Complex-analytic implies holomorphic: Difference between revisions

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(New page: ==Statement== Suppose <math>U \subset \mathbb{C}</math> is a domain and <math>f:U \to \mathbb{C}</math> is a complex-analytic function: for every point <math>z_0 \in U</math>, there e...)
 
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==Proof==
==Proof==


To execute this proof, we need to argue that
We guess that the power series actually represents the ''Taylor expansion''

Revision as of 22:29, 14 April 2008

Statement

Suppose UC is a domain and f:UC is a complex-analytic function: for every point z0U, there exists a real number r>0 and a power series n=0an(zz0)n such that the power series converges and agrees with f in the ball of radius r.

Then, f is a holomorphic function: it is complex-differentiable, and the complex differential is a continuous function. In fact, f is differentiable infinitely often.

Proof

We guess that the power series actually represents the Taylor expansion