I am studying the book of John B. Conway Functions of One Complex Variable(1978), and in the section of Riemann Zeta Function chapter 7 I couldn't solve the last exercise. Here it is:
Let $\zeta (z)$ be the Riemann zeta function, which is meromorphic on C with a simple pole at z = 1 and holomorphic elsewhere, and set $$\eta(z)=\frac{\zeta'(z)}{\zeta(z)}$$ for $\operatorname{Re} z > 1$.
Show that for any $z_0$ with $\operatorname{Re} z_0 \geq 1$ $\lim\limits_{z \to z_0}(z-z_0) \eta(z)=N$, where $N \in \mathbb{Z}$.
There are 3 more following questions, this is only part (a) but I didn't wanted ask them all since if I can understand this one I might be able do the rest. So any hints/help is appreciated.
Since $\zeta(z)$ is holomorphic on $\operatorname{Re}{z}>1$, it follows that $$\zeta(z)=(z-z_0)^m(a_0+a_1(z-z_0)+\cdots)$$ with $m\in\mathbb{Z}_{\geq0}$ and $a_0\neq0$.
Then it follows that \begin{align} \zeta'(z)&=m(z-z_0)^{m-1}(a_0+a_1(z-z_0)+\cdots)+(z-z_0)^m(a_1+\cdots)\\ &=(z-z_0)^{m-1}(ma_0+a_1(m+1)(z-z_0)+\cdots) \end{align}
Now we calculate the limit: \begin{align} \lim_{z\to z_0}(z-z_0)\eta(z)&=\lim_{z\to z_0}\frac{(z-z_0)^m(ma_0+a_1(m+1)(z-z_0)+\cdots)}{(z-z_0)^m(a_0+a_1(z-z_0)+\cdots)}\\ &=\lim_{z\to z_0}\frac{ma_0+a_1(m+1)(z-z_0)+\cdots}{a_0+a_1(z-z_0)+\cdots}=m \end{align}
Because $a_0\neq0$, this limit exists and is equal to an integer $m$.
However, it is the question how often you encounter the case where $m\neq0$, because that implies a zero of the Riemann zeta function with $\operatorname{Re}{z}>1$...