# Cauchy's integral theorem {displaystyle int _{C}f(z),dz=0.} Contents 1 Statement 1.1 Fundamental theorem for complex line integrals 1.1.1 Formulation on Simply Connected Regions 1.1.2 General Formulation 1.1.3 Main Example 2 Discussion 3 Proof 4 See also 5 References 6 External links Statement Fundamental theorem for complex line integrals If f(z) is a holomorphic function on an open region U, and {displaystyle gamma } is a curve in U from {displaystyle z_{0}} to {displaystyle z_{1}} then, {displaystyle int _{gamma }f'(z),dz=f(z_{1})-f(z_{0}).} Also, when f(z) has a single-valued antiderivative in an open region U, then the path integral {textstyle int _{gamma }f'(z),dz} is path independent for all paths in U.

Formulation on Simply Connected Regions Let {displaystyle Usubseteq mathbb {C} } be a simply connected open set, and let {displaystyle f:Uto mathbb {C} } be a holomorphic function. Let {displaystyle gamma :[a,b]to U} be a smooth closed curve. Then: {displaystyle int _{gamma }f(z),dz=0.} (The condition that {displaystyle U} be simply connected means that {displaystyle U} has no "holes", or in other words, that the fundamental group of {displaystyle U} is trivial.) General Formulation Let {displaystyle Usubseteq mathbb {C} } be an open set, and let {displaystyle f:Uto mathbb {C} } be a holomorphic function. Let {displaystyle gamma :[a,b]to U} be a smooth closed curve. If {displaystyle gamma } is homotopic to a constant curve, then: {displaystyle int _{gamma }f(z),dz=0.} (Recall that a curve is homotopic to a constant curve if there exists a smooth homotopy from the curve to the constant curve. Intuitively, this means that one can shrink the curve into a point without exiting the space.) The first version is a special case of this because on a simply connected set, every closed curve is homotopic to a constant curve.

Main Example In both cases, it is important to remember that the curve {displaystyle gamma } does not surround any "holes" in the domain, or else the theorem does not apply. A famous example is the following curve: {displaystyle gamma (t)=e^{it}quad tin left[0,2pi right],} which traces out the unit circle. Here the following integral: {displaystyle int _{gamma }{frac {1}{z}},dz=2pi ineq 0,} is nonzero. The Cauchy integral theorem does not apply here since {displaystyle f(z)=1/z} is not defined at {displaystyle z=0} . Intuitively, {displaystyle gamma } surrounds a "hole" in the domain of {displaystyle f} , so {displaystyle gamma } cannot be shrunk to a point without exiting the space. Thus, the theorem does not apply.

Discussion As Édouard Goursat showed, Cauchy's integral theorem can be proven assuming only that the complex derivative {displaystyle f'(z)} exists everywhere in {displaystyle U} . This is significant because one can then prove Cauchy's integral formula for these functions, and from that deduce these functions are infinitely differentiable.

The condition that {displaystyle U} be simply connected means that {displaystyle U} has no "holes" or, in homotopy terms, that the fundamental group of {displaystyle U} is trivial; for instance, every open disk {displaystyle U_{z_{0}}={z:left|z-z_{0}right|

Si quieres conocer otros artículos parecidos a Cauchy's integral theorem puedes visitar la categoría Augustin-Louis Cauchy.

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