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 cycles . 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
G
⊆
C
{\displaystyle G\subseteq \mathbb {C} }
D
{\displaystyle D}
D
¯
⊆
G
{\displaystyle {\bar {D}}\subseteq G}
f
:
G
→
C
{\displaystyle f\colon G\to \mathbb {C} }
f
(
z
)
=
1
2
π
i
∫
∂
D
f
(
w
)
w
−
z
,
d
w
{\displaystyle f(z)={\frac {1}{2\pi i}}\int _{\partial D}{\frac {f(w)}{w-z}},dw}
for each
z
∈
D
{\displaystyle z\in D}
By slightly enlarging the radius of the circular disk, we find an open circular disk
U
{\displaystyle U}
D
¯
⊆
U
⊆
G
{\displaystyle {\bar {D}}\subseteq U\subseteq G}
g
:
U
→
C
{\displaystyle g\colon U\to \mathbb {C} }
g
(
w
)
:=
{
f
(
w
)
−
f
(
z
)
w
−
z
w
≠
z
f
′
(
z
)
w
=
z
{\displaystyle g(w):=\left\{{\begin{array}{ll}{\frac {f(w)-f(z)}{w-z}}&w\neq z\\f'(z)&w=z\end{array}}\right.}
The function
g
{\displaystyle g}
U
{\displaystyle U}
U
−
z
{\displaystyle U-{z}}
Cauchy integral theorem on
U
{\displaystyle U}
0
=
∫
∂
D
g
(
w
)
,
d
w
=
∫
∂
D
f
(
w
)
w
−
z
,
d
w
−
f
(
z
)
∫
∂
D
d
w
w
−
z
{\displaystyle 0=\int _{\partial D}g(w),dw=\int _{\partial D}{\frac {f(w)}{w-z}},dw-f(z)\int _{\partial D}{\frac {dw}{w-z}}}
For
z
∈
D
{\displaystyle z\in D}
h
(
z
)
:=
∫
∂
D
d
w
w
−
z
{\displaystyle h(z):=\int _{\partial D}{\frac {dw}{w-z}}}
h
{\displaystyle h}
holomorphic with
h
′
(
z
)
=
−
∫
∂
D
d
w
(
w
−
z
)
2
{\displaystyle h'(z)=-\int _{\partial D}{\frac {dw}{(w-z)^{2}}}}
Since the integrand
d
w
(
w
−
z
)
2
{\displaystyle {\frac {dw}{(w-z)^{2}}}}
D
{\displaystyle D}
h
′
(
z
)
=
−
∫
∂
D
d
w
(
w
−
z
)
2
=
0
{\displaystyle h'(z)=-\int _{\partial D}{\frac {dw}{(w-z)^{2}}}=0}
Because
h
′
(
z
)
=
0
{\displaystyle h'(z)=0}
D
{\displaystyle D}
h
{\displaystyle h}
h
(
z
)
{\displaystyle h(z)}
h
(
z
0
)
{\displaystyle h(z_{0})}
D
{\displaystyle D}
h
(
z
0
)
=
2
π
i
{\displaystyle h(z_{0})=2\pi i}
0
=
∫
∂
D
f
(
w
)
w
−
z
,
d
w
−
f
(
z
)
∫
∂
D
d
w
w
−
z
⟺
f
(
z
)
=
1
2
π
i
∫
∂
D
f
(
w
)
w
−
z
,
d
w
{\displaystyle 0=\int _{\partial D}{\frac {f(w)}{w-z}},dw-f(z)\int _{\partial D}{\frac {dw}{w-z}}\iff f(z)={\frac {1}{2\pi i}}\int _{\partial D}{\frac {f(w)}{w-z}},dw}
This proves the statement.
For Cycles in Arbitrary Open Sets
edit
Let
G
⊆
C
{\displaystyle G\subseteq \mathbb {C} }
Γ
{\displaystyle \Gamma }
null-homologous cycle in
G
{\displaystyle G}
f
:
G
→
C
{\displaystyle f\colon G\to \mathbb {C} }
holomorphic . Then,
n
(
Γ
,
z
)
⋅
f
(
z
)
=
1
2
π
i
∫
Γ
f
(
w
)
w
−
z
,
d
w
{\displaystyle n(\Gamma ,z)\cdot f(z)={\frac {1}{2\pi i}}\int _{\Gamma }{\frac {f(w)}{w-z}},dw}
for each
z
∈
G
∖
s
p
u
r
(
Γ
)
{\displaystyle z\in G\setminus \mathrm {spur} (\Gamma )}
n
(
Γ
,
⋅
)
{\displaystyle n(\Gamma ,\cdot )}
winding number .
