Residue Theorem: Theory & Application

[mmzaj]

the Residue theorem states that :
\oint {f(z)dz} = 2\pi i\sum Res f(z)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

\oint \frac{f(z)dz}{z^{n+1}} = 2\pi i\sum Res f(z) a^{n}

 

[Pere Callahan]

What's the question? And you messed up the second math disply.

 

[mmzaj]

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 .

 

[mmzaj]

come on guys ... !

 

[mmzaj]

ok , now i got things going right . for a function f

\oint f(z)dz = 2\pi i \sum Res(f,z_k)

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

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

where \ z_k are the poles of f .

any help is appreciated .