The Cauchy integral formula, alongside the Cauchy's integral theorem, is one of the central statements in complex analysis. Here, we present two variants: the 'classical' formula for circular disks and a relatively general version for null-homologous Chain. Note that we will deduce the circular disk version from Cauchy's integral theorem, but for the general variant, we proceed in the opposite direction.
Let
be an open set,
a circular disk with
, and
holomorphic. Then, we have

for each
.
By slightly enlarging the radius of the circular disk, we find an open circular disk
such that
. Define
by

The function
is continuous on
and holomorphic on
. Thus, we can apply the Cauchy integral theorem on
and obtain

For
, define
. Then
is holomorphic with

Since the integrand
has a primitive in
, we find

Because
throughout
, it follows that
is constant. Thus,
always takes the same value as at the center
of the disk
, i.e.,
. Hence,

This proves the statement.
Let
be an open set,
a null-homologous cycle in
, and
holomorphic. Then,

for each
, where
denotes the winding number.
Define a function
by

defined.
We demonstrate the continuity in both variables. Let
with
, then
is given in the vicinity of
by the above formula and is trivially continuous.Now let
. We choose a
-neighborhood
and examine
auf
.
a) In the case
:
:

b) In the case
:
![{\displaystyle g(w,z)-g(z_{0},z_{0})={\frac {f(w)-f(z)}{w-z}}-f'(z_{0})={\frac {1}{w-z}}\int _{[z,w]}(f'(v)-f'(z_{0}))dv}](https://wikimedia.org/api/rest_v1/media/math/render/svg/60c85027e7063708facf238da1c0b467a38d693c)
Now, as a consequence of Cauchy's formulas for circles! the derivative
is continuous in
. For a given
we can choose
such that

for all
.
This implies, in case a:

and in case b:
![{\displaystyle |g(w,z)-g(z_{0},z_{0})|\leq {\frac {1}{|w-z|}}|w-z|\cdot \sup \limits _{w\in [w,z]}|f'(v)-f'(z_{0})|<\epsilon .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a255d3c8c63f8a44450a3bbd7145d9aaf2c205ad)
We now define

function is continuous on whole of
; we will show that it is even holomorphic. For this, we use Morera's theorem.
Let
be the oriented boundary of a triangle that lies entirely with in
. We must show

prove it is

because the integrations are commutable due to the continuity of the integrand on
For fixed ,
the function is
in the Variable
continuous in and holomorphic for
, hence holomorphic everywhere.
By Goursat's theorem, it follows that

this of course also mean that

so far we have not yet exploited the conditions above
. We will do so
.
Since on
the function
has a simpler form, namely

and since the function
is clearly holomorphic on the entire
, we can extend
to a holomorphic function
defined on the entire by

Now
is null-homologous in , and thus

i.e.
is an entire function.
For
we have
the notation:

where
, if
is.
contains the complement of a sufficiently large circle around 0. Therefore, the above inequality holds for all
in this region: it follows that
is bounded, and by Liouville's theorem, it must be constant. If we choose a sequence
such that
, the inequality (*) again implies that:

thus we conculude that
, and in particular
; this is what we wanted to prove
From the Cauchy integral formula, it follows that every holomorphic function is infinitely differentiable because the integrand in
is infinitely differentiable. We obtain the following results:
Let
be an open set,
a circular disk with
, and
holomorphic. Then
is infinitely differentiable, and for each
, we have

for each
.
Let
be an open set,
a null-homologous cycle, and
holomorphic. Then

for each
and
.
Moreover, every holomorphic function is analytic at every point, i.e., it can be expanded into a power series:
Let
be open, and
holomorphic. Let
and
such that
. Then
can be represented on
by a convergent power series

where the coefficients are given by
.
For
,
we have:

the real converges absolutely
and we obtain

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