# Integralsatz von Cauchy

{Anzeigestil int _{C}f(z),dz=0.} Inhalt 1 Aussage 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 Diskussion 3 Nachweisen 4 Siehe auch 5 Verweise 6 External links Statement Fundamental theorem for complex line integrals If f(z) is a holomorphic function on an open region U, und {Anzeigestil Gamma } is a curve in U from {Anzeigestil z_{0}} zu {Anzeigestil z_{1}} dann, {Anzeigestil int _{Gamma }f'(z),dz=f(z_{1})-f(z_{0}).} Ebenfalls, 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, und lass {Anzeigestil f:Uto mathbb {C} } sei eine holomorphe Funktion. Lassen {Anzeigestil Gamma :[a,b]to U} be a smooth closed curve. Dann: {Anzeigestil int _{Gamma }f(z),dz=0.} (The condition that {Anzeigestil U} be simply connected means that {Anzeigestil U} has no "Löcher", or in other words, that the fundamental group of {Anzeigestil U} is trivial.) General Formulation Let {displaystyle Usubseteq mathbb {C} } be an open set, und lass {Anzeigestil f:Uto mathbb {C} } sei eine holomorphe Funktion. Lassen {Anzeigestil Gamma :[a,b]to U} be a smooth closed curve. Wenn {Anzeigestil Gamma } is homotopic to a constant curve, dann: {Anzeigestil 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. Intuitiv, 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 {Anzeigestil Gamma } does not surround any "Löcher" in the domain, or else the theorem does not apply. A famous example is the following curve: {Anzeigestil Gamma (t)=e^{it}quad tin left[0,2pi right],} which traces out the unit circle. Here the following integral: {Anzeigestil int _{Gamma }{frac {1}{z}},dz=2pi ineq 0,} is nonzero. The Cauchy integral theorem does not apply here since {Anzeigestil f(z)=1/z} is not defined at {displaystyle z=0} . Intuitiv, {Anzeigestil Gamma } surrounds a "hole" in the domain of {Anzeigestil f} , Also {Anzeigestil Gamma } cannot be shrunk to a point without exiting the space. Daher, the theorem does not apply.

Discussion As Édouard Goursat showed, Cauchy's integral theorem can be proven assuming only that the complex derivative {Anzeigestil f'(z)} exists everywhere in {Anzeigestil 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 {Anzeigestil U} be simply connected means that {Anzeigestil U} has no "Löcher" oder, in homotopy terms, that the fundamental group of {Anzeigestil U} ist trivial; zum Beispiel, every open disk {Anzeigestil U_{z_{0}}={z:links|z-z_{0}Rechts|

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