Integral representation
Theorem. If \(f\) is holomorphic on a neighborhood of \(\overline{D(a,r)}\), then for \(|z-a|<r\), \[ f(z)=\frac{1}{2\pi i}\int_{|\zeta-a|=r}\frac{f(\zeta)}{\zeta-z}\,\mathrm{d}\zeta. \] Values inside the circle are determined by boundary values.
Derivative formula
Corollary. Under the same hypotheses, \[ f^{(n)}(z)=\frac{n!}{2\pi i}\int_{|\zeta-a|=r}\frac{f(\zeta)}{(\zeta-z)^{n+1}}\,\mathrm{d}\zeta. \] Thus holomorphic functions are automatically infinitely complex differentiable.
Proof mechanism
Proof. Subtract \(f(z)\) inside the numerator: \((f(\zeta)-f(z))/(\zeta-z)\) is holomorphic near \(z\), so its integral vanishes by Cauchy's theorem. The remaining integral is \(f(z)\int(\zeta-z)^{-1}\,\mathrm{d}\zeta=2\pi i f(z)\).
Mean value property
Corollary. For a circle centered at \(a\), Cauchy's formula implies \(f(a)=(2\pi)^{-1}\int_0^{2\pi}f(a+re^{it})\,\mathrm{d}t\). Holomorphic functions equal their circular averages.
Analyticity follows
Theorem. Expanding \((\zeta-z)^{-1}\) as a geometric series inside the contour gives a convergent power series for \(f\). Hence every holomorphic function is analytic.
Boundary dependence
Remark. The formula is local: the contour must lie in the domain. Singularities between the point and the contour change the representation by adding residue contributions.
Interior values from boundary data
Theorem (Cauchy integral formula). Let \(\Gamma\) be a positively oriented cycle null-homologous in \(\Omega\), and let \(f\) be holomorphic on \(\Omega\). For \(z\in\Omega\setminus\Gamma\), \[ \frac1{2\pi i}\int_\Gamma\frac{f(\zeta)}{\zeta-z}\,\mathrm{d}\zeta =\operatorname{Ind}(\Gamma,z)f(z). \]
Proof. Define \[ g(\zeta)= \begin{cases} \dfrac{f(\zeta)-f(z)}{\zeta-z},&\zeta\ne z,\\ f'(z),&\zeta=z. \end{cases} \] The removable value makes \(g\) holomorphic. Cauchy's theorem gives \(\int_\Gamma g\,\mathrm{d}\zeta=0\), hence \[ \int_\Gamma\frac{f(\zeta)}{\zeta-z}\,\mathrm{d}\zeta =f(z)\int_\Gamma\frac{\mathrm{d}\zeta}{\zeta-z} =2\pi i\,\operatorname{Ind}(\Gamma,z)f(z). \]
All derivatives at once
Theorem. If the circle \(|\zeta-z_0|=r\) and its interior lie in \(\Omega\), then for \(|z-z_0|<r\) and \(n\ge0\), \[ f^{(n)}(z) =\frac{n!}{2\pi i} \int_{|\zeta-z_0|=r} \frac{f(\zeta)}{(\zeta-z)^{n+1}}\,\mathrm{d}\zeta. \]
Proof. On a compact subdisc, the denominator stays uniformly away from zero, so differentiation under the contour integral is justified by uniform convergence. Repeatedly differentiating \((\zeta-z)^{-1}\) gives \(n!(\zeta-z)^{-n-1}\).
Taylor expansion and analytic rigidity
Corollary. Every holomorphic function is analytic. For \(|z-z_0|<r\), \[ f(z)=\sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n. \]
Proof. On the contour, \[ \frac1{\zeta-z} =\frac1{\zeta-z_0} \sum_{n\ge0}\frac{(z-z_0)^n}{(\zeta-z_0)^{n+1}}, \] with uniform convergence for \(z\) in a smaller disc. Integrate termwise and use the derivative formula for the coefficients.
Mean value and boundary dependence
Parametrizing a circle gives \[ f(z_0)=\frac1{2\pi}\int_0^{2\pi} f(z_0+re^{it})\,\mathrm{d}t. \] Taking real parts yields the harmonic mean-value property. More broadly, the formula says a holomorphic function inside a contour is determined by its boundary values, a nonlocal rigidity that has no analogue for arbitrary smooth functions.