Cauchy's Integral Formula/General Result

Theorem

Let $D = \left\{ {z \in \C: \left\lvert{z}\right\rvert \le r}\right\}$ 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 $D$:

$\displaystyle f^{\left({n}\right)} \left({a}\right) = \frac {n!} {2 \pi i} \int_{\partial D} \frac {f \left({z}\right)} {\left({z - a}\right)^{n + 1} } \rd z$

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

Corollary

Let $G \left({z}\right)$ be the generating function for the sequence $\left\langle{a_n}\right\rangle$.

Let the coefficient of $z^n$ extracted from $G \left({z}\right)$ be denoted:

$\left[{z^n}\right] G \left({z}\right) := a_n$

Let $G \left({z}\right)$ be convergent for $z = z_0$ and $0 < r < \left\lvert{z_0}\right\rvert$.

Then: $\left[{z^n}\right] G \left({z}\right) = \displaystyle \frac 1 {2 \pi i} \oint_{\left\lvert{z}\right\rvert \mathop = r} \dfrac {G \left({z}\right) \rd z} {z^{n + 1} }$

Proof

Proof by induction:

For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:

$\displaystyle f^{\left({n}\right)} \left({a}\right) = \frac {n!} {2 \pi i} \int_{\partial D} \frac {f \left({z}\right)} {\left({z - a}\right)^{n + 1} } \rd z$

Basis for the Induction

$P \left({0}\right)$ holds, as this is:

$\displaystyle f \left({a}\right) = \frac 1 {2 \pi i} \int_{\partial D} \frac {f \left({z}\right)} {\left({z - a}\right)} \rd z$

which is Cauchy's Integral Formula.

This is our basis for the induction.

Induction Hypothesis

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

So this is our induction hypothesis:

$\displaystyle f^{\left({k}\right)} \left({a}\right) = \frac {k!} {2 \pi i} \int_{\partial D} \frac {f \left({z}\right)} {\left({z - a}\right)^{k + 1} } \rd z$

Then we need to show:

$\displaystyle f^{\left({k + 1}\right)} \left({a}\right) = \frac {\left({k + 1}\right)!} {2 \pi i} \int_{\partial D} \frac {f \left({z}\right)} {\left({z - a}\right)^{k + 2} } \rd z$

Induction Step

This is our induction step:

\(\displaystyle \frac {\rd} {\rd a} f^{\left({k}\right)} \left({a}\right)\)

\(=\)

\(\displaystyle \frac {k!} {2 \pi i} \int_{\partial D} \frac {\rd} {\rd a} \frac {f \left({z}\right)} {\left({z - a}\right)^{k + 1} } \rd z\)

\(\displaystyle \)

\(=\)

\(\displaystyle \frac {k!} {2 \pi i}\int_{\partial D} \frac {\left({k + 1}\right) f \left({z}\right)}{\left({z - a}\right)^{k + 2} } \rd z\)

\(\displaystyle \)

\(=\)

\(\displaystyle \frac {\left({k + 1}\right)!} {2 \pi i} \int_{\partial D} \frac {f \left({z}\right)} {\left({z - a}\right)^{k + 2} } \rd z\)

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\displaystyle \forall n \in \N: f^{\left({n}\right)} \left({a}\right) = \frac {n!} {2 \pi i} \int_{\partial D} \frac {f \left({z}\right)} {\left({z - a}\right)^{n + 1} } \rd z$

$\blacksquare$

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.