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Region of convergence

  1. Jul 26, 2009 #1
    First of all I am a first time user this evening..... So bear with me.

    The question was: Find a Laurent series for (z-2)/(z+1) centered at z = -1 and specify the region in which it converges.


    Laurent Series will be employed 1/z [tex]\Sigma\frac{alpha}{z}[/tex][tex]^{k}[/tex]


    So far I have done long division and yielded

    (z-2)/(z+1) = 1 - [tex]\frac{3}{z+1}[/tex] = 1 - [ 3 / (1- ([tex]\frac{-1}{z}[/tex]))]

    which leads to the Laurent series

    1 + 3/z [tex]\Sigma[/tex](-1/z)^{k}


    (again I apologize if the Latex ref is wrong)

    I think I'm ok to this point however I am at a loss for where the region of convergence might be.... Just guessing something having to do with left of -1 or right of -1???

    cleannj
     
    Last edited: Jul 26, 2009
  2. jcsd
  3. Jul 26, 2009 #2

    Dick

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    A Laurent series expanded around z=(-1) is an expansion in powers of (z+1), isn't it? It would appear to me that you were done the minute you wrote 1-3/(z+1). And that makes it a finite series, it converges everywhere except at z=(-1). Or am I confused?
     
    Last edited: Jul 26, 2009
  4. Jul 27, 2009 #3
    Thank you for your help.
     
  5. Jul 27, 2009 #4
    Find integral over counterclockwise circle

    Much like the problem with residues I seem to be good with the big stuff just not the details.

    Another question would be:

    Let [tex]\gamma_{r}[/tex] be the counterclockwie circle w/ center at 0 and radius r. Find [tex]\int[/tex]dz/(z^2-2z-8) for r = 1,3,5

    I did the partial fraction decomposition and got

    1/(z^2-2z-8) = 1/(6(z-4)) - 1/(6(z+2)) so the singularitites are z = 4 and z = -2 right?

    However when I want to integrate those terms I assume I should get natural log stuff.... namely 1/6 [ln(z-4) - ln (z+2)]

    Where do the r = 1 , 3, 5 come into play?
     
  6. Jul 27, 2009 #5

    Dick

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    Right. You have simple poles at z=4 and z=(-2). You integrate around a curve by summing the residues (review the residue theorem again) of the poles INSIDE the curve, right? You don't need to worry about the antiderivatives. So what are the residues at z=4 and z=(-2)? And which of the poles are inside of the curves in each cases r=1, r=3 and r=5?
     
  7. Jul 27, 2009 #6
    Poles and Residues

    Quite honestly the only thing I can make out from the theorem is the constant 2[tex]\pi[/tex]i is multiplied by the sum of residues.....

    I cannot make out the right hand side of the formula to determine the residue itself.
    I seem to remember reading somewhere that the coefficient of the 1/z term is used....

    So 1/6 * 2 pi i = pi i / 3

    I looked through drawings of contours but am unsure how I am to distinguish ...
    Maybe I'm making it too hard...

    For r = 1 the isolated singularities are not interior (CORRECT???)
    For r = 3 only the z=-2 would be interior (so what does this tell me?)
    For r = 5 both the z = 4 and z = -2 would be interior

    Still don't know what to do with the info.
     
  8. Jul 27, 2009 #7

    Cyosis

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    Yes the residue is the coefficient of the 1/z term in the Laurent series. Alternatively the residue of a simple pole can be calculated by [itex]\text{Res}_{z=\alpha}(f)=\lim_{z \to \alpha} (z-\alpha)f(z)[/itex].

    As for the different contours. If r=1 the contour describes a circle of radius 1 centered at the origin in the complex plane. You have simple poles at z=-2 and z=4. Therefore |z|=2 and |z|=4 in other words the poles lie at a distance 2 and 4 from the origin. You can easily see that both these poles lie outside of the unit circle, because r<|z|. So yes you have the correct answer for all three cases.

