Cauchy integral theorem 

The theorem is usually formulated for closed paths as follows: let U be an open subset of C which is simply connected, let f : U -> C be a holomorphic function, and let γ be a rectifiable path[?] in U whose start point is equal to its end point. Then,

<math>\int_\gamma f(z) dz = 0 </math>

As was shown by Goursat, Cauchy's integral theorem can be proven assuming only that the complex derivative f '(z) exists everywhere in U. This is significant, because one can then prove Cauchy's integral formula for these functions, and from that one can deduce that these functions are in fact infinitely often continuously differentiable.

The condition that U be simply connected means that U have no "holes"; for instance, every open disk U = { z : |z - z₀| < r } qualifies. The condition is crucial; for example, if

<math>\gamma(t) = \exp( 2\pi it)</math>

where exp() is the exponential function traces out the unit circle, then the path integral

<math>\int_\gamma \frac{1}{z} dz = 2\pi i </math>

is non-zero; the Cauchy integral theorem doesn't apply here since f(z) = 1/z isn't defined (and certainly not holomorphic) at z = 0.

One important consequence of the theorem is that path integrals of holomorphic functions on simply connected domains can be computed in a manner familiar from the fundamental theorem of real calculus: let U be a simply connected open subset of C, let f : U -> C be a holomorphic function, and let γ be a piecewise continuously differentiable path[?] in U with start point a and end point b. If F is a complex antiderivative of f, then

∫_γ f(z) dz = F(b) - F(a).

The Cauchy integral theorem is valid in a slightly stronger form than given above. Suppose U is an open simply connected subset of C whose boundary is the image of the rectifiable path γ. If f is a function which is holomorphic on U and continuous on the closure of U, then

∫_γ f(z) dz = 0.

The Cauchy integral theorem is considerably generalized by the residue theorem.

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