Contour Integration: Find Out Non-Removable Singularities?

[Manchot]

Suppose one was trying to find out whether a function has exactly one non-removable singularity in a given region on the complex plane or not, but for some reason could not find out directly. Does it suffice to take a closed path integral around the boundary of the region, and see if it equals zero? That is, if one determines that the path integral around the region's boundary is zero, does one know for sure that there isn't exactly one singularity? Or, vice versa, if one finds out that the integral isn't zero, does one automatically know that there's at least one singularity? Thanks for the help.

 

[mathwonk]

well look at the residue theorem, which says the integral equalks the sum of the residues of the integrand at all singualrities inside the contour.

hence no singularities implies zero residue, hence non zero integral implies some singularities.

but zero integral does not imply much: you could have one singularity with zero residue, or two singularities with equal and opposite residues, etc...

in fact the residue theorem says that every meromorphic differential on a compact riemann surface has the sum of its residues at all points equal to zero.

 

[Manchot]

Ok, thanks for the information. I actually haven't taken a complex analysis course, so under usual circumstances I'd have no idea what you were talking about. Fortunately, I learned about residues just last week by reading Wikipedia and MathWorld in my spare time (a potent combination). I can see why you are a Homework Helper and Science Advisor.