Compute derivative of $\frac{f(z)}{f'(z)}$ at a root of $f(z)$. Let $h(z)=\frac{f(z)}{f'(z)}$ where $f$ is holomorphic and $0$ is a zero of $f$ of order $n\geq 1$. Then
$$h'(0)=\lim_{z\to a} \left(1-\frac{f(z)f''(z)}{f'(z)^2}\right).$$
I want to know the value of $h'(0)$ in terms of $n$ (I conjecture the value is $\frac{1}{n})$.
This came up while trying to find the residue of $\frac{g(z)f'(z)}{f(z)}$ at $0$.
What kind of approach should I take for this last step?
 A: The Taylor series of $f$ near $a$ is of the type $a_n(z-a)^n+a_{n+1}(z-a)^{n+1}+\cdots$, with $a_n\neq0$. So, the power series of $f'$ near $a$ is $na_n(z-a)^{n-1}+(n+1)a_{n+1}(z-a)^n+\cdots$. Therefore\begin{align}h(z)&=\frac{a_n(z-a)^n+a_{n+1}(z-a)^{n+1}+\cdots}{na_n(z-a)^{n-1}+(n+1)a_{n+1}(z-a)^n+\cdots}\\&=\frac1{z-a}\frac{a_n+a_{n+1}(z-a)+\cdots}{na_n+(n+1)a_{n+1}(z-a)+\cdots}\\&=\frac1{n(z-a)}\frac{1+\frac{a_{n+1}}{a_n}(z-a)+\cdots}{1+(n+a)\frac{a_{n+1}}{a_n}(z-a)+\cdots}\end{align}and so your conjecture is correct.
A: It's easiest to consider this case when the root of $f$ is at zero.  In this case,
$$
f(z)=z^ng(z)
$$
for some homomorphic function $g(z)$ which is nonzero at $0$.  Then
$$
f'(z)=nz^{n-1}g(z)+z^ng'(z).
$$
Therefore, the function that you are considering is
$$
h(z)=\frac{f(z)}{f'(z)}=\frac{z^ng(z)}{nz^{n-1}g(z)+z^ng'(z)}.
$$
Moreover,
$$
\frac{zg(z)}{ng(z)+zg'(z)}
$$
is a meromorphic continuation of $h(z)$ which is defined at $0$.  So, we'll call this function $h(z)$.  Then
$$
h'(z)=\frac{(g(z)+zg'(z))(ng(z)+zg'(z))-zg(z)(ng'(z)+g'(z)+zg''(z))}{(ng(z)+zg'(z))^2}
$$
At zero, this simplifies to
$$
h'(0)=\frac{ng^2(0)}{n^2g^2(0)}=\frac{1}{n},
$$
as desired.
