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 Post subject: removable singularities
PostPosted: Wed, 17 Nov 2010 15:53:08 UTC 
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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 \in  H (A(z_{0} , 0, R)) -annulus is bounded . Fix r \in (0,R) and show that

f(z) = \frac{1}{2 \pi i}  \int_{\partial_{D(z_{0} , r)}}  \frac {f(w)dw} {w-z}  - \frac{1}{2 \pi i}  \int_{\partial_{D(z_{0} , \rho )}}  \frac {f(w)dw} {w-z}

whenver 0 <  \rho <   | z - z_{0} | <  r . Now fix z and consider what happens as \rhotends 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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PostPosted: Wed, 17 Nov 2010 17:01:44 UTC 
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
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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 \in  H (A(z_{0} , 0, R)) -annulus is bounded . Fix r \in (0,R) and show that

f(z) = \frac{1}{2 \pi i}  \int_{\partial_{D(z_{0} , r)}}  \frac {f(w)dw} {w-z}  - \frac{1}{2 \pi i}  \int_{\partial_{D(z_{0} , \rho )}}  \frac {f(w)dw} {w-z}

whenver 0 <  \rho <   | z - z_{0} | <  r . Now fix z and consider what happens as \rhotends to 0.

Ok, how do get this integral formulation? After this , how should we proceed to to understand z_{0} 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 z_0. Then the closed loop (outer circle)-(p)-(inner circle reversed)-(p reversed) has winding number 1 about the point z_0, 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.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject:
PostPosted: Thu, 18 Nov 2010 12:55:08 UTC 
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Joined: Mon, 15 Nov 2010 19:28:26 UTC
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Ok, I got the idea. :o So, how do we proceed then? We need to have a f( z_ {0} ), should we define it as :

f(z_{0} )  = \frac{1}{2 \pi i}  \int_{ \partial_{D(z_{0} , r) }} \frac {f(w)dw}{w - z_{0} }


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 Post subject:
PostPosted: Thu, 18 Nov 2010 14:16:09 UTC 
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6633
Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
hermanni wrote:
Ok, I got the idea. :o So, how do we proceed then? We need to have a f( z_ {0} ), should we define it as :

f(z_{0} )  = \frac{1}{2 \pi i}  \int_{ \partial_{D(z_{0} , r) }} \frac {f(w)dw}{w - z_{0} }


That is what you morally should do, yes, except you will need to check analyticity at z_0.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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