If $\zeta(1+it_0)=0$ then $|\zeta (\sigma)^3 \zeta (\sigma +it_0)^4 \zeta(\sigma + 2it_0)| \longrightarrow 0$ as $\sigma \longrightarrow 1$

Question

I need to prove that if $\zeta(1+it_0)=0$ then $|\zeta (\sigma)^3 \zeta (\sigma +it_0)^4 \zeta(\sigma + 2it_0)| \longrightarrow 0$ as $\sigma \longrightarrow 1$, where $\zeta$ is the Riemann Zeta function. I guess I need to show that $\zeta(\sigma+it_0)^4$ approaches $0$ faster than $\zeta (\sigma)^3\zeta (\sigma + 2it)$ approaches $\infty$ when $\sigma \longrightarrow 1$. But don't know how to get around that.

Edit: The purpose of this computation is to show, by contradiction, that $\zeta$ has no zeros at the line $1+it$.

Answer 1

$\zeta(1+it_0)=0$, thus if $n$ denotes the multiplicity of $1+it_0$ as a zero of $\zeta$, there exists $g$ holomorphic such that $\zeta(\sigma+it_0)=(\sigma-1)^ng(\sigma)$ and $g(1)\neq 0$ around $\sigma=1$. We thus have $$ |\zeta(\sigma)^3\zeta(\sigma+it_0)^4\zeta(\sigma+2it_0)|=(\sigma-1)^3\zeta(\sigma)^3|\cdot(\sigma-1)^{4n-3}|g(\sigma)|\cdot|\zeta(\sigma+2it_0)|\underset{\sigma\rightarrow 1^+}{\longrightarrow}0 $$ because $\lim\limits_{\sigma\rightarrow 1^+}(\sigma-1)^3\zeta(\sigma)^3=1$ and $4n-3\geqslant 1$. (Which btw leads to a contradiction since $|\zeta(\sigma)^3\zeta(\sigma+it_0)^4\zeta(\sigma+2it_0)|\geqslant 1$ for all $\sigma>1$)

Answer 2

This is a standard computation. By Euler's product identity, for all complex numbers with $\operatorname{Re}(s):=\sigma>1$ \begin{align*} \log(\zeta(s))=\sum_{n,m}\frac{1}{m}\frac{1}{p^{ms}_n} \end{align*} where $\{p_n:n\in\mathbb{N}\}$ is the enumeration of the prime numbers listed according to the usual order in $\mathbb{N}$. Then, for any real $\tau\neq0$ and $\sigma>1$ \begin{align*} \log\Big(\big|\zeta^3(\sigma)\zeta^4(\sigma+i\tau)\zeta(\sigma+i2\tau)\big|\Big) = \sum_{n,m}\frac{1}{m}\frac{1}{p^{\sigma m}_n} \operatorname{Re}\Big(3+\frac{4}{p^{4m\tau i}_n}+\frac{1}{p^{2m\tau i}_n}\Big)\geq0 \end{align*} since $\operatorname{Re}(3+4e^{i\theta}+e^{2i\theta})=2(1+\cos\theta)^2\geq0$ for all $\theta\in\mathbb{R}$. It follows that $$\begin{align} |(\sigma-1)\zeta(\sigma)|^3\Big|\frac{\zeta(\sigma+i\tau)}{\sigma-1}\Big|^4|\zeta(\sigma+i2\tau)| \geq\frac{1}{\sigma-1} \tag{1}\label{riemann-at-1} \end{align} $$ If $\zeta(1+it_0)=0$ for some $t_0\in\mathbb{R}$, the left--hand--side of~\eqref{riemann-at-1} converges to $|\zeta'(1+i\tau)|^4|\zeta(1+i2\tau)|$ as $\sigma\searrow1$, where $\tau=t_0$. The right--hand--side of \eqref{riemann-at-1} however, tends to $\infty$ and so, we get a contradiction. Therefore, $\zeta$ has no zeroes along the line $1+i\tau$, $\tau\in\mathbb{R}$