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

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SUMMARY

The integral of the function f(z) = z5 / (1 - z3) around the circle |z|=2 can be evaluated using Cauchy's Residue Theorem. The discussion reveals that the denominator can be factored into (1 - z)(z2 + z + 1), leading to three poles within the contour. The participants conclude that the residues must be calculated to determine the integral's value, emphasizing that if all residues are zero, the integral evaluates to zero. The conversation highlights the importance of recognizing poles and applying the theorem correctly.

PREREQUISITES
  • Cauchy's Residue Theorem
  • Complex analysis fundamentals
  • Understanding of poles and residues
  • Factorization of polynomials in complex functions
NEXT STEPS
  • Study the application of Cauchy's Residue Theorem in complex integrals
  • Learn how to identify and evaluate residues for complex functions
  • Explore the factorization of polynomials in complex analysis
  • Investigate examples of integrals with poles at infinity
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Students and professionals in mathematics, particularly those studying complex analysis, as well as anyone interested in evaluating complex integrals using residue techniques.

PsychoDash
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Homework Statement



"Evaluate the integral of f(z)=\frac{z^5}{1-z^3} 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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Ok, I've had an idea. If I let x=z3, then the denominator becomes 1-x. Nice, since the series representation of \frac{1}{1-x} 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: \frac{x^{5/3}}{1-x}

But now there are no fractional terms of the series, so no residue. Does that just mean the value of the integral is zero?
 
Last edited:
All 3 poles are inside the contour. Write the denominator as a product of 3 terms. The residues should then be easily evaluated.
 
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.
 
I edited my remark before seeing your response. Is my new remark more helpful?
 
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. 1-z^3=(1-z)(z^2+z+1) but the second term is not factorable.
 
Of course it is. Just not with real coefficients. Fundamental Theorem of something or other.
 
Oh bah. And humbug. And whatever other appropriate phrases one uses when they've forgotten something basic.
 
Haha.
 

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