Analyticity of holomorphic functions explained

f

of a complex variable

z

:

a

if it is differentiable at every point within some open disk centered at

a

, and

a

if in some open disk centered at

a

it can be expanded as a convergent power series f(z)=\sum_^\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 and vice versa. Among the corollaries of this theorem are

S

with an accumulation point inside the intersection of their domains also agree everywhere in every connected open subset of their domains that contains the set

S

, and

a

to the nearest non-removable 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.

Proof

The argument, first given by Cauchy, hinges on Cauchy's integral formula and the power series expansion of the expression

1
w-z

.

Let

D

be an open disk centered at

a

and suppose

f

is differentiable everywhere within an open neighborhood containing the closure of

D

. Let

C

be the positively oriented (i.e., counterclockwise) circle which is the boundary of

D

and let

z

be a point in

D

. Starting with Cauchy's integral formula, we have

\begin{align}f(z)&{}={1\over2\pii}\intC{f(w)\overw-z}dw\\[10pt] &{}={1\over2\pii}\intC{f(w)\over(w-a)-(z-a)}dw\\[10pt] &{}={1\over2\pii}\intC{1\overw-a}{1\over1-{z-a\overw-a}}f(w)dw\\[10pt] &{}={1\over2\pii}\intC{1\over

infty\left({z-a
w-a}{\sum
n=0

\overw-a}\right)n}f(w)dw

infty{1
\\[10pt] &{}=\sum
n=0

\over2\pii}\intC{(z-a)n\over(w-a)n+1

} f(w)\,\mathrmw.\end

Interchange of the integral and infinite sum is justified by observing that

f(w)/(w-a)

is bounded on

C

by some positive number

M

, while for all

w

in

C

\left|z-a
w-a

\right|\leqr<1

for some positive

r

as well. We therefore have

\left|{(z-a)n\over(w-a)n+1}f(w)\right|\leMrn,

on

C

, and as the Weierstrass M-test shows the series converges uniformly over

C

, the sum and the integral may be interchanged.

As the factor

(z-a)n

does not depend on the variable of integration

w

, it may be factored out to yield
infty
f(z)=\sum
n=0

(z-a)n{1\over2\pii}\intC{f(w)\over(w-a)n+1

} \,\mathrmw,

which has the desired form of a power series in

z

:
infty
f(z)=\sum
n=0
n
c
n(z-a)

with coefficients

cn={1\over2\pii}\intC{f(w)\over(w-a)n+1

} \,\mathrmw.

Remarks

f

.

z

is any point that is closer to the center

a

than is any singularity of

f

. Therefore, the radius of convergence of the Taylor series cannot be smaller than the distance from

a

to the nearest singularity (nor can it be larger, since power series have no singularities in the interiors of their circles of convergence).

U

of

a

, then they coincide on the open disk

Bd(a)

, where

d

is the distance from

a

to the nearest singularity