Cauchy's integral formula

Introduction

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.

For Circular Disks

Statement

Let $G\subseteq \mathbb {C}$ be an open set, $D$ a circular disk with ${\bar {D}}\subseteq G$, and $f\colon G\to \mathbb {C}$ holomorphic. Then, we have

$f(z)={\frac {1}{2\pi i}}\int _{\partial D}{\frac {f(w)}{w-z}},dw$

for each $z\in D$.

Proof 1

By slightly enlarging the radius of the circular disk, we find an open circular disk $U$ such that ${\bar {D}}\subseteq U\subseteq G$. Define $g\colon U\to \mathbb {C}$ by

$g(w):=\left\{{\begin{array}{ll}{\frac {f(w)-f(z)}{w-z}}&w\neq z\\f'(z)&w=z\end{array}}\right.$

Proof 2

The function $g$ is continuous on $U$ and holomorphic on $U-{z}$. Thus, we can apply the Cauchy integral theorem on $U$ and obtain

$0=\int _{\partial D}g(w),dw=\int _{\partial D}{\frac {f(w)}{w-z}},dw-f(z)\int _{\partial D}{\frac {dw}{w-z}}$

For $z\in D$, define $h(z):=\int _{\partial D}{\frac {dw}{w-z}}$. Then $h$ is holomorphic with

$h'(z)=-\int _{\partial D}{\frac {dw}{(w-z)^{2}}}$

Proof 3

Since the integrand ${\frac {dw}{(w-z)^{2}}}$ has a primitive in $D$, we find

$h'(z)=-\int _{\partial D}{\frac {dw}{(w-z)^{2}}}=0$

Proof 4

Because $h'(z)=0$ throughout $D$, it follows that $h$ is constant. Thus, $h(z)$ always takes the same value as at the center $h(z_{0})$ of the disk $D$, i.e., $h(z_{0})=2\pi i$. Hence,

$0=\int _{\partial D}{\frac {f(w)}{w-z}},dw-f(z)\int _{\partial D}{\frac {dw}{w-z}}\iff f(z)={\frac {1}{2\pi i}}\int _{\partial D}{\frac {f(w)}{w-z}},dw$

This proves the statement.

For Cycles in Arbitrary Open Sets

Statement

Let $G\subseteq \mathbb {C}$ be an open set, $\Gamma$ a null-homologous cycle in $G$, and $f\colon G\to \mathbb {C}$ holomorphic. Then,

$n(\Gamma ,z)\cdot f(z)={\frac {1}{2\pi i}}\int _{\Gamma }{\frac {f(w)}{w-z}},dw$

for each $z\in G\setminus \mathrm {spur} (\Gamma )$, where $n(\Gamma ,\cdot )$ denotes the winding number.

Proof 1

Define a function $g\colon G^{2}\to \mathbb {C}$ by

$g(z,w):=\left\{{\begin{array}{ll}{\frac {f(w)-f(z)}{w-z}}&w\neq z\\f'(z)&w=z\end{array}}\right.$

defined.

Proof 2: continuous

We demonstrate the continuity in both variables. Let $(w_{0},z_{0})\in U\times U$ with $z_{0}\neq w_{0}$, then $g$ is given in the vicinity of $(w_{0},z_{0})$ by the above formula and is trivially continuous.Now let $z_{0}=w_{0}$. We choose a $\delta$-neighborhood $U_{\delta }(z_{0})\subset \subset G$ and examine $g(w,z)-g(z_{0},z_{0})$ auf $U_{\delta }(z_{0})\times U_{\delta }(z_{0})$

. a) In the case $w=z$:
:

$g(z,z)-g(z_{0},z_{0})=f'(z)-f'(z_{0})$

Proof 3

b) In the case $w\neq z$:

$g(w,z)-g(z_{0},z_{0})={\frac {f(w)-f(z)}{w-z}}-f'(z_{0})={\frac {1}{w-z}}\int _{[z,w]}(f'(v)-f'(z_{0}))dv$

Now, as a consequence of Cauchy's formulas for circles! the derivative $f'$ is continuous in $z_{0}$. For a given $\epsilon >0$ we can choose $\delta >0$ such that

$|f'(v)-f'(z_{0})|<\epsilon$

for all $v\in U_{\delta }(z_{0})$.

Proof 4

This implies, in case a:

$|g(z,z)-g(z_{0},z_{0})|<\epsilon ;$

and in case b:

$|g(w,z)-g(z_{0},z_{0})|\leq {\frac {1}{|w-z|}}|w-z|\cdot \sup \limits _{w\in [w,z]}|f'(v)-f'(z_{0})|<\epsilon .$

We now define

$h_{0}(z)=\int _{\Gamma }g(w,z)dw.$

$h_{0}$ function is continuous on whole of $G$ ; we will show that it is even holomorphic. For this, we use Morera's theorem.

Proof 5

Let $\gamma$ be the oriented boundary of a triangle that lies entirely with in $G$ . We must show

$\int _{\gamma }h_{0}(z)dz=0$

prove it is

$\int _{\gamma }h_{0}(z)dz=\int _{\Gamma }\int _{\gamma }g(w,z)dz\,dw=0.$

because the integrations are commutable due to the continuity of the integrand on $G\times G$ For fixed , $w$ the function is $g(w,z)$ in the Variable $z$ continuous in and holomorphic for $w\neq z$, hence holomorphic everywhere.

Proof 6

By Goursat's theorem, it follows that

$\int _{\gamma }g(w,z)dz=0.$

this of course also mean that

$\int _{\gamma }h_{0}(z)dz=\int _{\Gamma }\int _{\gamma }g(w,z)dz\,dw=0.$

so far we have not yet exploited the conditions above $\Gamma$. We will do so

$G_{0}=\{z\in \mathbb {C} :n(\Gamma ,z)=0\}$.

