Residue theorem and laurent expansion

In summary, The conversation is about calculating the residue of a function at infinity. The teacher's method involves expanding the function in a Laurent expansion and deducing the value from that. However, the person finds this method to be unnecessarily difficult. They ask if there is an easier way and mention that letting z approach infinity does not yield the desired result. They specifically need to find the residue at infinity for the function f(z)=\frac{z^{2}}{\sqrt{z^{2}-1}(z-t)} and are looking for a different method to do so. The other person mentions that for nicer functions, there are some tricks, but when the Laurent expansion does not terminate, using a series expansion is the only option.
  • #1
elimenohpee
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Homework Statement


I need to calculate the residue of a function at infinity. My teacher does this by expanding the function in a laurent expansion and deduces the value from that. That seems much harder than it needs to be. For example, in the notes he calculates the residue at infinity of:
[tex]g(z)=\frac{\sqrt{z^{2}-1}}{z-t}=...=-t[/tex]

Is there an easier way than resorting to a laurent series? If I let z approach infinity in the function above, I get 1 not -t, so I'm assuming you can't evaluate the residue in that way?

Specifically I need to find the residue at infinity of
[tex]f(z)=\frac{z^{2}}{\sqrt{z^{2}-1}(z-t)}[/tex]
but I'm looking for a method to do so.
 
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  • #2
If you have nicer functions than that there are a few tricks, but when the Laurent expansion doesn't terminate I don't think that there is an easier way than to use a series expansion.
 

Related to Residue theorem and laurent expansion

1. What is the Residue theorem and how is it used in complex analysis?

The Residue theorem is a powerful tool in complex analysis that allows us to calculate the value of a complex integral using the residues (singular points) of a function. It states that if a function is analytic within a closed contour, then the value of the integral around that contour is equal to the sum of the residues inside the contour. This theorem is useful in evaluating integrals that would otherwise be difficult to solve using traditional methods.

2. How is the Residue theorem related to the Cauchy integral formula?

The Cauchy integral formula is a special case of the Residue theorem, where the contour used is a circle centered at the singularity of the function. The formula states that the value of the integral is equal to 2πi multiplied by the value of the function at the singularity. This is a direct application of the Residue theorem, as the singularity acts as the only residue inside the contour.

3. What is the Laurent expansion of a function?

The Laurent expansion is a representation of a function as a sum of a series of terms involving positive and negative powers of the independent variable. This expansion is valid for functions that are analytic in an annulus (a region between two concentric circles) and can be used to evaluate the function at points outside its domain. It is similar to a Taylor series expansion, but the Laurent expansion includes negative powers as well.

4. How is the Laurent expansion used to calculate residues?

The Laurent expansion can be used to calculate residues by finding the coefficient of the term with a negative power in the series. This coefficient is the residue, and it can then be used in the Residue theorem to evaluate the value of a complex integral. The Laurent expansion provides a straightforward way to determine the residues of a function without having to use complex contour integration techniques.

5. Can the Residue theorem be used to calculate integrals with multiple poles?

Yes, the Residue theorem can be used to calculate integrals with multiple poles. In this case, the theorem states that the integral is equal to the sum of the residues of all the poles inside the contour. This allows us to evaluate integrals that would be difficult or impossible to solve using other methods. However, it is important to note that the poles must be isolated and not located on the contour for the theorem to be valid.

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