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Residue theorem

  1. Apr 14, 2008 #1
    the Residue theorem states that :
    [tex]\oint {f(z)dz} [/tex] = 2[tex]\pi i[/tex][tex]\sum Res f(z) [/tex]

    and the summation is taken for all the poles of f(z) enclosed by the counter at which the integration is performed .

    now i have read somewhere that

    [tex]\oint \frac{f(z)dz}{z^{n+1}}[/tex] = 2[tex]\pi i[/tex][tex]\sum Res f(z) [tex]a^{n}[/tex]
    Last edited: Apr 14, 2008
  2. jcsd
  3. Apr 14, 2008 #2
    What's the question? And you messed up the second math disply. :smile:
  4. Apr 14, 2008 #3
    i'm sorry !! it took me half an hour writing up , and i don't know how it got posted , but it really looks bad :) .
    anyway ... my question is : for the second integral - the one with z raised to n+1 in the denominator - is it possible to evaluate it using the Residue theorem ? what i have read that it can be evaluated using a series in which each pole is raised to n and multiplied with it's residue .
    again , i'm very sorry , but latex needs to improved deeply .
  5. Apr 15, 2008 #4
    come on guys .... !!
  6. Apr 15, 2008 #5
    ok , now i got things going right .

    for a function f

    [tex]\oint f(z)dz[/tex] = 2[tex]\pi i [/tex] [tex]\sum Res(f,z_k)[/tex]

    if f is a rational function , does the following relation hold ??

    [tex]\oint z^n f(z)dz[/tex] = 2[tex]\pi i [/tex] [tex]\sum Res(f,z_k)[/tex] [tex]\ {z_k}^n[/tex]

    where [tex] \ z_k [/tex] are the poles of f .

    any help is appreciated .
    Last edited: Apr 15, 2008
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