How can I calculate the residues of this function

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In summary: That's what I was looking for.In summary, the residues of a function at a pole of order n can be found by using the residue formula with an integration around a closed path that contain z= 0 but not z= 1 or z= 1 but not z= 0.
  • #1
asi123
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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

 

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  • #2


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?
 
  • #3


Tom Mattson said:
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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  • #4


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)!
 
  • #5


asi123 said:
Well, where can I find that?

In your book? :confused: 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.
 
  • #6


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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  • #7


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.
 
  • #8


HallsofIvy said:
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.

Thanks.
 

Related to How can I calculate the residues of this function

1. What is the definition of a residue in calculus?

A residue is the value of a function at a specific point where the function is discontinuous or undefined. It is typically calculated using the limit of the function as it approaches the point in question.

2. How do I identify the poles of a function?

Poles are the points where a function becomes undefined or infinite. To identify the poles of a function, you can set the denominator of the function equal to zero and solve for the variable. The resulting values will be the poles of the function.

3. Can residues be negative?

Yes, residues can be negative. The sign of a residue is determined by the direction in which the limit of the function is taken in the calculation. A positive limit will result in a positive residue and a negative limit will result in a negative residue.

4. What is the Cauchy residue theorem?

The Cauchy residue theorem states that for a function that is analytic within a closed contour, the integral of the function around the contour is equal to the sum of the residues of the function at its poles within the contour.

5. How do I calculate residues using the Cauchy residue theorem?

To calculate residues using the Cauchy residue theorem, you must first identify the poles of the function within the contour. Then, you can use the formula: Res(f, z_i) = (1/(n-1)!)*lim(z->z_i) [(d^(n-1)/dz^(n-1))(z-z_i)^n * f(z)] to calculate the residue at each pole. Finally, you can sum up all the residues to get the total value of the integral around the contour.

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