Complex Analysis/Cauchy's Integral Theorem for Disks

The Cauchy Integral Formula (named after Augustin-Louis Cauchy) is one of the fundamental results of Complex Analysis, a branch of mathematics. In its weakest form, it states that the values of a holomorphic function $f$ inside a disk are completely determined by its values on the boundary of that disk. A powerful generalization of this is the Residue theorem.

Cauchy Integral Formula for Disks

Statement

Let $G\subseteq ℂ$ be open, $f:G\to ℂ$ holomorphic, $z_0\in G$ a point in $G$, and $U:=D_r(z_0)\subset G$ a bounded disk in $G$. Then for all $z\in D_r(z_0)$ (i.e., for all $z$ with $|z-z_0|<r$), the following holds: :$f(z)=\frac{1}{2\pi \mathrm{i}}{\oint }_{\partial U}\frac{f(\zeta )}{\zeta -z}\mathrm{d}\zeta$ Here, $\partial U$ denotes the positively oriented curve $t\mapsto z_0+r{e}^{\mathrm{i}t}$ for $t\in [0,2\pi ]$ along the boundary of the disk $U$.

Proof 1

For a fixed $z\in U$, the function $g:U\to ℂ$ defined by $w\mapsto \frac{f(w)-f(z)}{w-z}$ for $w\ne z$ und $w\mapsto f^{\prime }(z)$ for $w=z$. $g$ is steadily on $U$ and holomorphic on $U\setminus \{z\}$. By the Cauchy Integral Theorem, we now have: :$0={\oint }_{\partial U}g(w)dw={\oint }_{\partial U}\frac{f(\zeta )}{\zeta -z}\mathrm{d}\zeta -f(z){\oint }_{\partial U}\frac{\mathrm{d}\zeta }{\zeta -z}$.

Proof 2

The function $h:U\to ℂ$, $w\mapsto {\oint }_{\partial U}\frac{\mathrm{d}\zeta }{\zeta -w}$ is holomorphic with the derivative $h^{\prime }(w)={\oint }_{\partial U}\frac{\mathrm{d}\zeta }{{(\zeta -w)}^2}$, which vanishes since the integrand has an antiderivative (namely $\zeta \mapsto -\frac{1}{\zeta -w}$). Therefore, $h$ is constant, and since $h(a)=2\pi i$, we have $h(z)=2\pi i$.

Consequences of the Cauchy Integral Theorem

The Cauchy Integral Theorem (CIS) leads to the following corollaries:

Representation of the Function at the Center of the Disk

For every holomorphic function, the function value at the center of a circle is the average of the function values on the circle's boundary. Use $\zeta (t)=z_o+r{e}^{\mathrm{i}t},\mathrm{d}\zeta =\mathrm{i}r{e}^{\mathrm{i}t}\mathrm{d}t$. Test: :$\begin{array}{ccc}f|U(z_o)& =\frac{1}{2\pi \mathrm{i}}\oint \partial U\frac{f(\zeta )}{\zeta -z_o}\mathrm{d}\zeta =\frac{1}{2\pi \mathrm{i}}\underset{0}{\overset{2\pi }{\int }}\frac{f(z_o+r{e}^{\mathrm{i}t})}{r{e}^{\mathrm{i}t}}\mathrm{i}r{e}^{\mathrm{i}t},\mathrm{d}t& =\frac{1}{2\pi }\underset{0}{\overset{2\pi }{\int }}f(z_o+r{e}^{\mathrm{i}t}),\mathrm{d}t\end{array}$

Derivatives - Cauchy Integral Formula - CIF

Every holomorphic function is infinitely complex differentiable, and each of these derivatives is also holomorphic. Expressed using the integral formula, this means for $|z-z_o|<r$ and $n\in \mathbb{N}0$: :$f^{(n)}(z)=\frac{n!}{2\pi \mathrm{i}}\oint \partial U\frac{f(\zeta )}{(\zeta -z{)}^{n+1}}\mathrm{d}\zeta .$

Local Developability in Power Series

Every holomorphic function can be locally expanded into a power series for $|z-z_o|<r$.

$f(z)=\sum _{n=0}^{\mathrm{\infty }}\left(\frac{1}{2\pi \mathrm{i}}{\oint }_{\partial U}\frac{f(\zeta )}{(\zeta -z_o{)}^{n+1}}\mathrm{d}\zeta \right)(z-z_o{)}^n=\sum _{n=0}^{\mathrm{\infty }}{a}_n(z-z_o{)}^n.$

Using the integral formula for $f^{(n)}$, it immediately follows that the coefficients $a_n$ are exactly the Taylor coefficients.

