Cauchy Integral Formula

Let $f$ be analytic inside and on a simple closed positively oriented contour $\gamma$, and let $z_0$ be any point in the interior of $\gamma$. Then:

$f(z_0)=\frac{1}{2\pi i}{\oint }_{\gamma }\frac{f(z)}{z-z_0}dz$

The formula recovers the value of $f$ at any interior point entirely from its values on the boundary, a consequence of Cauchy’s Theorem and the structure of analytic functions.

Generalized Form for Derivatives

The formula extends to all higher-order derivatives. For $n\ge 0$:

$f^{(n)}(z_0)=\frac{n!}{2\pi i}{\oint }_{\gamma }\frac{f(z)}{(z-z_0{)}^{n+1}}dz$

The $n=0$ case recovers the basic formula. This is proved by differentiating under the integral sign, which is justified by uniform convergence on $\gamma$.

Infinite Differentiability

If $f$ is analytic on an open set $U$, then $f$ is infinitely differentiable on $U$, and every derivative $f^{(n)}$ is itself analytic on $U$. This stands in sharp contrast to real analysis, where a function can be differentiable exactly $k$ times and no more. Analyticity in the complex sense is a far stronger condition.

Connection to Residues

The integrand $f(z)/(z-z_0)$ has a simple pole at $z_0$ with Residue:

${\mathrm{Res}}_{z=z_0}\frac{f(z)}{z-z_0}=f(z_0)$

The Cauchy Integral Formula is therefore a direct application of the residue theorem: the contour integral equals $2\pi i$ times the sum of residues of enclosed poles. With $f$ analytic inside $\gamma$, the only singularity of the integrand is the simple pole at $z_0$, giving exactly $2\pi i\cdot f(z_0)$.

More generally, the generalized formula extracts the coefficient of $(z-z_0{)}^n$ in the Laurent expansion of $f(z)/(z-z_0{)}^{n+1}$, which is the residue at the order-$(n+1)$ pole.