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It is well known that given a meromorphic function $f(z)$ then the sum of the residues of all the poles of $f(z)$ is zero. If $f(z)$ has some essential singularities, is it still true? If we drew a contour which enclosed all of the poles and didn't have essential singularities within it would the integral of $f$ on this contour be zero?

Edit: I think my question was not clear enough so I will try and make it clearer. If a function has both poles and essential singularties which are isolated then one can pick a disk on which the function is holomorphic, draw a contour around it and then conclude that the sum of all the integrals of $f$ around contours containing one isolated singularity of $f$ is zero. There are two contributions to this sum - residues of $f$ at poles and integrals around closed contours containing essential singularities. My question is if both of these contributions each seperately sum to zero. I see no reason for this to be the case, but it doesn't seem incredible that it might be true, so I was just wondering.

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  • $\begingroup$ If the singularities of the function in the Riemann sphere are only isolated ones, then the sum of the residues on all of them is zero. $\endgroup$ – logarithm May 30 '19 at 18:17
  • $\begingroup$ $\frac{e^z}{z}$ has a pole at zero and an essential singularity at infinity and obviously the residue at the one pole is non-zero $\endgroup$ – Conrad Jun 2 '19 at 17:33
  • $\begingroup$ Dear @Conrad, thank you. Sorry, I'm still new to this so I'm a bit slow but your answer helped a lot. If you wish to add your comment as an answer I will accept $\endgroup$ – R Mary Jun 2 '19 at 17:49
  • $\begingroup$ no problem - done $\endgroup$ – Conrad Jun 2 '19 at 17:52
  • $\begingroup$ It is not true that you always can encompass all poles of a meromorphic function with a contour. There can be a pole at infinity! And this makes a huge difference. The correct statement is that the sum of the residues over all singularities (i.e poles and essential singularities including those at infinity) is $0$. There is no general reason that a sum of residues over a proper subset of the singularities (including the subset of all poles) is $0$. $\endgroup$ – user Apr 15 at 15:23
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$\frac{e^z}{z}$ has a pole at zero and an essential singularity at infinity and obviously the residue at the one pole is non-zero

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Sure if $f$ is analytic on $\Bbb{C}$ minus a few points $p_1,\ldots,p_J$, pick $R >|p_j|, \delta < \min_{i,j} |p_i-p_j|$ then (by the Cauchy integral theorem) the sum of residue at the $p_j$ is $$ \sum_{j=1}^J \frac{1}{2i\pi}\int_{|z-p_j|=\delta} f(z)dz =\frac{1}{2i\pi}\int_{|z|=R}f(z)dz$$

Then $-\lim_{R\to \infty}\frac{1}{2i\pi}\int_{|z|=R}f(z)dz$ is called "the residue at $\infty$" making the sum of all residues $=0$

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  • $\begingroup$ Dear Reuns, thank you for your answer however I see a couple of issues - first, if you have essential singularities you are not guaranteed that your residue at infinity is zero. Secondly, I believe when essential singularities are involved the idea of "residue" is no longer applicable. If I'm not wrong, your answer is an answer to the question for when the function is meromorphic which is unfortunately not the case here. I will try and edit my question to make it clearer. $\endgroup$ – R Mary Jun 2 '19 at 14:04
  • $\begingroup$ No and no. The residue applies to isolated singularities (poles or removable or essential), it is defined as $\frac{1}{2i\pi}\int_{|z-a| = \epsilon} f(z)dz$, moreover $f$ has a Laurent series around $z=a$ and the residue is the coef of $(z-a)^{-1}$. I never said the residue at $\infty$ is zero, in general it is not, I have shown we define the residue at $\infty$ as $-\frac{1}{2i\pi}\int_{|z|= R}f(z)dz$ to obtain (for a function analytic on $\Bbb{C}$ minus finitely many points) the sum of residue is $0$. $\endgroup$ – reuns Jun 2 '19 at 19:59
  • $\begingroup$ I apologize, I thought the term "residue" only applied to poles but I can see now I was mistaken. Thanks for your answer! $\endgroup$ – R Mary Jun 3 '19 at 8:31

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