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.
Best Answer
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 $=0$