Define a function
g
:
G
2
→
C
{\displaystyle g\colon G^{2}\to \mathbb {C} }
g
(
z
,
w
)
:=
{
f
(
w
)
−
f
(
z
)
w
−
z
w
≠
z
f
′
(
z
)
w
=
z
{\displaystyle g(z,w):=\left\{{\begin{array}{ll}{\frac {f(w)-f(z)}{w-z}}&w\neq z\\f'(z)&w=z\end{array}}\right.}
defined.
We demonstrate the continuity in both variables. Let
(
w
0
,
z
0
)
∈
U
×
U
{\displaystyle (w_{0},z_{0})\in U\times U}
z
0
≠
w
0
{\displaystyle z_{0}\neq w_{0}}
g
{\displaystyle g}
(
w
0
,
z
0
)
{\displaystyle (w_{0},z_{0})}
z
0
=
w
0
{\displaystyle z_{0}=w_{0}}
δ
{\displaystyle \delta }
U
δ
(
z
0
)
⊂⊂
G
{\displaystyle U_{\delta }(z_{0})\subset \subset G}
g
(
w
,
z
)
−
g
(
z
0
,
z
0
)
{\displaystyle g(w,z)-g(z_{0},z_{0})}
U
δ
(
z
0
)
×
U
δ
(
z
0
)
{\displaystyle U_{\delta }(z_{0})\times U_{\delta }(z_{0})}
w
=
z
{\displaystyle w=z}
g
(
z
,
z
)
−
g
(
z
0
,
z
0
)
=
f
′
(
z
)
−
f
′
(
z
0
)
{\displaystyle g(z,z)-g(z_{0},z_{0})=f'(z)-f'(z_{0})}
b) In the case
w
≠
z
{\displaystyle w\neq z}
g
(
w
,
z
)
−
g
(
z
0
,
z
0
)
=
f
(
w
)
−
f
(
z
)
w
−
z
−
f
′
(
z
0
)
=
1
w
−
z
∫
[
z
,
w
]
(
f
′
(
v
)
−
f
′
(
z
0
)
)
d
v
{\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}
Now, as a consequence of Cauchy's formulas for circles! the derivative
f
′
{\displaystyle f'}
z
0
{\displaystyle z_{0}}
ϵ
>
0
{\displaystyle \epsilon >0}
δ
>
0
{\displaystyle \delta >0}
|
f
′
(
v
)
−
f
′
(
z
0
)
|
<
ϵ
{\displaystyle |f'(v)-f'(z_{0})|<\epsilon }
for all
v
∈
U
δ
(
z
0
)
{\displaystyle v\in U_{\delta }(z_{0})}
This implies, in case a:
|
g
(
z
,
z
)
−
g
(
z
0
,
z
0
)
|
<
ϵ
;
{\displaystyle |g(z,z)-g(z_{0},z_{0})|<\epsilon ;}
and in case b:
|
g
(
w
,
z
)
−
g
(
z
0
,
z
0
)
|
≤
1
|
w
−
z
|
|
w
−
z
|
⋅
sup
w
∈
[
w
,
z
]
|
f
′
(
v
)
−
f
′
(
z
0
)
|
<
ϵ
.
{\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 .}
We now define
h
0
(
z
)
=
∫
Γ
g
(
w
,
z
)
d
w
.
{\displaystyle h_{0}(z)=\int _{\Gamma }g(w,z)dw.}
h
0
{\displaystyle h_{0}}
G
{\displaystyle G}
Let
γ
{\displaystyle \gamma }
G
{\displaystyle G}
∫
γ
h
0
(
z
)
d
z
=
0
{\displaystyle \int _{\gamma }h_{0}(z)dz=0}
prove it is
∫
γ
h
0
(
z
)
d
z
=
∫
Γ
∫
γ
g
(
w
,
z
)
d
z
d
w
=
0.
{\displaystyle \int _{\gamma }h_{0}(z)dz=\int _{\Gamma }\int _{\gamma }g(w,z)dz\,dw=0.}
because the integrations are commutable due to the continuity of the integrand on
G
×
G
{\displaystyle G\times G}
w
{\displaystyle w}
g
(
w
,
z
)
{\displaystyle g(w,z)}
z
{\displaystyle z}
w
≠
z
{\displaystyle w\neq z}
By Goursat's theorem, it follows that
∫
γ
g
(
w
,
z
)
d
z
=
0.
{\displaystyle \int _{\gamma }g(w,z)dz=0.}
this of course also mean that
∫
γ
h
0
(
z
)
d
z
=
∫
Γ
∫
γ
g
(
w
,
z
)
d
z
d
w
=
0.
