Cauchy's Integral Formula/General Result

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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 $D$, 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.


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 $0 < r < \cmod {z_0}$.


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


Proof by induction:

For all $n \in \N$, 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:

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

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 $z \in \partial D$:

$\ds \cmod {\frac {\paren {k + 1} \map f z} {\paren {z - \alpha}^{k + 2} } } = \frac {\paren {k + 1} \cmod {\map f z} }{r^{k + 2} }$

Since $f$ is continuous on the compact set $\partial D$, it follows that $\ds \sup_{z \mathop \in \partial D} \cmod {\map f z} < \infty$.

Thus the arc length integral:

$\ds \oint_{\partial D} \frac {\paren {k + 1} \cmod {\map f z} }{r^{k + 2}} ds(z) \le \paren {k + 1} \map \ell {\partial D} \sup_{z \mathop \in \partial D}\frac{\cmod {\map f z} }{r^{k + 2} } < \infty$

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.


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


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.