A quick way to calculate residues of logarithmic derivatives Assume that $U\subset \mathbb{C}$ is open an let $f(s)$ be a meropmorphic function on $U$. Consider the logarithmic derivative of $f(s)$ where I mean the meromorphic function $\frac{f'(s)}{f(s)}$. I am currently working on a proof of the following statement:


Statement: If $s_0 \in U$ is a zero of $f(s)$ of order $k$, then $s_0$ is a pole of $\frac{f'(s)}{f(s)}$ of order $k$ with residue $k$.


I have no problems in proving this. My approach here is to consider the expansions
$$f(s)=\sum_{n=k}^{\infty}a_n(s-s_0)^n,\phantom{aa}f'(s)=\sum_{n=k}^{\infty}na_n(s-s_0)^{n-1},\phantom{aa}\frac{1}{f(s)}=\sum_{m=1}^{k} \frac{b_m}{(s-s_0)^m} + h(s)$$
with $h(s)$ analytic in a region containing $s_0$. I use those expansions to compute $\frac{f'(s)}{f(s)}$ which shows that the residue is $ka_kb_k$. I then show that $a_kb_k=1$ by considering the equation $1=\frac{f(s)}{f(s)}$ where the right-hand-side is again computed by using the expansions from above.


The approach to the problem I described above is straightforward and not really difficult but requires some annoying formal calculations (double sums, manipulating indices, etc.). Is there a quicker way to prove the statement?
 A: Your approach is not the simplest or easier as you are considering the expansion of the reciprocal of an analytic function, what requires annoying computations.
Instead you can just note that as $f$ have a zero of order $k$ in $z_0$ then if $f$ is not identically zero then there exists a neighborhood of $z_0$, say $U$, where $f(z)=(z-z_0)^k g(z)$ for some analytic function $g$ such that $g\neq 0$ in $U$. Therefore
$$
\frac{f'(z)}{f(z)}=\frac{k(z-z_0)^{k-1}g(z)+(z-z_0)^k g'(z)}{(z-z_0)^k g(z)}=\frac{k}{z-z_0}+h(z)
$$
where $h$ is an analytic function in $U$, and the conclusion follows immediately.∎
A: Your approach is the quick way. WLOG, we can assume $s_0 = 0$. Write $f(z) = a_kz^k + a_{k + 1}z^{k + 1} + \dots$ with $a_k \neq 0$, for $z$ in a neighborhood of $0$. We have
$$\frac{f'(z)}{f(z)} = \frac{ka_{k}z^{k - 1} + O(z^k)}{a_kz^k + O(z^{k + 1})} = \frac{k + O(z)}{z + O(z^2)} = \frac{1}{z}\frac{k + O(z)}{1 + O(z)}.$$
This immediately yields that $\frac{f'(z)}{f(z)}$ has a pole of order $1$ (not of order $k$ as you claim) at $0$ and that the residue is $k$.
