Complex Integral Evaluation: Solving an Integral Using Cauchy's Residue Theorem

In summary: I was going to say something similar.In summary, the integral of f(z)=\frac{z^5}{1-z^3} around the circle |z|=2 in the positive sense is difficult, and may not yield any residues.
  • #1
PsychoDash
16
1

Homework Statement



"Evaluate the integral of f(z)=[tex]\frac{z^5}{1-z^3}[/tex] around the circle |z|=2 in the positive sense.


Homework Equations



Cauchy's residue theorem?


The Attempt at a Solution



Truthfully, I don't know where to begin. I've done others of these using Cauchy's theorem and either a partial fraction expansion of the function or a series expansion, but this one seems odd. I realize that not posting work is frowned upon, but I'm stuck at the beginning without any work to show. I don't know how to begin.
 
Last edited:
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  • #2
Ok, I've had an idea. If I let x=z3, then the denominator becomes 1-x. Nice, since the series representation of [tex]\frac{1}{1-x}[/tex] is well known. Now I'm just not sure what this change of variable does to the numerator. z5=z3*z2, but I'm not sure how to write z2 in terms of z3. My brain fails me.

edit: [tex]\frac{x^{5/3}}{1-x}[/tex]

But now there are no fractional terms of the series, so no residue. Does that just mean the value of the integral is zero?
 
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  • #3
All 3 poles are inside the contour. Write the denominator as a product of 3 terms. The residues should then be easily evaluated.
 
  • #4
Berko said:
If all the residues are zero, then the integral gives zero.

In general. the integral is 2pi-i-times the sum of the residues, if I remember correctly.

I realize that. My question is, am I right in saying that there are no residues? I'm thinking this is some form of a "residue at infinity" example, in which case there are new rules, but I don't understand this enough to tell.
 
  • #5
I edited my remark before seeing your response. Is my new remark more helpful?
 
  • #6
Berko said:
All 3 poles are inside the contour. Write the denominator as a product of 3 terms. The residues should then be easily evaluated.

3 terms? I see only two. [tex]1-z^3=(1-z)(z^2+z+1)[/tex] but the second term is not factorable.
 
  • #7
Of course it is. Just not with real coefficients. Fundamental Theorem of something or other.
 
  • #8
Oh bah. And humbug. And whatever other appropriate phrases one uses when they've forgotten something basic.
 
  • #9
Haha.
 

1. What is a complex integral?

A complex integral is a mathematical concept that involves finding the area under a curve in the complex plane. It is a generalization of the familiar concept of a real-valued integral and is used in various branches of mathematics and physics.

2. How is a complex integral evaluated?

A complex integral is typically evaluated using the Cauchy integral formula, which states that the value of an integral around a closed contour is equal to the sum of the values of the function at all points inside the contour. This formula is derived from the Cauchy-Goursat theorem.

3. What are the key properties of complex integrals?

Some key properties of complex integrals include linearity, additivity, and the use of the residue theorem to evaluate integrals involving singularities. Additionally, the contour of integration can be deformed without changing the value of the integral as long as it does not cross any singularities.

4. What is the significance of Cauchy's residue theorem in complex integral evaluation?

The residue theorem, also known as Cauchy's residue theorem, is a powerful tool for evaluating complex integrals. It states that the value of a complex integral around a closed contour is equal to 2πi times the sum of the residues of the function at all of its isolated singularities inside the contour.

5. How are complex integrals used in real-world applications?

Complex integrals have numerous applications in mathematics, physics, and engineering. Some examples include calculating the electric potential of a charged body in electrostatics, computing the flow of fluids in fluid mechanics, and determining the path of light in optics. They are also used in the branch of mathematics known as complex analysis.

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