_0} + O(1) \qquad (z \to z_0)$$ This is because $f(z) - \frac{m}{z-z_0}$ has a removable singularity at $z_0$, so it is bounded in a neighborhood of $z_0$.

When $s_0 = 1+it$ is a zero of multiplicity $m_t$ of $\zeta$, one has that $s_0$ is a pole of $\zeta'/\zeta$ of order $1$ and residue $m_t$, so $$-\frac{\zeta'}{\zeta}(s) = -\frac{m_t}{s-s_0} + O(1) \qquad (s \to s_0)$$ Now let $s = \sigma + it$ and let $\sigma \to 1$ to obtain $$-\frac{\zeta'}{\zeta}(\sigma + it) = -\frac{m_t}{\sigma - 1} + O(1) \qquad (\sigma \to 1)$$

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