Cauchy integral formula

The formulas.

Let be an open disk in thecomplex plane, and let $f(z)$ be a holomorphic functiondefined on some open domain that contains $D$ and its boundary. Then,for every $z\in D$ we have

$f(z)$$=$${\displaystyle \frac{1}{2\pi i}}{\displaystyle {\oint }_C}{\displaystyle \frac{f(\zeta )}{\zeta -z}}𝑑\zeta$

$f^{\prime }(z)$$=$${\displaystyle \frac{1}{2\pi i}}{\displaystyle {\oint }_C}{\displaystyle \frac{f(\zeta )}{{(\zeta -z)}^2}}𝑑\zeta$

$\mathrm{⋮}$

$f^{(n)}(z)$$=$${\displaystyle \frac{n!}{2\pi i}}{\displaystyle {\oint }_C}{\displaystyle \frac{f(\zeta )}{{(\zeta -z)}^{n+1}}}𝑑\zeta$

Here $C=\partial D$ is the corresponding circular boundary contour,oriented counterclockwise, with the most obvious parameterizationgiven by

$$\zeta =z_0+R{e}^{it},0\le t\le 2\pi .$$

Discussion.

The first of the above formulas underscores the “rigidity” ofholomorphic functions. Indeed, the values of the holomorphic functioninside a disk $D$ are completely specified by its values on theboundary of the disk. The second formula is useful, because it givesthe derivative in terms of an integral, rather than as the outcome ofa limit process.

Generalization.

The following technical generalization of the formula is needed forthe treatment of removable singularities. Let $S$ be a finite subsetof $D$, and suppose that $f(z)$ is holomorphic for all $z\notin S$, but also that$f(z)$ is bounded near all $z\in S$. Then, theabove formulas are valid for all $z\in D\setminus S$.

Using the Cauchy residue theorem, one can further generalize the integral formula to the situation where $D$ is any domain and $C$ is any closed rectifiable curve in $D$; in this case, the formula becomes

$$\eta (C,z)f(z)=\frac{1}{2\pi i}{\oint }_C\frac{f(\zeta )}{\zeta -z}𝑑\zeta$$

where $\eta (C,z)$ denotes the winding number of $C$. It is valid for all points $z\in D\setminus S$ which are not on the curve $C$.