Cauchy integral formula

If \(f\) is holomorphic on a simple closed contour \(C\) and inside it, traversed counterclockwise, and \(z_0\) lies inside \(C\), then \(\oint_C\frac{f(z)}{z-z_0}dz=2\pi i\,f(z_0)\). The proof uses Cauchy-Goursat: deform \(C\) to a small circle around \(z_0\), where \(f\) is nearly constant and the integral of \(\frac{1}{z-z_0}\) equals \(2\pi i\).

It is a computational rule: given an integral with a factor \(\frac{1}{z-z_0}\), you identify \(f\) as the rest of the integrand and evaluate \(f(z_0)\). Thus \(\oint_{|z|=2}\frac{e^z}{z-1}dz=2\pi i\,e\). If the integrand has several trouble points, as in \(\frac{1}{z^2-1}\), split it with partial fractions and apply the formula to each term. If \(z_0\) lies outside \(C\), the integral is zero.

The result is striking: the values of \(f\) on the boundary determine those inside. On a circle centered at \(z_0\) it gives the mean value property, \(f(z_0)=\frac1{2\pi}\int_0^{2\pi}f(z_0+re^{it})dt\). Differentiating with respect to \(z_0\) under the integral also shows that \(f\) has derivatives of all orders, \(f^{(n)}(z_0)=\frac{n!}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^{n+1}}dz\); hence every holomorphic function is analytic.

Typical mistakes: applying it when \(f\) has a singularity inside \(C\) (then \(f\) is not holomorphic on the region), forgetting the factor \(2\pi i\), and neglecting orientation, which flips the sign. It is the key tool behind Liouville's theorem and the fundamental theorem of algebra.