## residues

1. The problem statement, all variables and given/known data

Hey guys.
How can I calculate the residues of this function (in the pic) in all of its singularity points?
I'm kind of a newbie in this this residues stuff and I can really use an example.

2. Relevant equations

3. The attempt at a solution
Attached Thumbnails

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 Recognitions: Gold Member Homework Help Science Advisor Staff Emeritus Well, you've got to show us that you've tried something. Can you please state the formula for finding the residue of a function at a pole of order n?

 Quote by Tom Mattson Well, you've got to show us that you've tried something. Can you please state the formula for finding the residue of a function at a pole of order n?
Well, where can I find that?

All I know is how to use the residue with this formula (in the pic).

Thanks a lot.
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## residues

That formula is normally used to evaluate an integral after you have found the residues but it can be used the other way.

Notice that if you integrate $z^n$, for n not equal to -1, around a circle of radius R, you can take $z= Re^{i\theta}$ so $dz= Ri e^{i\theta}d\theta$ and the integral is
$$Ri \int_0^{2\pi} e^{i(n+1)\theta}d\theta= \frac{R}{n+1}e^{i(n+1)x}$$
evaluated at 0 and $2\pi$ so it is 0. If n= -1, n+1= 0 so we can't use that integral but we have
$$i\int_0^{2\pi}d\theta= 2\pi i$$

A function f(z) has a pole of order n at $z= z_0$ if and only if it can be written as a power series with negative integer powers down to -n, say f(z)= $a_{-n}z^{-n}+ a_{-n+1}z^{-n}+ \cdot\cdot\cdot+ a_{-1}z^{-1}+ a_0+ a_1z+ \cdot\cdot\cdot$. If we integrate that term by term, we get 0 for every term except the $z^{-1}$ term which gives $2\pi i a_{-1}$: The "residue" at z= $z_0$ IS the coefficient of $z^{-1}$.

So do this: use "partial fractions" to write this function as [tex]\frac{Az+ B}{z^2}+ \frac{C}{z-1}= \frac{Az}{z^2}+ \frac{B}{z^2}+ \frac{C}{z- 1}[/itex] and integrate first around a small circle z= 0 to find the residue at z= 0 and then a small circle around z= 1 to find the residue at z=1. Or just look at the coefficients of 1/z and 1/(z-1)!

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 Quote by asi123 Well, where can I find that?
In your book? If that doesn't work, then do a Google search for "residue of a pole". The first hit contains the formula. Once you've located the formula then we can talk about how to use it.

 Ok, this is what I did (in the pic). Now I need to sum the residues to get the answer? Thanks. Attached Thumbnails
 Recognitions: Gold Member Science Advisor Staff Emeritus The answer to what question? In your original post you only asked about the residues themselves. If the problem is to find the integral around a closed path having both z= 0 and z= 1 in its interior then the integral is the sum of the residues times $2\pi i$. If the problem is to find the integral around a closed path that contains z= 0 but not z= 1 in its interior then the integral is the residue at z= 0 times $2\pi i$. If the problem is to find the integral around a closed path that contain z= 1 but not z= 0 in its interior then the integral is the residue at z= 1 times $2\pi i$. To be complete, if the problem is to find the integral around a closed path that contains neither z= 0 nor z= 1 in its interior then the integral is 0, of course.

 Quote by HallsofIvy The answer to what question? In your original post you only asked about the residues themselves. If the problem is to find the integral around a closed path having both z= 0 and z= 1 in its interior then the integral is the sum of the residues times $2\pi i$. If the problem is to find the integral around a closed path that contains z= 0 but not z= 1 in its interior then the integral is the residue at z= 0 times $2\pi i$. If the problem is to find the integral around a closed path that contain z= 1 but not z= 0 in its interior then the integral is the residue at z= 1 times $2\pi i$. To be complete, if the problem is to find the integral around a closed path that contains neither z= 0 nor z= 1 in its interior then the integral is 0, of course.
Thanks.