Compute the residue at a pole of a meromorphic function

Question

I am working through a larger computation, and I have gotten stuck at a step where I need to compute the residue of $\frac{f(\zeta)}{(\zeta-z)\zeta^n}$ at the point $\zeta=0$. We are given that $f$ is a holomorphic function and I know that the residue should be equal to $$\frac{1}{(n-1)!}\lim_{\zeta\to0}\frac{d^{n-1}}{d\zeta^{n-1}}\frac{\zeta^nf(\zeta)}{(\zeta-z)\zeta^n}$$ But I have no idea how to get an expression for this.

Answer 1

I will assume that $z\neq 0$ in the following. Then, $ \frac{1}{(n-1)!} \frac{d^{n-1}}{d\zeta^{n-1}} \frac{f(\zeta)}{(\zeta - z)}$ is already a holomorphic function for $\zeta \neq z$ and we can obtain a result just by plugging in $\zeta = 0$.

To write it in terms of derivatives of $f$, you can use the product rule with binomial coefficients $\begin{pmatrix} n \\ k \end{pmatrix}$: $$ \frac{d^n}{d\zeta^n} g(\zeta) f(\zeta) = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} g^{(k)}(\zeta) f^{(n-k)}(\zeta). $$ For derivatives of $g$, one obtains $$ g(\zeta) = \frac{1}{(\zeta - z)} ~\Rightarrow~ g^{(k)}(\zeta) \Big|_{\zeta = 0} = \frac{(-1)^k k!}{(\zeta - z)^{k+1}} \Big|_{\zeta = 0} = -\frac{k!}{z^{k+1}}, $$ which leads to a fairly explicit result in terms of $f^{(k)}$: \begin{align} &\text{Res}_{\zeta = 0} \frac{f(\zeta)}{(\zeta - z)\zeta^n} = \frac{1}{(n-1)!} \frac{d^{n-1}}{d\zeta^{n-1}} \frac{f(\zeta)}{(\zeta - z)}\Big|_{\zeta = 0} \\ =& \frac{1}{(n-1)!} \sum_{k=0}^{n-1} \begin{pmatrix} n-1 \\ k \end{pmatrix} \frac{(-k!)}{z^{k+1}} f^{(n-1-k)}(0) = - \sum_{k=0}^{n-1} \frac{f^{(n-1-k)}(0)}{(n-1-k)!z^{k+1}} . \end{align} Let me know if there are any questions or I made any mistakes :)