Estimation of the Taylor Series Coefficients

For the coefficients, the following estimate holds when $|f(z)|\le M$ for $|z-a|<r\iff z\in U_r(a)$: :$|a_n|\le \frac{M}{r^n}$ The Liouville Theorem (every holomorphic function bounded on the entire complex plane is constant) can be easily proven using the integral formula. This can then be used to easily prove the Fundamental Theorem of Algebra (every polynomial in $ℂ$ factors into linear factors). Here's the translation with the specified conditions:

Proof 1

The Cauchy integral formula is differentiated partially, allowing differentiation and integration to be swapped:

$\begin{array}{cc}{f}^{(n)}{|}_U(z)& =\frac{{\partial }^{n}f}{\partial z^n}{|}_U(z)=\frac{1}{2\pi \mathrm{i}}\frac{{\partial }^n}{\partial z^n}{\oint }_{\partial U}\frac{f(\zeta )}{\zeta -z}\mathrm{d}\zeta \\ & =\frac{1}{2\pi \mathrm{i}}{\oint }_{\partial U}f(\zeta )\underset{n!/(\zeta -z{)}^{1+n}}{\underbrace{\frac{{\partial }^n}{\partial z^n}\frac{1}{\zeta -z}}}\mathrm{d}\zeta =\frac{n!}{2\pi \mathrm{i}}{\oint }_{\partial U}\frac{f(\zeta )}{(\zeta -z{)}^{1+n}}\mathrm{d}\zeta \end{array}$

Proof 2a: Cauchy Kernel

Developing $\frac{1}{\zeta -z}$ in the Cauchy integral formula using the geometric series gives (Cauchy kernel):

$\frac{1}{1-\frac{z-z_o}{\zeta -z_o}}=\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{z-z_o}{\zeta -z_o}\right)}^n$

Proof 2: Cauchy Kernel - Taylor Series

$\begin{array}{cc}f{|}_U(z)& =\frac{1}{2\pi \mathrm{i}}{\oint }_{\partial D_r(z_o)}\frac{f(\zeta )}{\zeta -z}\mathrm{d}\zeta =\frac{1}{2\pi \mathrm{i}}{\oint }_{\partial D_r(z_o)}\frac{f(\zeta )}{\zeta -z_o-(z-z_o)}\mathrm{d}\zeta \\ & =\frac{1}{2\pi \mathrm{i}}{\oint }_{\partial D_r(z_o)}\frac{f(\zeta )}{\zeta -z_o}\cdot \frac{1}{1-\frac{z-z_o}{\zeta -z_o}}\mathrm{d}\zeta \\ & \stackrel{|\frac{z-z_o}{\zeta -z_o}|<1}{=}\frac{1}{2\pi \mathrm{i}}{\oint }_{\partial D_r(z_o)}\frac{f(\zeta )}{\zeta -z_o}\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{z-z_o}{\zeta -z_o}\right)}^n\mathrm{d}\zeta \\ & =\sum _{n=0}^{\mathrm{\infty }}\underset{a_n}{\underbrace{\left(\frac{1}{2\pi \mathrm{i}}{\oint }_{\partial D_r(z_o)}\frac{f(\zeta )}{(\zeta -z_o{)}^{n+1}}\mathrm{d}\zeta \right)}}(z-z_o{)}^n\end{array}$

Proof 2b: Cauchy Kernel

Since the geometric series converges uniformly for $|z-z_o|<|\zeta -z_o|=r$, one can integrate term by term, i.e., swap the sum and the integral. The development coefficients are:

$\begin{array}{cc}{a}_n& =\frac{1}{n!}{f}^{(n)}{|}_U(z_o)=\frac{1}{2\pi \mathrm{i}}{\oint }_{\partial U_r(a)}\frac{f(\zeta )}{(\zeta -z_o{)}^{n+1}}\mathrm{d}\zeta \\ & =\frac{1}{2\pi \mathrm{i}}\underset{0}{\overset{2\pi }{\int }}\frac{f(z_o+r{e}^{\mathrm{i}t})}{(r{e}^{\mathrm{i}t}{)}^{n+1}}\mathrm{i}r{e}^{\mathrm{i}t}\mathrm{d}t=\frac{1}{2\pi r^n}\underset{0}{\overset{2\pi }{\int }}f(z_o+r{e}^{\mathrm{i}t})e^{-\mathrm{i}nt}\mathrm{d}t\end{array}$

Proof 3: Estimation of the Coefficients

For the coefficients $a_n\in ℂ$, the following estimate holds. There exists a $M>0$ such that $|f(z)|\le M$ for $|z-z_o|=r$. Then, for $n\in \mathbb{N}0$, we have:

