Use the Laurent series to show that the residue at a pole of the order $m$ can be evaluated by:

$$\operatorname{Res}_{z=z_0}f(z)=\frac{1}{(m-1)!}\lim_{z\to z_0}\frac{d^{m-1}}{dz^{m-1}}((z-z_0)^mf(z))$$

I have no clue how to approach this problem. First, it does look extremely related to Cauchy's Integral Theorem, or at least some generalization of it. However, I'm not exactly sure what I should be expanding and manipulating to show this.

If it's just straight forward plug in, then how do I simplify?

The definition I've been given is

$$f(z)=\sum_{k=0}^\infty a_k(z-z_0)^k+\sum_{k=1}^{\infty}a_{-k}(z-z_0)^{-k}$$

which I know can just be written as:

$$f(z)=\sum_{k=-\infty}^\infty a_k(z-z_0)^k$$

Using this directly we get

$$\operatorname{Res}_{z=z_0}f(z)=\frac{1}{(m-1)!}\lim_{z\to z_0}\frac{d^{m-1}}{dz^{m-1}}((z-z_0)^m\sum_{k=-\infty}^\infty a_k(z-z_0)^k)\implies$$

$$\operatorname{Res}_{z=z_0}f(z)=\frac{1}{(m-1)!}\sum_{k=-\infty}^\infty\lim_{z\to z_0}\frac{d^{m-1}}{dz^{m-1}}(a_k(z-z_0)^{k+m})$$

But I don't know how to proceed beyond this point...

  • $\begingroup$ First of all, you need to use the fact that the pole is of order $m$, in which case the Laurent expansion should look like $ f(z) = \sum_{k=-m}^{\infty}c_{k}(z-z_0)^k $. Then, what happens to all but the lowest-order term after you've taken $m-1$ derivatives and evaluated the limit $z\to z_0$? $\endgroup$
    – march
    Apr 27, 2022 at 3:40
  • $\begingroup$ @march If we start at $-m$ and we take $m-1$ derivatives, the lowest order term gets an alternating multiplication factorial $m$, but as we sweep through the derivatives, each successive term will have the form $a_i(z-z_0)^m$ at some point. I'm not exactly sure what happens in the limit. It seems like only the $\infty$ term is nonzero? $\endgroup$
    – help
    Apr 27, 2022 at 4:01

1 Answer 1


It's a basic result of complex analysis, here an approach.

Suppose that $f$ is analytic insde and on a simple closed curve $\gamma$, so indeed $f$ can be written by Laurent series as $$f(z)=\sum_{k=-\infty}^{+\infty}a_{k}(z-z_{0})^{k}$$ But we know that $f$ has a pole $z_{0}$ of order $m$ (we need the pole inside $\gamma$), so $f$ can be written as $$f(z)=\sum_{k=0}^{+\infty}a_{k}(z-z_{0})^{k}+\sum_{k=1}^{m}a_{-k}(z-z_{0})^{-k}$$ $$\implies (z-z_{0})^{m}f(z)=a_{-m}+a_{-m+1}(z-z_{0})+\cdots+a_{-1}(z-z_{0})^{m-1}+a_{0}(z-z_{0})^{m}+\cdots$$ $$\implies \frac{{\rm d}^{m-1}}{{\rm d}z^{m-1}}\left( (z-z_{0})^{m}f(z)\right)=(m-1)!a_{-1}+m(m-1)\cdots 2a_{0}(z-z_{0})+\cdots $$ $$\implies \lim_{z\to z_{0}} \frac{{\rm d}^{m-1}}{{\rm d}z^{m-1}}\left( (z-z_{0})^{m}f(z)\right)=\lim_{z\to z_{0}}(m-1)!a_{-1}+m(m-1)\cdots 2a_{0}(z-z_{0})+\cdots $$ $$\implies \lim_{z\to z_{0}} \frac{{\rm d}^{m-1}}{{\rm d}z^{m-1}}\left( (z-z_{0})^{m}f(z)\right)=(m-1)!a_{-1}$$ $$\implies a_{-1}:={\rm Res}(f,z_{0})= \lim_{z\to z_{0}} \frac{1}{(m-1)!} \frac{{\rm d}^{m-1}}{{\rm d}z^{m-1}}\left( (z-z_{0})^{m}f(z)\right)$$ Therefore, $$\boxed{{\rm Res}(f,z_{0})= \frac{1}{(m-1)!} \lim_{z\to z_{0}} \frac{{\rm d}^{m-1}}{{\rm d}z^{m-1}}\left( (z-z_{0})^{m}f(z)\right) }$$

  • $\begingroup$ The first implication is a bit unclear to me. $(z-z0)^mf(z) = a_m + \cdots$. Are you getting this by fulling expanding the expansion for $f$ in the line above? $\endgroup$
    – help
    Apr 27, 2022 at 4:42
  • $\begingroup$ First exand $f(z)$ then multiplying both sided by $(z-z_{0})^{m}$ you obtained that line. $\endgroup$
    – A. P.
    Apr 27, 2022 at 4:45

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