I would like to know under which circumstances the two following procedures for calculating the residue of a function $f$ at a pole of order $N>1$ are equivalent.

i) evaluating the residue using the standard formula which requires taking $N-1$ derivatives of $f$;

ii) considering that the multiple pole is a set of $N$ simple poles $z_1,...,z_N$ very close to each other. Then you compute the sum of the residues of $f$ at these simple poles and pass to the limit $z_2,z_3,...z_N$ tending to $z_1$.

I have looked for any result like this but I could not find this second method as a form of calculating residues at higher order poles. Under which circumstances it can be applied? Is there a general result like this in complex analysis?


Let $f(z,t)$ be a function from $U \times [0,\epsilon)$ to the Riemann sphere (extended complex plane), where $U$ is a neighbourhood of $z_0$ in $\mathbb C$, and let $\Gamma$ be a simple closed contour in $U$ enclosing $z_0$. Suppose

  1. $f(\cdot, t)$ is meromorphic in $U$ for every $t \in [0,\epsilon)$, with no poles on $\Gamma$.
  2. $f$ is continuous on $\Gamma \times [0,\epsilon)$.

Then the sum of the residues of $f(\cdot, t)$ at the poles inside $\Gamma$ is a continuous function of $t$ (since it is given by a contour integral of a continuous function around $\Gamma$).

In particular, if $f(z,0)$ has only one pole of order $m$ at $z_0$ inside $\Gamma$ while $f(z, t)$ has $m$ simple poles inside $\Gamma$ for $t > 0$, the residue of $f(z,0)$ at that pole is the limit as $t \to 0+$ of the sum of the residues of the poles of $f(z,t)$ inside $\Gamma$.

For example, suppose you want to find the residue of $g(z) = \dfrac{e^z}{\sin^2(z)}$ at $z = 0$. Take $$ f(z,t) = \dfrac{e^z}{\sin(z) \sin(z-t)}$$ so that $g(z) = f(z,0)$. For $0 < t < 1$, $f(\cdot,t)$ has simple poles inside the unit circle at $z=0$ and $z=t$, with corresponding residues $-1/\sin(t)$ and $e^t/\sin(t)$. Thus the residue of $g(z)$ at $z=0$ is

$$ \lim_{t \to 0+} \dfrac{-1+e^t}{\sin(t)} = 1 $$

  • $\begingroup$ Thank you very much, Robert Israel. This result is exactly what I was looking for. $\endgroup$ – Ciffolk Aug 27 '16 at 1:18

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