$\begin{array}{cc}|a_n|& =|\frac{1}{2\pi r^n}\underset{0}{\overset{2\pi }{\int }}f(z_o+r{e}^{\mathrm{i}t})e^{-\mathrm{i}nt}\mathrm{d}t|\\ & \le \frac{1}{2\pi r^n}\underset{0}{\overset{2\pi }{\int }}\underset{\le M}{\underbrace{|f(z_o+r{e}^{\mathrm{i}t})|}}\mathrm{d}t\le \frac{M}{r^n}\end{array}$

Proof 4: Liouville's Theorem

If $f$ is holomorphic on all of $ℂ$ and bounded, i.e., $|f(z)|=|\sum _{n=0}^{\mathrm{\infty }}{a}_n{z}^n|\le M$ for all $z\in ℂ$, then, as before, for all $r>0$, we have:

$|a_n|\le \frac{M}{r^n}$

Since $r$ was arbitrary, it follows that $a_n=0$ for all $n\in \mathbb{N}$. Therefore, from the boundedness of $f$, we conclude:

$f(z)=a_0$

Thus, every bounded holomorphic function on all of $ℂ$ is constant (Liouville's theorem).

Example

Using the integral formula, integrals can also be computed:

${\oint }_{\partial U_2(0)}\frac{e^{2\zeta }}{(\zeta +1{)}^4}\mathrm{d}\zeta =\frac{2\pi \mathrm{i}}{3!}\frac{{\mathrm{d}}^3}{\mathrm{d}{z}^3}{e}^{2z}{|}_{z=-1}=\frac{8\pi \mathrm{i}}{3e^2}$

Cauchy Integral Formula for Cycles

A generalization of the integral formula for circular contours is the version for cycles:

Let $G\subseteq ℂ$ be a domain, $f:G\to ℂ$ holomorphic, and $\mathrm{\Gamma }$ a zero homologous cycle in $D$. Then, for all $z\in D$ not on $\mathrm{\Gamma }$, the following integral formula holds:

$n(\mathrm{\Gamma },z)\cdot f(z)=\frac{1}{2\pi \mathrm{i}}\underset{\mathrm{\Gamma }}{\int }\frac{f(\zeta )}{\zeta -z}\mathrm{d}\zeta$

Here, $n(\mathrm{\Gamma },z)$ denotes the winding number or revolution of $\mathrm{\Gamma }$ around $z$.

Cauchy Integral Formula for Polycycles

The Cauchy integral formula has been generalized to the multidimensional complex space ${ℂ}^n$. Let $U_1,\dots ,U_n$ be disk domains in $ℂ$, then $U:=\prod _{i=1}^n{U}_i$ is a Polycylinder in ${ℂ}^n$. Let $f:U\to ℂ$ be a holomorphic function and $\xi \in U$. The Cauchy integral formula is given by

$f(z_1,\dots ,z_n)=\frac{1}{(2\pi \mathrm{i}{)}^n}{\oint }_{\partial U_n}\cdots {\oint }_{\partial U_1}\frac{f({\xi }_1,\dots ,{\xi }_n)}{({\xi }_1-z_1)\cdots ({\xi }_n-z_n)}\mathrm{d}{\xi }_1\cdots \mathrm{d}{\xi }_n$

Restrictions in Multidimensional Space

Since the Cauchy integral theorem does not hold in higher-dimensional space, this formula cannot be derived analogously to the one-dimensional case. Therefore, this integral formula is derived using induction from the Cauchy integral formula for disk domains. Using the multi-index notation, the formula can be simplified to:

$f(z)=\frac{1}{(2\pi \mathrm{i}{)}^n}{\oint }_{\partial U}\frac{f(\xi )}{(\xi -z)},\mathrm{d}\xi$

with $\partial U=\partial U_1\times \cdots \times \partial U_n$.

Polycycles

Polycycles are defined using a vector of radii, where $M:=\underset{\xi \in U}{\max }|f(\xi )|$ and $r=(r_1,\dots ,r_n)$ is the radius of the polycycle $U:=\prod _{i=1}^n{U}_i$.<ref> for the derivatives of the holomorphic Function $f$ as well as Cauchy's inequality

$|D^{k}f(z)|\le \frac{M\cdot k!}{r^k},$

See also

• Cycle

• Cauchy Integral Theorem

• null-homologous

Literature

• Kurt Endl, Wolfgang Luh: Analysis. Volume 3: Function Theory, Differential Equations. 6th revised edition. Aula-Verlag, Wiesbaden 1987, ISBN 3-89104-456-9, p. 153, Theorem 4.9.1.
• Wolfgang Fischer, Ingo Lieb: Function Theory. 7th improved edition. Vieweg, Braunschweig, 1994, ISBN 3-528-67247-1, p. 60, Chapter 3, Theorem 2.2 (Vieweg-Studium. Advanced Mathematics Course 47).

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