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.