## Residue theorem

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$$\pi i$$$$\sum Res f(z) [tex]a^{n}$$

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

## Residue theorem

come on guys .... !!

 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 .