How do I find the residue of 1/(z^2 sinz) ? It seems to me that it has a 3rd order singularity at z=0, but I cant find the residue. Thanks for any help rendered.
It has a pole of order 3 at z = 0. And there is a standard formula (it will be in your textbook or class notes) for finding the residue at a pole ....
Alternatively, you could find and use the first few terms of the Laurent series of (from which it immediately follows that the residue is equal to 1/6).
I tried using the formula but I got this equation eventually (attached along this reply). I could not eliminate the sinz at the denominatior, causing my ans to tend to inf as z tends to 0. Have I used the formula wrongly or is there another formula to get the residue?
Following up from my earlier reply:
Note that has a pole of order 1 at z = 0 since is finite (and is equal to 1). Furthermore, .
Therefore
Therefore .
To get the value of b, substitute into and re-arrange:
Expand and equate coefficients:
Coefficient of z: a = 0.
Coefficient of : .
What can't Wolfram Alpha do?