Tough math problem: residue at essential singularity

So here’s the situation: I’m a TA for a grad-level course in Mathematical Methods of Physics. One of my duties is to write up solutions for the homework problems. On the last problem set (which the students have already handed in), the prof assigned the following extra-credit problem:

“Find the residue of f(z) = z[sup]2[/sup] e[sup]1/sin z[/sup] at the isolated essential singularity z = pi.”

The prof has assigned this problem every time he’s taught this course, he says, and nobody has been able to solve it analytically. I can get a numerical answer of about -7.576, by picking a simple contour that circumnavigates the singularity and plugging it in to Mathematica; but I’m stymied as to how one could get a handle on this problem analytically.

So, Dopers, I turn to you: do you have any ideas on how this problem can be solved?

Here’s my answer: Huh?

A holomoprhic function can be undefined at an isolated point in one of three cases. It has a removable singularity, it has a pole of finite order, or it has an essential singularity. I only recall a definition of the term residue for a pole. What would it mean for an essential singularity to have a residue?

The way that I’ve always seen it defined is that the function f(z) can be written as a Laurent series in a neighbourhood of any point z[sub]0[/sub]:

f(z) = (sum from n = -infinity to infinity) of (a[sub]n[/sub] (z - z[sub]0[/sub]) [sup]n[/sup])

If none of the negative-power coefficients are non-zero, then you have a regular function at z[sub]0[/sub]; if a finite number of them are non-zero, then you have a pole of finite order; if an infinite number are non-zero, you have an essential singularity. But in all three cases, the residue is the coefficient a[sub]-1[/sub].

See this answer at math.SE for an answer to this question.

Always nice to get an answer, even if six and a half years have passed. I’ll note that Mathews & Walker doesn’t cover either Struve functions or Lagrange inversion, so I’m a little skeptical that this was the method they intended. Still, it’s nice to know that it can be done.