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Residue Theorem

  • #1

Homework Statement



Suppose that f is continuous and that there exist constants A,B ≥ 0 and k>1 such that |f(z)|≤A|z|−k for all z such that |z|>B. let CR denote the semicircle given by |z| = R, Re(z) ≥ 0. Prove that limR→∞∫f(z)dz=0

Homework Equations





The Attempt at a Solution

I dont understand what i'm prooving here. I think it has something to do with jordans lemma but I havent a clue! any ideas...please...
 

Answers and Replies

  • #2
gabbagabbahey
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It is against forum rules to create multiple threads for the same problem.

Have you learned the estimation lemma yet?
 
  • #3
Hi!
Sorry about the multiple thread but I called it jordans lemma then thought, it might have nothing to do with it but I couldnt delete the thread after creating it!

I havent done the estimation lemma but I looked it up
Let f : U[itex]\rightarrow[/itex]C be continuous (where U is some subset of C), let [itex]\gamma[/itex] be a path in U, and suppose |f(z)| < M for all z [itex]\in \lambda[/itex]. Let length( [itex]\lambda[/itex])= L. Then

|[itex]\int[/itex]f(z)dz| [itex]\leq[/itex] ML.
 
  • #4
vela
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I deleted the other thread. Next time, just hit the report button and ask a mentor to edit or delete the thread for you.
 
  • #5
gabbagabbahey
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I havent done the estimation lemma but I looked it up
Let f : U[itex]\rightarrow[/itex]C be continuous (where U is some subset of C), let [itex]\gamma[/itex] be a path in U, and suppose |f(z)| < M for all z [itex]\in \lambda[/itex]. Let length( [itex]\lambda[/itex])= L. Then

|[itex]\int[/itex]f(z)dz| [itex]\leq[/itex] ML.
It is probably in your textbook, but just not named. Many texts (c.f. Mathematical Methods for Physicists A Concise Introduction Tai L. Chow p. 255) simply list it as a property of complex integrals and give a short proof of it.

What is the arclength [itex]L[/itex] of your semicircular contour CR? If [itex]|f(z)| \leq A|z|^{-k}[/itex] (which is what I assume you meant when you wrote |f(z)|≤A|z|−k) and [itex]|z|>B>0[/itex], what is an upper bound for [itex]|f(z)|[/itex] on your semi-circular contour?

What does that tell you about [itex]\left| \int_{CR} f(z) dz \right|[/itex]?
 
  • #6
The arc lenght is[itex]\pi r[/itex]

[itex] |f(z)| \leq A|z|^{-k} [/itex] the semicircle is given by |z| = R so for k>1, if the value of R gets bigger, then the value of [itex] \frac{A}{|z|^{k}} [/itex] gets smaller so as R[itex]\rightarrow\infty[/itex] then [itex] |f(z)| \rightarrow [/itex] 0
I guess thats what ther're saying but i guess i better try and word it better
 
  • #7
gabbagabbahey
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The arc length is[itex]\pi r[/itex]

[itex] |f(z)| \leq A|z|^{-k} [/itex] the semicircle is given by |z| = R so for k>1, if the value of R gets bigger, then the value of [itex] \frac{A}{|z|^{k}} [/itex] gets smaller so as R[itex]\rightarrow\infty[/itex] then [itex] |f(z)| \rightarrow [/itex] 0
Right or, using the estimation lemma with the fact that [itex]|f(z)| \leq A|R|^{-k} [/itex], you have

[tex]\left| \lim_{R \to \infty} \int_{CR} f(z)dz \right| = \lim_{R \to \infty} \left| \int_{CR} f(z)dz \right| \leq \lim_{R \to \infty} \left( A|R|^{-k} \right) = 0[/tex]
 

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