Residue theorem

  • Thread starter mmzaj
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Main Question or Discussion Point

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:

Answers and Replies

What's the question? And you messed up the second math disply. :smile:
 
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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 .
 
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come on guys .... !!
 
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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 .
 
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