Editing Cauchy's integral theorem (section)

=== Fundamental theorem for complex line integrals ===
If {{math|''f''(''z'')}} is a holomorphic function on an open [[region (mathematical analysis)|region]] {{mvar|U}}, and <math>\gamma</math> is a curve in {{mvar|U}} from <math>z_0</math> to <math>z_1</math> then,
<math display="block">\int_{\gamma}f'(z) \, dz = f(z_1)-f(z_0).</math>

Also, when {{math|''f''(''z'')}} has a single-valued antiderivative in an open region {{mvar|U}}, then the path integral <math display="inline">\int_{\gamma}f(z) \, dz</math> is path independent for all paths in {{mvar|U}}.

==== Formulation on simply connected regions ====

Let <math>U \subseteq \Complex</math> be a [[Simply connected space|simply connected]] [[open subset|open]] set, and let <math>f: U \to \Complex</math> be a [[holomorphic function]]. Let <math>\gamma: [a,b] \to U</math> be a smooth closed curve. Then:
<math display="block">\int_\gamma f(z)\,dz = 0. </math>
(The condition that <math>U</math> be [[simply connected]] means that <math>U</math> has no "holes", or in other words, that the [[fundamental group]] of <math>U</math> is trivial.)

==== General formulation ====

Let <math>U \subseteq \Complex</math> be an [[open subset|open set]], and let <math>f: U \to \Complex</math> be a [[holomorphic function]]. Let <math>\gamma: [a,b] \to U</math> be a smooth closed curve. If  <math>\gamma</math> is [[Homotopy|homotopic]] to a constant curve, then:
<math display="block">\int_\gamma f(z)\,dz = 0. </math>where z є ''U''

(Recall that a curve is [[Homotopy|homotopic]] to a constant curve if there exists a smooth [[homotopy]] (within <math>U</math>) 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 space|simply connected]] set, every closed curve is [[Homotopy|homotopic]] to a constant curve.

==== Main example ====

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