    Now the residue theorem says that the integral around the contour equals the sum of the residues of the poles that lie within the contour. For r=1 there are no poles inside the contour so the integral is?

    For r=2 we have one pole within the contour namely z=-2 so the integral becomes:

    [tex]
    \int_{\gamma(r=3)} \frac{dz}{z^2-2z-8}=2 \pi i \text{Res}_{z=-2}(f)
    [/tex]

    Now calculate the residue with one of the methods I have described above.
     
    Last edited: Jul 27, 2009
  9. Jul 27, 2009 #8

    HallsofIvy

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    [itex]1/(z+1)= (z+1)^{-1}[/itex] IS the "Laurent series" for 1/(z+1). Now, (z-2)/(z+1)= ((z+1)- 3)/(z+1)= 1- 3/(z+1).

    That should make it obvious what the residue is.
     
  10. Jul 27, 2009 #9
    Cyosis continue

    For r = 1 the integral must be zero right?

    For r = 3 only z = -2 lies in the interior

    So the Res f = lim (z--2) as z --> -2 so this is zero also???

    AND for r = 5 z = -2 and z = 4 lie in the interior

    so lastly Res f = lim (z-4) as z --> 4 is zero???

    This all seems wasted??

    I apologize (it has been 7 years since I originally took the class) and I'm comprehensive testing tomorrow....
     
  11. Jul 27, 2009 #10

    Dick

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    If f(z)=1/(z^2-2*z-8) the residue at z=(-2) is the limit (z+2)*f(z) as z->(-2). It's not zero.
     
  12. Jul 27, 2009 #11

    Cyosis

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    I suggest you write [itex]z^2-2z-8[/itex] as [itex](z-4)(z+2)[/itex] when computing the limit.
     
  13. Jul 27, 2009 #12



    Very well, please explain (even if you need to use another rational function example) EXACTLY how to compute the residue.....

    Your short answers are not helping -- the limit of (z +2 ) * anything if z approaches -2 will be zero since -2 + 2 = 0.
     
  14. Jul 27, 2009 #13

    Dick

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    Use Cyosis' suggestion. Cancel the (z+2) factor. You could also multiply (z+2) times your partial fraction expression. It's not true that (z+2)*anything approaches zero.
     
  15. Jul 27, 2009 #14

    Cyosis

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    I wouldn't call these answers short. Instead of desperately trying to get an answer asap, you may want to take your time, read the comments and use the hints. This is the only way you will actually start understanding what you're doing. For one you ignored the hint I gave you in post 11.
    Calculating the residue at this point is not the problem, the problem is that you can't compute the limit. Once more I will redirect you to post 11.

    As for your argument that (z+2)f(z), when z goes to -2 is always zero let me provide you with a counter example. Define f(z)=1/z we want to compute z f(z) as z tends to 0, according to you this will yield 0. I suggest you plot this function so you can see that this is wrong.
    The reason why taking the limit goes wrong is because you pretend that taking a limit is nothing more than entering the value into zf(z) with the additional assumption that 0/0=0.

    [tex]
    \lim_{z->0}zf(z)=\lim_{z->0}\frac{z}{z}=\lim_{z->0}1=1 \neq \frac{0}{0} \neq 0
    [/tex]
     
  16. Jul 27, 2009 #15
    Then you are saying

    (z+2)* [tex]\frac{1}{(z+2)(z-4)}[/tex] is what is necessary

    then



    1/ (z-4) plug in -2 and get the residue to be 1 / (-2-4) = -1/6
     
    Last edited: Jul 27, 2009
  17. Jul 27, 2009 #16

    Cyosis

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    Use the formula for the residue found in post 7

    Write it out in full preferably on this forum. That is write out f(z) fully.

    ps. seeing as you've not been doing maths for seven years it would be advisable to brush up on the basics, in particular limits. There is no way you will be able to understand complex analysis properly without understanding basic concepts from real analysis.
     
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