Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be a holomorphic function and suppose that there is a point $z_0 \in \mathbb{C}$ and a real $R > \lvert z_0 \rvert$ such that
- $f(z_0) = 0$
- $f(z) \ne 0$ for all $z \ne z_0$ with $\lvert z \rvert < R$
- $f^\prime(z_0) \ne 0$
Derive an asymptotic expression for $[z^n] \frac{1}{f(z)}$ up to a multiplicative error of $1+\mathcal{O}(1/n)$.
I understand that we may write $$\frac{1}{f(z)} = \frac{1}{f^\prime(z_0)(z-z_0)} + H(z)$$
, for $H(z)$ holomorphic.
Proof: Using the Taylor series expansion of $f$ around $z_0$ we have
\begin{equation} H(z) := \frac{1}{f(z)} - \frac{1}{f^\prime(z_0)(z-z_0)} = \frac{1}{f^\prime(z_0)(z-z_0)+(z-z_0)^2G(z)} - \frac{1}{f^\prime(z_0)(z-z_0)} \end{equation}
where $G(z) = \sum_{n \ge 2} f^{(n)}(z) (z-z_0)^{n-2}$. So $H(z)$ is a combination of holomorphic functions it is thus itself holomorphic.
However, I do not see how to continue from here. Could you please give me a hint?
As OP observed, we may decompose $1/f(z)$ as
$$ \frac{1}{f(z)} = \frac{1}{f'(z_0)(z - z_0)} + H(z) $$
for some holomorphic function $H(z)$ on $|z| < R$. Now by expanding the right-hand side for $|z| < |z_0|$,
$$ \frac{1}{f(z)} = \sum_{n=0}^{\infty} \left( [w^n]H(w) - \frac{1}{f'(z_0)z_0^{n+1}} \right) z^n, $$
Now Cauchy–Hadamard theorem gives a bound on the asymptotic behavior of the coefficients $[w^n]H(w)$ in terms of the radius of convergence $R'$ of $H(w)$. This and the obvious relation $R \leq R'$ together, we get
$$ [w^n]H(w) = \mathcal{O}(R^{-n + o(n)}). $$
Therefore we obtain an asymptotic formula of the form
$$ [z^n]\frac{1}{f(z)} = - \frac{1}{f'(z_0)z_0^{n+1}} + \mathcal{O}(R^{-n + o(n)}). \tag{*} $$
The error term in the equality above decays much faster than $\frac{1}{n|z_0|^n}$, hence $\text{(*)}$ is the desired expression.
Remark. This is an elementary example of Darboux method.