Evaluate: Integral((exp(1/z)/(z-1)dz)), where the curve is IzI =2, oriented positively.
The function f(z) = 1/[(2z-1)(Log(2z)] has a singularity at z = 1/2 where the denominator is zero. It also has a branch point singularity at z = 0, even though the function can be defined there such that it is continuous (you have to define f(0) = 0). Then to define the logarithm, we need to choose a branch cut and the function is then discontinuous on that branch cut. The branch cut is any arbitrary non self-interesecting curve that starts at z = 0 and moves to infinity. Finally, the point z = infinity is also a singular point. To see this, put w = 1/z, then in the w-plane the point w = 0 corresponds to z = infinity. The function in terms of w is:
1/[2 w^(-1) - 1)log(2w^(-1))] = -w/[(2 - w) log(w/2)]
We can now see that w = 0 is a branch point singularity of the logarithm. In general, you have branch cuts that connect two branch point singularities. But the most well known case of Log(z) is atypical in that one branch point singularity is at infinity and the branch cut runs toward that point at infinity. If you then don't consider the point at infinity, it looks like you only have one ...
We explain how to find the singularities and residues of the given function and we show how to compute the contour integral of the function with an essential singularity.