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Poles and residues

  1. Feb 18, 2015 #1
    1. The problem statement, all variables and given/known data
    Show that all the singular points of the following functions are poles.
    Determine the order of each pole and the corresponding residue.

    z/cosz

    2. Relevant equations
    maybe cosz = 1/2 ez+e-z is relevant but i dont use it here

    3. The attempt at a solution
    the pole i found was z = π/2 + 2nπ and -π/2 +2nπ

    i attempted to find the residues using limits.
    limz→π/2 + 2nπ (z-(π/2 + 2nπ)) z/cosz

    doing some annoying algebra and just evaluating at π/2 + 2nπ i get 0/0. so i can use L'hopitals rule, right?
    doing so i get my limit to equal π/2 + 6nπ

    is that right?

    i know i can use lorent series but i honestly dont know where to start with that mess haha
     
  2. jcsd
  3. Feb 18, 2015 #2

    Dick

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    That's pretty bad. cosz = (1/2)(eiz+e-iz) so your relevant equation is wrong and you do want to use it to show the only poles are on the real axis. Given that you do have the right poles on the real axis. And yes, you can use l'hopital. But show how you got that limit.
     
  4. Feb 18, 2015 #3
    oops i forgot the i's. but doesnt just putting the poles into the original equation shows that theyre poles?

    i didn't realize it was "pretty bad" LOL
     
  5. Feb 18, 2015 #4

    Dick

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    It shows they are poles, it doesn't show they are the only poles. That's not my only complaint. The limit to compute the residue is bad and I can't tell why because you didn't show your work.
     
  6. Feb 18, 2015 #5
    lim z → (z-(π/2 + 2nπ)) z/cosz

    z2-π/2z - 2nπz

    then evaluate at z = π/2 + 2nπ

    π2/4 + π/2(2nπ) +π/2(2nπ) + 4n2π22/4- π/2(2nπ) π/2(2nπ) - 4n2π2 = 0

    cos(π/2 + 2nπ) = 0

    0/0

    l'hopital rule
    2z - π/2 - 2nπ/ - sinz evaluate at π/2 + 2nπ

    2(π/2) + 4nπ -π/2 -2nπ / -sin(π/2 + 6nπ)

    = -π/2 - 2nπ

    so my first calculation was different but its still the same method.
     
  7. Feb 18, 2015 #6

    Dick

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    "Different" is what I wanted since the first one was bad. That one looks ok (with some typos). Now what about the -π/2+2nπ series and what about showing that the ONLY poles are on the real axis?
     
  8. Feb 18, 2015 #7
    im gonna have to do more reading because i dont know what youre talking about with "-π/2+2nπ series" i thought i take the limit and the limit is the residue.

    i also need to do some more reading because i dont know how to find any other poles.

    im happy ive made it this far though, i had no idea what the hell i was looking at 2 hours ago.
     
  9. Feb 18, 2015 #8
    thank you though, even though youre kind of a dick. ;-)
     
  10. Feb 19, 2015 #9

    Dick

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    You said "the pole i found was z = π/2 + 2nπ and -π/2 +2nπ". So far you've only found the residue at π/2 + 2nπ. What about -π/2 +2nπ?
     
  11. Feb 19, 2015 #10
    sorry, i went to sleep then to class lol.
    the next one should be -π/2 +2nπ.

    using the same messy method
     
  12. Feb 19, 2015 #11

    Dick

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    Right again. It's not really all that bad is it? Now can you show they are the only poles using your relevant equation? And what's the order of the poles? Yes, I'm kind of a dick :-;
     
  13. Feb 19, 2015 #12
    i think theyre both order 1.

    im not sure how to use the equation to show that. im guessing it has something to do with using the ln|z| + iarg though. not sure if thats right.
     
  14. Feb 19, 2015 #13

    Dick

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    You are correct about the order, can you say why? As to why they are the only poles, if ##cos(z)=0## then ##e^{iz}=-e^{-iz}##, right? Put ##z=a+ib## and show ##b## must be 0.
     
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