# Cauchy's Integral Formula/General Result

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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 $U$ such that $D \subseteq U$.

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

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

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

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

### Corollary

Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.

Let the coefficient of $z^n$ extracted from $\map G z$ be denoted:

- $\sqbrk {z^n} \map G z := a_n$

Let $\map G z$ be convergent for $z = z_0$ and $0 < r < \cmod {z_0}$.

Then:

- $\sqbrk {z^n} \map G z = \displaystyle \frac 1 {2 \pi i} \oint_{\cmod z \mathop = r} \dfrac {\map G z \d z} {z^{n + 1} }$

## Proof

Proof by induction:

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

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

### Basis for the Induction

$\map P 0$ holds, as this is:

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

which is Cauchy's Integral Formula.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:

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

Then we need to show:

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

### Induction Step

This is our induction step:

\(\ds \frac {\rd} {\rd a} \map {f^{\paren k} } a\) | \(=\) | \(\ds \frac {k!} {2 \pi i} \oint_{\partial D} \frac {\rd} {\rd a} \frac {\map f z} {\paren {z - a}^{k + 1} } \rd z\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \frac {k!} {2 \pi i}\oint_{\partial D} \frac {\paren {k + 1} \map f z} {\paren {z - a}^{k + 2} } \rd z\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \frac {\paren {k + 1}!} {2 \pi i} \oint_{\partial D} \frac {\map f z} {\paren {z - a}^{k + 2} } \rd z\) |

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To justify the passage of the derivative under the integral sign in line 1, note that for $\alpha$ near $a$, we have the following bound for all $z \in \partial D$:

- $\ds \left|\frac{(k + 1)f(z)}{(z - \alpha)^{k + 2}}\right| = \frac{(k + 1)|f(z)|}{r^{k + 2}}.$

Since $f$ is continuous on the compact set $\partial D$, it follows that $\sup_{z \in \partial D}|f(z)| < \infty$. Thus the arc length integral

- $\ds \oint_{\partial D}\frac{(k + 1)|f(z)|}{r^{k + 2}}ds(z) \leq (k + 1)\ell(\partial D)\sup_{z \in \partial D}\frac{|f(z)|}{r^{k + 2}} < \infty.$

Hence an argument involving the fundamental theorem of calculus followed by the dominated convergence theorem justifies the swap. So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:

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

$\blacksquare$

## Also known as

This result can also be referred to as Cauchy's Integral Formula for Derivatives.

## Source of Name

This entry was named for Augustin Louis Cauchy.