{\displaystyle \int _{\gamma }h_{0}(z)dz=\int _{\Gamma }\int _{\gamma }g(w,z)dz\,dw=0.}
so far we have not yet exploited the conditions above
Γ
{\displaystyle \Gamma }
G
0
=
{
z
∈
C
:
n
(
Γ
,
z
)
=
0
}
{\displaystyle G_{0}=\{z\in \mathbb {C} :n(\Gamma ,z)=0\}}
Since on
G
∩
G
0
{\displaystyle G\cap G_{0}}
h
0
{\displaystyle h_{0}}
h
0
(
z
)
=
∫
Γ
f
(
w
)
w
−
z
d
w
=
h
1
(
z
)
,
{\displaystyle h_{0}(z)=\int _{\Gamma }{\frac {f(w)}{w-z}}dw=h_{1}(z),}
and since the function
h
1
{\displaystyle h_{1}}
G
0
{\displaystyle G_{0}}
h
0
{\displaystyle h_{0}}
h
{\displaystyle h}
G
∪
G
0
{\displaystyle G\cup G_{0}}
h
(
z
)
=
{
h
0
(
z
)
z
∈
G
h
1
(
z
)
z
∈
G
0
{\displaystyle h(z)=\left\{{\begin{array}{ll}h_{0}(z)&z\in G\\h_{1}(z)&z\in G_{0}\end{array}}\right.}
Now
Γ
{\displaystyle \Gamma }
G
∪
G
0
=
C
,
{\displaystyle G\cup G_{0}=\mathbb {C} ,}
i.e.
h
{\displaystyle h}
For
h
{\displaystyle h}
G
0
{\displaystyle G_{0}}
|
h
(
z
)
|
=
|
h
1
(
z
)
|
≤
1
d
i
s
t
(
z
,
Γ
)
L
(
Γ
)
max
Γ
|
f
|
(
∗
)
;
{\displaystyle |h(z)|=|h_{1}(z)|\leq {\frac {1}{dist(z,\Gamma )}}L(\Gamma )\max _{\Gamma }|f|\;\;\;(*);}
where
L
(
Γ
)
=
∑
|
n
k
|
L
(
γ
k
)
{\displaystyle L(\Gamma )=\sum |n_{k}|L(\gamma _{k})}
Γ
=
n
k
⋅
γ
k
{\displaystyle \Gamma =n_{k}\cdot \gamma _{k}}
G
0
{\displaystyle G_{0}}
z
{\displaystyle z}
h
{\displaystyle h}
z
ν
∈
G
0
{\displaystyle z_{\nu }\in G_{0}}
|
z
ν
|
≥
ν
{\displaystyle |z_{\nu }|\geq \nu }
lim
ν
→
∞
(
z
ν
)
=
0
,
{\displaystyle \lim \limits _{\nu \to \infty }(z_{\nu })=0,}
thus we conculude that
h
≡
0
{\displaystyle h\equiv 0}
h
0
≡
0
{\displaystyle h_{0}\equiv 0}
From the Cauchy integral formula, it follows that every holomorphic function is infinitely differentiable because the integrand in
z
{\displaystyle z}
Let
G
⊆
C
{\displaystyle G\subseteq \mathbb {C} }
D
{\displaystyle D}
D
¯
⊆
G
{\displaystyle {\bar {D}}\subseteq G}
f
:
G
→
C
{\displaystyle f\colon G\to \mathbb {C} }
f
{\displaystyle f}
n
∈
N
{\displaystyle n\in \mathbb {N} }
f
(
n
)
(
z
)
=
n
!
2
π
i
∫
∂
D
f
(
w
)
(
w
−
z
)
n
+
1
,
d
w
{\displaystyle f^{(n)}(z)={\frac {n!}{2\pi i}}\int _{\partial D}{\frac {f(w)}{(w-z)^{n+1}}},dw}
for each
z
∈
D
{\displaystyle z\in D}
Let
G
⊆
C
{\displaystyle G\subseteq \mathbb {C} }
Γ
∈
C
(
G
)
{\displaystyle \Gamma \in C(G)}
null-homologous cycle, and
f
:
G
→
C
{\displaystyle f\colon G\to \mathbb {C} }
holomorphic . Then
n
(
Γ
,
z
)
⋅
f
(
n
)
(
z
)
=
n
!
