How can I calculate the residues of this function

[asi123]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

 

[quantumdude]

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, 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.

 

[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 $z^n$, 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)![/tex]

 

[quantumdude]

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.

 

[asi123]

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

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

Thanks.

 

[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.

 

[asi123]

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.