Calculating Residue of cos(z)/z in Contour Integral

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In summary, the problem at hand is to calculate the contour integral \oint cos(z)/z around the circle abs(z)=1.5. The integral follows from the Cauchy-Integral Formula and the Residue theorem, with the pole at z=0. The residue can be found using the formula lim z->c of (z-c)*f(z). In this case, the residue is 1, giving a final answer of 2*Pi*i.
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Homework Statement


So the problem at hand is to calculate the contour integral [tex]\oint[/tex] cos(z)/z around the circle abs(z)=1.5 .


Homework Equations


The integral is going to follow from the Cauchy-Integral Formula and the Residue theorem. The problem I am having is figuring out what the residue is going to be.


The Attempt at a Solution



So I know the pole is at z = 0, which lies inside of the contour. So the integral reduces to I=2*pi*i*residue @ 0. What I can't figure out is how to determine the residue. If I use maple, I know that the residue is 1, but I want to figure out where it comes from it. Any help?
 
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If f(z) has a simple pole at z=c then the residue is lim z->c of (z-c)*f(z), isn't it? What does that give you?
 
  • #3
Ah, thank you. I just needed someone to write it out clearly for me.

So the formal answer =2*Pi*i
 
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1. What is the purpose of calculating the residue of cos(z)/z in contour integral?

The residue of a function is used to evaluate the integral of the function over a closed contour. In the case of cos(z)/z, calculating the residue allows us to find the value of the integral by using the Residue Theorem, which states that the integral is equal to 2πi times the sum of the residues of the function at its singularities.

2. How do you find the residue of cos(z)/z?

To find the residue of cos(z)/z, we first need to determine the singularities of the function. In this case, the function has a singularity at z = 0. Then, we can use the formula for calculating the residue at a simple pole, which is given by Res(f,z0) = limz→z0 (z-z0) * f(z). In the case of cos(z)/z, we get Res(cos(z)/z, 0) = limz→0 (z-0) * cos(z) = 1.

3. What is the difference between simple and non-simple poles when calculating residues?

A simple pole is a singularity of a function where the function has a simple root, meaning the function can be written as (z-z0) * f(z). Non-simple poles, on the other hand, have higher order roots and require different formulas to calculate their residues. In the case of cos(z)/z, the pole at z = 0 is a simple pole, so we can use the formula mentioned in the previous question.

4. Can the residue of cos(z)/z have a complex value?

Yes, the residue of cos(z)/z can have a complex value. This is because the residue is calculated using a complex limit, and the value of the residue can be affected by the behavior of the function around the singularity. In the case of cos(z)/z, the residue is equal to 1, which is a real number.

5. How is the value of the contour integral affected by the residue of cos(z)/z?

According to the Residue Theorem, the value of the contour integral of a function is equal to 2πi times the sum of the residues of the function at its singularities. In the case of cos(z)/z, the residue at z = 0 is equal to 1, so the value of the contour integral will be 2πi * 1 = 2πi. This means that knowing the residue allows us to easily evaluate the integral without having to perform any complex calculations.

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