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.