Editing Cauchy's integral formula (section)

==Theorem==
Let {{math|''U''}} be an [[open subset]] of the [[complex plane]] {{math|'''C'''}}, and suppose the closed disk {{math|''D''}} defined as
<math display="block">D = \bigl\{z:|z - z_0| \leq r\bigr\}</math>
is completely contained in {{math|''U''}}.  Let {{math|''f'' : ''U'' → '''C'''}} be a [[holomorphic function]], and let {{math|''γ''}} be the [[circle]], oriented [[Curve orientation|counterclockwise]], forming the [[boundary (topology)|boundary]] of {{math|''D''}}. Then for every {{math|''a''}} in the [[interior (topology)|interior]] of {{math|''D''}},
<math display="block">f(a) = \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{z-a}\,dz.\,</math>

The proof of this statement uses the [[Cauchy integral theorem]] and like that theorem, it only requires {{math|''f''}} to be [[complex differentiable]].  Since <math>1/(z-a)</math> can be expanded as a [[power series]] in the variable <math>a</math>
<math display="block">\frac{1}{z-a} = \frac{1+\frac{a}{z}+\left(\frac{a}{z}\right)^2+\cdots}{z}</math>
it follows that [[holomorphic functions are analytic]], i.e. they can be expanded as convergent power series.
In particular {{math|''f''}} is actually infinitely differentiable, with
<math display="block">f^{(n)}(a) = \frac{n!}{2\pi i} \oint_\gamma \frac{f(z)}{\left(z-a\right)^{n+1}}\,dz.</math>

This formula is sometimes referred to as '''Cauchy's differentiation formula'''.

The theorem stated above can be generalized. The circle {{math|''γ''}} can be replaced by any closed [[rectifiable curve]] in {{math|''U''}} which has [[winding number]] one about {{math|''a''}}. Moreover, as for the Cauchy integral theorem, it is sufficient to require that {{math|''f''}} be holomorphic in the open region enclosed by the path and continuous on its [[closure (topology)|closure]].

Note that not every continuous function on the boundary can be used to produce a function inside the boundary that fits the given boundary function. For instance, if we put the function  {{math|''f''(''z'') {{=}} {{sfrac|1|''z''}}}}, defined for {{math|1={{abs|''z''}} = 1}}, into the Cauchy integral formula, we get zero for all points inside the circle. In fact, giving just the real part on the boundary of a holomorphic function is enough to determine the function [[up to]] an imaginary constant — there is only one imaginary part on the boundary that corresponds to the given real part, up to addition of a constant. We can use a combination of a [[Möbius transformation]] and the [[Stieltjes_transformation|Stieltjes inversion formula]] to construct the holomorphic function from the real part on the boundary. For example, the function {{math|1=''f''(''z'') = ''i'' − ''iz''}} has real part {{math|1=Re ''f''(''z'') = Im ''z''}}. On the unit circle this can be written {{math|{{sfrac|{{sfrac|''i''|''z''}} − ''iz''|2}}}}. Using the Möbius transformation and the Stieltjes formula we construct the function inside the circle. The {{math|{{sfrac|''i''|''z''}}}} term makes no contribution, and we find the function {{math|−''iz''}}. This has the correct real part on the boundary, and also gives us the corresponding imaginary part, but off by a constant, namely {{math|''i''}}.