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

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Homework Help Overview

The problem involves evaluating the integral of the function f(z)=z^5/(1-z^3) around the circle |z|=2, with a focus on applying Cauchy's residue theorem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster expresses uncertainty about how to start the problem, mentioning previous experiences with Cauchy's theorem but feeling stuck. Another participant suggests a change of variable to simplify the denominator but questions how it affects the numerator. There is discussion about the poles of the function and the evaluation of residues, with some participants questioning the existence of residues and the implications for the integral's value.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and the implications of their findings. Some guidance has been offered regarding the poles and residues, but there is no explicit consensus on the evaluation of the integral or the nature of the residues.

Contextual Notes

Participants note the presence of three poles within the contour and discuss the factorization of the denominator, indicating a potential misunderstanding of the number of terms involved. There is also mention of a "residue at infinity" scenario, suggesting a complexity in the evaluation process.

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?
 
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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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