Proof 7

Since on $G\cap G_{0}$ the function $h_{0}$ has a simpler form, namely

$h_{0}(z)=\int _{\Gamma }{\frac {f(w)}{w-z}}dw=h_{1}(z),$

and since the function $h_{1}$ is clearly holomorphic on the entire $G_{0}$, we can extend $h_{0}$ to a holomorphic function $h$ defined on the entire by$G\cup G_{0}$

$h(z)=\left\{{\begin{array}{ll}h_{0}(z)&z\in G\\h_{1}(z)&z\in G_{0}\end{array}}\right.$

Now $\Gamma$ is null-homologous in , and thus

$G\cup G_{0}=\mathbb {C} ,$

i.e. $h$ is an entire function.

Proof 8

For $h$ we have$G_{0}$ the notation:

$|h(z)|=|h_{1}(z)|\leq {\frac {1}{dist(z,\Gamma )}}L(\Gamma )\max _{\Gamma }|f|\;\;\;(*);$

where $L(\Gamma )=\sum |n_{k}|L(\gamma _{k})$, if $\Gamma =n_{k}\cdot \gamma _{k}$ is.

$G_{0}$ contains the complement of a sufficiently large circle around 0. Therefore, the above inequality holds for all $z$ in this region: it follows that $h$ is bounded, and by Liouville's theorem, it must be constant. If we choose a sequence $z_{\nu }\in G_{0}$ such that $|z_{\nu }|\geq \nu$, the inequality (*) again implies that:

$\lim \limits _{\nu \to \infty }(z_{\nu })=0,$

thus we conculude that $h\equiv 0$, and in particular $h_{0}\equiv 0$; this is what we wanted to prove

Conclusions

From the Cauchy integral formula, it follows that every holomorphic function is infinitely differentiable because the integrand in $z$ is infinitely differentiable. We obtain the following results:

For Circular Disks

Let $G\subseteq \mathbb {C}$ be an open set, $D$ a circular disk with ${\bar {D}}\subseteq G$, and $f\colon G\to \mathbb {C}$ holomorphic. Then $f$ is infinitely differentiable, and for each $n\in \mathbb {N}$, we have

$f^{(n)}(z)={\frac {n!}{2\pi i}}\int _{\partial D}{\frac {f(w)}{(w-z)^{n+1}}},dw$

for each $z\in D$.

For Cycles

Let $G\subseteq \mathbb {C}$ be an open set, $\Gamma \in C(G)$ a null-homologous cycle, and $f\colon G\to \mathbb {C}$ holomorphic. Then

$n(\Gamma ,z)\cdot f^{(n)}(z)={\frac {n!}{2\pi i}}\int _{\Gamma }{\frac {f(w)}{(w-z)^{n+1}}},dw$

for each $z\in G\setminus \mathrm {Trace} (\Gamma )$ and $n\in \mathbb {N}$.

Analyticity

Moreover, every holomorphic function is analytic at every point, i.e., it can be expanded into a power series:

Statement

Let $G\subseteq \mathbb {C}$ be open, and $f\colon G\to \mathbb {C}$ holomorphic. Let $z_{0}\in G$ and $r>0$ such that ${\bar {B}}_{r}(z_{0})\subseteq G$. Then $f$ can be represented on $B_{r}(z_{0})$ by a convergent power series

$f(z)=\sum _{n=0}^{\infty }a_{n}(z-z_{0})^{n}$

where the coefficients are given by

$a_{n}={\frac {1}{2\pi i}}\int _{\partial B_{r}(z_{0})}{\frac {f(w)}{(w-z_{0})^{n+1}}},dw$.

Proof 1

For $z\in B_{r}(z_{0})$, $w\in \partial B_{r}(z_{0})$ we have:

${\begin{array}{rl}\displaystyle {\frac {1}{w-z}}&=\displaystyle {\frac {1}{(w-z_{0})-(z-z_{0})}}\\&=\displaystyle {\frac {1}{w-z_{0}}}\cdot {\frac {1}{1-{\frac {z-z_{0}}{w-z_{0}}}}}\\&=\displaystyle {\frac {1}{w-z_{0}}}\sum _{n=0}^{\infty }{\frac {(z-z_{0})^{n}}{(w-z_{0})^{n}}}.\end{array}}$

Proof 2

the real converges absolutely $|z-z_{0}|<r=|w-z_{0}|$ and we obtain

${\begin{array}{rl}f(z)&=\displaystyle {\frac {1}{2\pi i}}\int _{\partial B_{r}(z_{0})}{\frac {f(w)}{w-z}}\,dw\\&=\displaystyle {\frac {1}{2\pi i}}\int _{\partial B_{r}(z_{0})}{\frac {1}{w-z_{0}}}{\frac {f(w)(z-z_{0})^{n}}{(w-z_{0})^{n}}}\,dw\\&=\displaystyle {\frac {1}{2\pi i}}\sum _{n=0}^{\infty }\int _{\partial B_{r}(z_{0})}{\frac {f(w)}{(w-z_{0})^{n+1}}}\,dw\cdot (z-z_{0})^{n}\\&=\displaystyle {\frac {1}{2\pi i}}\sum _{n=0}^{\infty }\int _{\partial B_{r}(z_{0})}{\frac {f(w)}{(w-z_{0})^{n+1}}}\,dw\cdot (z-z_{0})^{n}\end{array}}$

See also

• Category:Complex analysis

• Maximum principle

• Cycle

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