# Thread: Question regarding covert multiple poles when calculating residues

1. ## Question regarding covert multiple poles when calculating residues

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?

2. Originally Posted by Alexrey
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 $\displaystyle z = k \pi$ is a double pole. So just use the limit formula for higher order poles: http://en.wikipedia.org/wiki/Residue_(complex_analysis)

3. Surely that limit formula only works for overt multiple, and not covert multiple poles?

4. Originally Posted by Alexrey
Surely that limit formula only works for overt multiple, and not covert multiple poles?
What's with the 'surely' ....? As far as I'm aware, it works for all poles of all order (and I have never heard of poles being catgegorised as covert and overt and seperate treatments being required for each).

5. 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.

6. Originally Posted by Alexrey
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.
I am very confused and agree with Mr.F

For example if you have

$\displaystyle f(z)=\frac{1}{\sin^2(z)}$ this a has pole of order two at $\displaystyle \pm k\pi$

So the Residue can be calculated as

$\displaystyle \text{Res}(f,0)=\frac{1}{1!}\lim_{z \to 0}\frac{d}{dz}\frac{z^2}{\sin^2(z)}=$

$\displaystyle \lim_{z \to 0}\frac{2z\sin(z)-2z^2\cos(z)}{(\sin^3(z))}=0$

The hard part is resolving the above limit

7. Hey guys, my apologies, that method looked very similar to the method we use for solving multiple overt poles, I did not notice that it was actually a limit.