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Is there a simple and fast way of computing the residue at an essential singularity ?

I mean if we have a pole of order $n$ at $c$ we can use the formula :

$$\mathrm{Res}(f,c) = \frac{1}{(n-1)!} \lim_{z \to c} \frac{d^{n-1}}{dz^{n-1}}\left( (z-c)^{n}f(z) \right) $$

Is there a similar formula for an essential singularity (without using Laurent series) ?

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  • $\begingroup$ It would be pretty nice if someone could answer this, but I'm quite skeptical. So I'm just going to go with the old fashion answer: integrate (which clearly is equivalent to finding the corresponding term in the Laurent series). $\endgroup$
    – hjhjhj57
    Nov 7 '14 at 4:58
  • $\begingroup$ Trust me, Laurent series all the way. You dont wanna have to deal with those derivatives, especially when taking a timed test $\endgroup$
    – Shemafied
    Dec 4 '15 at 23:49

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