Editing Morera's theorem (section)

{{Short description|Integral criterion for holomorphy}}
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[[File:Morera's Theorem.png|thumb|right|If the integral along every ''C'' is zero, then ''f'' is [[Holomorphic function|holomorphic]] on ''D''.]]
In [[complex analysis]], a branch of [[mathematics]], '''Morera's theorem''', named after [[Giacinto Morera]], gives a criterion for proving that a [[function (mathematics)|function]] is [[holomorphic function|holomorphic]].

Morera's theorem states that a [[continuous function|continuous]], [[complex number|complex]]-valued function ''f'' defined on an  [[open set]] ''D'' in the [[complex plane]] that satisfies
<math display="block">\oint_\gamma f(z)\,dz = 0</math>
for every closed piecewise ''C''<sup>1</sup> curve <math>\gamma</math> in ''D'' must be holomorphic on ''D''.

The assumption of Morera's theorem is equivalent to ''f'' having an [[antiderivative (complex analysis)|antiderivative]] on&nbsp;''D''.

The converse of the theorem is not true in general. A holomorphic function need not possess an antiderivative on its domain, unless one imposes additional assumptions. The converse does hold e.g. if the domain is [[simply connected]]; this is [[Cauchy's integral theorem]], stating that the [[line integral]] of a holomorphic function along a [[closed curve]] is zero.  

The standard counterexample is the function {{math|1=''f''(''z'') = 1/''z''}}, which is holomorphic on '''C'''&nbsp;−&nbsp;{0}.  On any simply connected neighborhood U in '''C'''&nbsp;−&nbsp;{0}, 1/''z'' has an antiderivative defined by {{math|1=''L''(''z'') = ln(''r'') + ''iθ''}}, where {{math|1=''z'' = ''re''<sup>''iθ''</sup>}}. Because of the ambiguity of ''θ'' up to the addition of any integer multiple of 2{{pi}}, any continuous choice of ''θ'' on ''U'' will suffice to define an antiderivative of 1/''z'' on ''U''.  (It is the fact that ''θ'' cannot be defined continuously on a simple closed curve containing the origin in its interior that is the root of why 1/''z'' has no antiderivative on its entire domain '''C'''&nbsp;−&nbsp;{0}.) And because the derivative of an additive constant is 0, any constant may be added to the antiderivative and the result will still be an antiderivative of 1/''z''.

In a certain sense, the 1/''z'' counterexample is universal: For every analytic function that has no antiderivative on its domain, the reason for this is that 1/''z'' itself does not have an antiderivative on '''C'''&nbsp;−&nbsp;{0}.