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 Post subject: removable singularitiesPosted: Wed, 17 Nov 2010 15:53:08 UTC
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Joined: Mon, 15 Nov 2010 19:28:26 UTC
Posts: 26
Hi all,
I'm trying to solve this question:

Prove:
If a holomorphic function is bounded near an isolated singularity then the singularity is removable.
Using Cauchy's integral formula , homology version as follows:

Suppose that f -annulus is bounded . Fix and show that

whenver . Now fix z and consider what happens as tends to 0.

Ok, how do get this integral formulation? After this , how should we proceed to to understand [tex]z_{0} [\tex] is a removable singularity? I can't think of any solution , any help is appreciated.

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 Post subject: Re: removable singularitiesPosted: Wed, 17 Nov 2010 17:01:44 UTC
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 7003
Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
hermanni wrote:
Hi all,
I'm trying to solve this question:

Prove:
If a holomorphic function is bounded near an isolated singularity then the singularity is removable.
Using Cauchy's integral formula , homology version as follows:

Suppose that f -annulus is bounded . Fix and show that

whenver . Now fix z and consider what happens as tends to 0.

Ok, how do get this integral formulation? After this , how should we proceed to to understand is a removable singularity? I can't think of any solution , any help is appreciated.

I wouldn't call a punctured disc an annulus.

Look up Riemann's removable singularity criterion if the question doesn't actually need you to use Cauchy's integral formula. It is a much nicer proof.

As for how you get this version of Cauchy integral formula, pick a point each from the outer circle and the inner circle, join them by a simple C^1 path p not passing through . Then the closed loop (outer circle)-(p)-(inner circle reversed)-(p reversed) has winding number 1 about the point , and the integrals along p and p-reversed cancel each other out, or equivalently, the outer circle has winding number 1 about every point of the open annulus, and the inner circle has winding number 0 about these points.

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 Post subject: Posted: Thu, 18 Nov 2010 12:55:08 UTC
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Joined: Mon, 15 Nov 2010 19:28:26 UTC
Posts: 26
Ok, I got the idea. So, how do we proceed then? We need to have a , should we define it as :

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 Post subject: Posted: Thu, 18 Nov 2010 14:16:09 UTC
 Moderator

Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 7003
Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
hermanni wrote:
Ok, I got the idea. So, how do we proceed then? We need to have a , should we define it as :

That is what you morally should do, yes, except you will need to check analyticity at .

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