Complex-analytic implies holomorphic: Difference between revisions

Latest revision as of 16:08, 12 September 2008

This article gives the statement, and possibly proof, of a basic fact in complex analysis.
View a complete list of basic facts in complex analysis

Statement

Suppose $U\subset \mathbb {C}$ is a domain and $f:U\to \mathbb {C}$ is a complex-analytic function: for every point $z_{0}\in U$ , there exists a real number $r>0$ and a power series $\sum _{n=0}^{\infty }a_{n}(z-z_{0})^{n}$ 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.

Facts used

Power series is infinitely differentiable in disk of convergence

Proof

The proof follows directly from fact (1).

@@ Line 6: / Line 6: @@
 Then, <math>f</math> is a [[holomorphic function]]: it is complex-differentiable, and the complex differential is a continuous function. In fact, <math>f</math> is differentiable infinitely often.
-==Proof==
+==Facts used==
-We will show the following somewhat stronger fact: if <math>f</math> is a complex-analytic function at <math>z_0</math> with a power series:
+# [[uses::Power series is infinitely differentiable in disk of convergence]]
-<math>f(z) := \sum_{n=0}^\infty a_n(z - z_0)^n</math>
+==Proof==
-By basic facts of power series, <math>f</math> converges absolutely, and uniformly on every disc of radius strictly smaller than its radius of convergence. We'll show that <math>f'</math> exists on the open disc of the radius of convergence, and is given by the following absolutely convergent power series:
-<math>f'(z) := \sum_{n=0}^\infty na_n(z - z_0)^{n-1}</math>
-Note that the power series on the right clearly has the same radius of convergence as the power series for <math>f</math>, so we can concentrate on showing that it equals <math>f</math>. Observe that if <math>s_n</math> is the partial sum till the <math>n^{th}</math> term of <math>f</math>, then <math>s_n'(z)</math> indeed equals the sum, till the <math>n^{th}</math> term, for <math>f'(z)</math>. Thus, what we really need to show is that the error term goes to zero.
+The proof follows directly from fact (1).