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Homework Help: Hard Laurent series. A little lost.

  1. May 9, 2010 #1
    1. The problem statement, all variables and given/known data
    find the Laurent series for

    [tex]\frac{z+2}{z^{5}-8z^{2}}[/tex] in 2<|z|<[tex]\infty[/tex]


    2. Relevant equations



    3. The attempt at a solution
    Well, I factored out z[tex]^{5}[/tex] in the denominator, which left me with a geometric sum (since |z|>2). I've come up with [tex]\frac{z+2}{z^{5}}[/tex][tex]\sum[/tex][tex]\frac{8}{z^{3}}[/tex][tex]^{n}[/tex]

    I'm not sure if I'm going in the right direction, but if I am, could someone please give me a hint at what to look at next.
     
  2. jcsd
  3. May 9, 2010 #2

    vela

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    Looks good so far. Just multiply out and collect terms now.
     
  4. May 9, 2010 #3
    When you say multiply out and collect terms, do you mean I need to write it out as:

    8(z+2)/z^8 + 8^2(z+2)/z^11 + .... (tried to type it in latex, but it calculated it back to f(z)... impressive!)

    and that is the Laurent series?
     
  5. May 9, 2010 #4

    vela

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    You need to multiply it out completely so you're just left with a constant times a power of z for each term.
     
  6. May 10, 2010 #5
    I think I've got it.

    Sum from n=3 -> infinity of 2^n/z^(n+4) - Sum from n=3 -> infinity of 2^(3n+2)/z^(3n+6)

    because when I multiply and combine terms I get 2^3/z^7+2^4/z^8+2^6/z^10+2^7/z^11+...

    I'm sorry about my format, latex is too smart for me.
     
  7. May 10, 2010 #6
    Now that I've got the infinite sum, should I write it out as the difference of two summations? Or is it typically acceptable to write it out as the sum of terms with a clear pattern? I will go and ask my professor about his preferred notation anyhow.
     
  8. May 10, 2010 #7

    vela

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    It's a judgment call, so asking your professor is a good idea in case he or she wants it expressed in a certain way.
     
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