2
π
i
∫
Γ
f
(
w
)
(
w
−
z
)
n
+
1
,
d
w
{\displaystyle n(\Gamma ,z)\cdot f^{(n)}(z)={\frac {n!}{2\pi i}}\int _{\Gamma }{\frac {f(w)}{(w-z)^{n+1}}},dw}
for each
z
∈
G
∖
T
r
a
c
e
(
Γ
)
{\displaystyle z\in G\setminus \mathrm {Trace} (\Gamma )}
n
∈
N
{\displaystyle n\in \mathbb {N} }
Moreover, every holomorphic function is analytic at every point, i.e., it can be expanded into a power series:
Let
G
⊆
C
{\displaystyle G\subseteq \mathbb {C} }
f
:
G
→
C
{\displaystyle f\colon G\to \mathbb {C} }
z
0
∈
G
{\displaystyle z_{0}\in G}
r
>
0
{\displaystyle r>0}
B
¯
r
(
z
0
)
⊆
G
{\displaystyle {\bar {B}}_{r}(z_{0})\subseteq G}
f
{\displaystyle f}
B
r
(
z
0
)
{\displaystyle B_{r}(z_{0})}
f
(
z
)
=
∑
n
=
0
∞
a
n
(
z
−
z
0
)
n
{\displaystyle f(z)=\sum _{n=0}^{\infty }a_{n}(z-z_{0})^{n}}
where the coefficients are given by
a
n
=
1
2
π
i
∫
∂
B
r
(
z
0
)
f
(
w
)
(
w
−
z
0
)
n
+
1
,
d
w
{\displaystyle a_{n}={\frac {1}{2\pi i}}\int _{\partial B_{r}(z_{0})}{\frac {f(w)}{(w-z_{0})^{n+1}}},dw}
For
z
∈
B
r
(
z
0
)
{\displaystyle z\in B_{r}(z_{0})}
w
∈
∂
B
r
(
z
0
)
{\displaystyle w\in \partial B_{r}(z_{0})}
1
w
−
z
=
1
(
w
−
z
0
)
−
(
z
−
z
0
)
=
1
w
−
z
0
⋅
1
1
−
z
−
z
0
w
−
z
0
=
1
w
−
z
0
∑
n
=
0
∞
(
z
−
z
0
)
n
(
w
−
z
0
)
n
.
{\displaystyle {\begin{array}{rl}\displaystyle {\frac {1}{w-z}}&=\displaystyle {\frac {1}{(w-z_{0})-(z-z_{0})}}\\&=\displaystyle {\frac {1}{w-z_{0}}}\cdot {\frac {1}{1-{\frac {z-z_{0}}{w-z_{0}}}}}\\&=\displaystyle {\frac {1}{w-z_{0}}}\sum _{n=0}^{\infty }{\frac {(z-z_{0})^{n}}{(w-z_{0})^{n}}}.\end{array}}}
the real converges absolutely
|
z
−
z
0
|
<
r
=
|
w
−
z
0
|
{\displaystyle |z-z_{0}|<r=|w-z_{0}|}
f
(
z
)
=
1
2
π
i
∫
∂
B
r
(
z
0
)
f
(
w
)
w
−
z
d
w
=
1
2
π
i
∫
∂
B
r
(
z
0
)
1
w
−
z
0
f
(
w
)
(
z
−
z
0
)
n
(
w
−
z
0
)
n
d
w
=
1
2
π
i
∑
n
=
0
∞
∫
∂
B
r
(
z
0
)
f
(
w
)
(
w
−
z
0
)
n
+
1
d
w
⋅
(
z
−
z
0
)
n
=
1
2
π
i
∑
n
=
0
∞
∫
∂
B
r
(
z
0
)
f
(
w
)
(
w
−
z
0
)
n
+
1
d
w
⋅
(
z
−
z
0
)
n
{\displaystyle {\begin{array}{rl}f(z)&=\displaystyle {\frac {1}{2\pi i}}\int _{\partial B_{r}(z_{0})}{\frac {f(w)}{w-z}}\,dw\\&=\displaystyle {\frac {1}{2\pi i}}\int _{\partial B_{r}(z_{0})}{\frac {1}{w-z_{0}}}{\frac {f(w)(z-z_{0})^{n}}{(w-z_{0})^{n}}}\,dw\\&=\displaystyle {\frac {1}{2\pi i}}\sum _{n=0}^{\infty }\int _{\partial B_{r}(z_{0})}{\frac {f(w)}{(w-z_{0})^{n+1}}}\,dw\cdot (z-z_{0})^{n}\\&=\displaystyle {\frac {1}{2\pi i}}\sum _{n=0}^{\infty }\int _{\partial B_{r}(z_{0})}{\frac {f(w)}{(w-z_{0})^{n+1}}}\,dw\cdot (z-z_{0})^{n}\end{array}}}
![{\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)
![{\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)
