It is simple to show that is a double pole. So just use the limit formula for higher order poles: http://en.wikipedia.org/wiki/Residue_(complex_analysis)
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?
It is simple to show that is a double pole. So just use the limit formula for higher order poles: http://en.wikipedia.org/wiki/Residue_(complex_analysis)
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.