If $f(z)$ has both poles and essential singularities - do the residues of the poles still sum to zero?

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.

• 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. – logarithm May 30 '19 at 18:17
• $\frac{e^z}{z}$ has a pole at zero and an essential singularity at infinity and obviously the residue at the one pole is non-zero – Conrad Jun 2 '19 at 17:33
• 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 – R Mary Jun 2 '19 at 17:49
• no problem - done – Conrad Jun 2 '19 at 17:52
• 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 $0$. – user Apr 15 at 15:23

$$\frac{e^z}{z}$$ has a pole at zero and an essential singularity at infinity and obviously the residue at the one pole is non-zero
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 $$=0$$
• 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 $0$. – reuns Jun 2 '19 at 19:59