Proof that holomorphic functions are analytic

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In complex analysis, a field of mathematics, a complex-valued function f of a complex variable

is holomorphic at a point a iff it is differentiable at every point within some open disk centered at a, and

is analytic at a if in some open disk centered at a it can be expanded as a convergent power series

$f(z)=\sum_{n=0}^\infty c_n(z-a)^n$

(this implies that the radius of convergence is positive).

One of the most important theorems of complex analysis is that holomorphic functions are analytic. Among the corollaries of this theorem are

the fact that two holomorphic functions that agree at every point of an infinite set with an accumulation point inside the intersection of their domains also agree everywhere in some open set, and

the fact that holomorphic functions are differentiable not just once, but infinitely often, and

the fact that the radius of convergence is always the distance from the center a to the nearest singularity; if there are no singularities (i.e., if f is an entire function), then the radius of convergence is infinite. Strictly speaking, this is not a corollary of the theorem but rather a by-product of the proof.

1 Proof
2 The generalized Cauchy integral formula
3 A by-product of the proof
4 External links

[edit] Proof

Suppose f is differentiable everywhere within some open disk centered at a. Let z be within that open disk. Let C be a positively oriented (i.e., counterclockwise) circle centered at a, lying within that open disk but farther from a than z is. Then, using Cauchy's integral formula, we get

$\begin{align}f(z) &{}= {1 \over 2\pi i}\int_C {f(w) \over w-z}\,dw \\ \\ &{}= {1 \over 2\pi i}\int_C {1 \over w-a}\cdot{w-a \over w-z}f(w)\,dw \\ \\ &{}= {1 \over 2\pi i}\int_C {1 \over w-a}\cdot{w-a \over (w-a)-(z-a)}f(w)\,dw \\ \\ &{}={1 \over 2\pi i}\int_C {1 \over w-a}\cdot{1 \over 1-{z-a \over w-a}}f(w)\,dw \\ \\ &{}={1 \over 2\pi i}\int_C {1 \over w-a}\cdot{\sum_{n=0}^\infty\left({z-a \over w-a}\right)^n} f(w)\,dw \\ \\ &{}=\sum_{n=0}^\infty{1 \over 2\pi i}\int_C {(z-a)^n \over (w-a)^{n+1}} f(w)\,dw.\end{align}$

The interchange of the sum and the integeral is justified by the uniform convergence of the geometric series within subsets of its disk of convergence that are bounded away from the boundary. Since the factor (z − a)ⁿ does not depend on the variable of integration w, it can be pulled out:

$=\sum_{n=0}^\infty (z-a)^n {1 \over 2\pi i}\int_C {f(w) \over (w-a)^{n+1}} \,dw.$

And now the integral and the factor of 1/(2πi) do not depend on z, i.e., as a function of z, that whole expression is a constant c_n, so we can write:

$=\sum_{n=0}^\infty c_n(z-a)^n$

and that is the desired power series.

[edit] The generalized Cauchy integral formula

The fact that the coefficient is given by

$c_n={1 \over 2\pi i}\int_C{f(w) \over (w-a)^{n+1}}\,dw$

is a generalization of Cauchy's integral formula, since the latter is just the case in which n = 0.

[edit] A by-product of the proof

The argument works if z is any point that is closer to the center than is any singularity of f. Therefore the radius of convergence of the power series cannot be smaller than the distance from the center to the nearest singularity (nor can it be larger, since power series have no singularities in the interiors of their circles of convergence).