Cauchy's Integral Formula

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Theorem

Let $D = \set {z \in \C: \cmod z \le r}$ be the closed disk of radius $r$ in $\C$.

Let $f: U \to \C$ be holomorphic on some open set containing $D$.


Then for each $a$ in the interior of $D$:

$\ds \map f a = \frac 1 {2 \pi i} \oint_{\partial D} \frac {\map f z} {z - a} \rd z$

where $\partial D$ is the boundary of $D$, and is traversed anticlockwise.


General Result

Let $n \in \N$ be a natural number.


Then for each $a$ in the interior of $D$, the $n$-th derivative of $f$ at $a$ exists and can be written as:

$\ds \map {f^{\paren n} } a = \frac {n!} {2 \pi i} \oint_{\partial D} \frac {\map f z} {\paren {z - a}^{n + 1} } \rd z$

where $\partial D$ is the boundary of $D$, and is traversed anticlockwise.


Proof

Let $C$ be any arbitrary closed curve which defines a region $R$ where the function $\map f z$ is holomorphic on $R$.

Let $z_0$ be any point in the region $R$ such that:

$\dfrac {\map f z} {z - z_0}$ is holomorphic on $R \setminus \set {z_0}$

We draw a circle $C_r$ with center at $z_0$ and radius $r$ such that $r \to 0$.

This makes $C$ and $C_r$ a multiply connected region for a sufficiently small $r > 0$.


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According to Cauchy's Integral Theorem for a multiply connected region:


\(\ds I\) \(:=\) \(\ds \oint_C \frac {\map f z} {z - z_0} \rd z\)
\(\ds \) \(=\) \(\ds \oint_{C_r} \frac {\map f z} {z - z_0} \rd z\)
\(\ds \) \(=\) \(\ds \oint_{C_r} \frac {\map f {z_0} + \paren {\map f z - \map f {z_0} } } {z - z_0} \rd z\)
\(\ds \) \(=\) \(\ds \map f {z_0} \oint_{C_r} \frac {\rd z} {z - z_0} + \oint_{C_r} \frac {\map f z - \map f {z_0} } {z - z_0} \rd z\)


Let:

\(\ds z - z_0\) \(=\) \(\ds r e^{i \theta}\)
\(\ds \leadsto \ \ \) \(\ds \d z\) \(=\) \(\ds i r e^{i \theta} \rd \theta\)
\(\ds \leadsto \ \ \) \(\ds \oint_{C_r} \frac {\rd z} {z - z_0}\) \(=\) \(\ds \int_0^{2 \pi} \frac {i r e^{i \theta} } {r e^{i \theta} } \rd \theta\)
\(\ds \) \(=\) \(\ds i \int_0^{2 \pi} \rd \theta\)
\(\ds \) \(=\) \(\ds 2 \pi i\)


Now:

$\ds I = 2 \pi i \map f {z_0} + \oint_{C_r} \frac {\map f z - \map f {z_0} } {z - z_0} \rd z$

From Complex-Differentiable Function is Continuous, $f$ is continous.

According to Epsilon-Delta definition, for every $\epsilon \in \R_{>0}$ there exists a $\delta \in \openint 0 r$ such that:

$\forall z \in \C: \cmod {z - z_0} < \delta \implies \cmod {\map f z - \map f {z_0} } < \epsilon$

Hence:

\(\ds \cmod {\oint_{C_r} \frac {\map f z - \map f {z_0} } {z - z_0} \rd z}\) \(=\) \(\ds \cmod {\oint_{C_\delta} \frac {\map f z - \map f {z_0} } {z - z_0} \rd z}\)
\(\ds \) \(\le\) \(\ds \oint_{C_\delta} \frac {\cmod {\map f z - \map f {z_0} } } {\cmod {z - z_0} } \cmod {\d z}\)
\(\ds \) \(\le\) \(\ds \frac \epsilon \delta \oint_{C_\delta} \cmod {\d z}\)
\(\ds \) \(=\) \(\ds 2 \pi \epsilon\)


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As $\epsilon \to 0$:

$\ds \oint_{C_r} \frac {\map f z - \map f {z_0} } {z - z_0} \rd z = 0$


So:

\(\ds I\) \(=\) \(\ds \map f {z_0} \oint_{C_r} \frac {\rd z} {z - z_0} + \oint_{C_r} \frac {\map f z - \map f {z_0} } {z - z_0} \rd z\)
\(\ds \) \(=\) \(\ds 2 \pi i \map f {z_0} + 0\)
\(\ds \leadsto \ \ \) \(\ds \oint_C \frac {\map f z} {z - z_0} \rd z\) \(=\) \(\ds 2 \pi i \, \map f {z_0}\)

$\blacksquare$


Also see


Source of Name

This entry was named for Augustin Louis Cauchy.


Historical Note

Cauchy's Integral Formula was developed by Augustin Louis Cauchy during his work to establish the groundwork of the discipline of complex analysis.

Karl Weierstrass independently discovered it during his own exercise to rebuild the theory from first principles.


Sources

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