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 Exercise [07.04] 
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Joined: 12 Mar 2008, 10:57
Posts: 69
Location: India
Post Exercise [07.04]
\oint{f(z)}dz
=\sum{\oint{\frac{h(z)}{(z-\alpha_{k})^n}}dz}(where n is a function of k.)
=\sum{\frac{2\pi\iota\ h^{n-1}(\alpha_k)}{(n-1)!}}(Cauchy's formula)
=2\pi\iota\sum{\frac{h^{n-1}(\alpha_k)}{(n-1)!}}


16 May 2008, 13:32
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Joined: 07 Jun 2008, 08:21
Posts: 235
Post Re: Exercise [07.04]
Sameed
I think your proof needs to make it clear which contour the \oint{f(z)}dz is referring to in each case. So, for example, in the first case I think you should write:
\oint_\Gamma {f(z)}dz to indicate the closed contour \Gammawhich encloses all the poles (as stated in the exercise), and then:
=\sum{\oint_{\Gamma_k}{\frac{h(z)}{(z-\alpha_{k})^n}}dz}(where n is a function of k.)
with an explanation of what the \Gamma_k are (maybe with a diagram?).

Also an explanation of how travelling round the \Gamma_k for all k is equivalent to travelling round \Gamma would be very useful to other members of the forum when trying to understand your proof.


31 Dec 2008, 12:55

Joined: 13 Aug 2009, 00:08
Posts: 13
Post Re: Exercise [07.04]
Oh! Just spotted a typo here -- very tiny, but the contour by the integral sign is actually a lower-case "r", when it should be an upper-case gamma \Gamma.

But thanks for the solution -- I couldn't quite think of the right way to start this one.


16 Aug 2009, 21:16

Joined: 14 May 2011, 20:28
Posts: 1
Post Re: Exercise [07.04]
vasco wrote:
Also an explanation of how travelling round the \Gamma_k for all k is equivalent to travelling round \Gamma would be very useful to other members of the forum when trying to understand your proof.

Hi,

I did a simple figure to illustrate this. See below.


Attachments:
File comment: Figure showing the equivalence of different closed contours.
contour.png
contour.png [ 11.24 KiB | Viewed 1923 times ]
14 May 2011, 20:34
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