Hey guys, I was told by my lecturer, that if we come across a covert multiple pole when trying to find the residue of a complex function, we have to either make it into a simpler form (i.e. covert simple or overt multiple forms) or find the coefficient of z^-1 by finding its Laurent expansion about its poles. This is all fine and good, but when I came across something like 1/sin^2(z), I had a bit of a problem finding its residues through simplification of its poles. I know that this function has a double covert pole at z= k(pi), but I don't really like finding Laurent expansions as they are quite involved when compared to normal residue finding methods. I was wondering if this would be possible:
Make sin^2(z) = (1/2i)e^2iz - (1/2i)e^-2iz, and then find the residues using simple covert methods. Would this be regarded as a covert simple form or is it still in its double covert form?
Okay well my lecturer, and his lecture notes say that we CANNOT use the limit formula for covert multiple poles, so I tend to trust what he says. But I'll try it out on a few multiple covert cases and see what happens, would be really cool if it works. An overt pole is something of the form (z-a)^m whereas a covert pole can be any function (e.g. e^z, z^2 + 1, sin(z) etc.).
EDIT: these functions obviously have to be on the denominator to be a pole otherwise they would be categorized as zeros. I tried it out and the limit formula did not work as expected. From the wiki page is looks like it has to be in the form (z-a)^n f(z) which is the form of an overt simple pole when n < 0.