residues

asi123

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.

Thanks in advance.


2. Relevant equations



3. The attempt at a solution

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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?

asi123

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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HallsofIvy

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

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

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)!

Tom Mattson

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.

asi123

Ok, this is what I did (in the pic).

Now I need to sum the residues to get the answer?

Thanks.

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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 [itex]2\pi i[/itex].

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 [itex]2\pi i[/itex].

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 [itex]2\pi i[/itex].

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.

asi123

Thanks.