Cauchy's integral formula

Statement

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

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

for each ${\displaystyle z\in D}$.

Proof 1

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

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

Proof 2

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

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

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

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

Proof 3

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

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

Proof 4

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

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

This proves the statement.

Statement

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

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

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

Proof 1

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

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

defined.

Proof 2: continuous

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

. a) In the case ${\displaystyle w=z}$:
:

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

Proof 3

b) In the case ${\displaystyle w\neq z}$:

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

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

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

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

Proof 4

This implies, in case a:

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

and in case b:

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

We now define

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

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

Proof 5

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

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

prove it is

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

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

Proof 6

By Goursat's theorem, it follows that

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

this of course also mean that

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

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

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

Proof 7

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

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

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

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

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

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

i.e. ${\displaystyle h}$ is an entire function.

Proof 8

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

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

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

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

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

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

For Circular Disks

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

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

for each ${\displaystyle z\in D}$.

For Cycles

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

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

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

Statement

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

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

where the coefficients are given by

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

Proof 1

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

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

Proof 2

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

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

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