# Approximation of $\zeta$(s) at $s=1$

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.

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 $$-\frac{\zeta'}{\zeta}(\sigma + it) = -\frac{m_t}{\sigma - 1} + O(1) \qquad (\sigma \to 1)$$