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Theorem

Let U be an open subset of the complex plane C, and suppose the closed disk D defined as D = { z: | z − z 0 | ≤ r } {\displaystyle D={\bigl \{}z:|z-z_{0}|\leq r{\bigr \}}} is completely contained in U. Let f : U → C be a holomorphic function, and let γ be the circle, oriented counterclockwise, forming the boundary of D. Then for every a in the interior of D, f (a) = 1 2 π i ∮ γ f (z) z − a d z. {\displaystyle f(a)={\frac {1}{2\pi i}}\oint _{\gamma }{\frac {f(z)}{z-a}}\,dz.\,}

The proof of this statement uses the Cauchy integral theorem and like that theorem, it only requires f to be complex differentiable. Since 1 / (z − a) {\displaystyle 1/(z-a)} can be expanded as a power series in the variable a {\displaystyle a} 1 z − a = 1 + a z + (a z) 2 + ⋯ z {\displaystyle {\frac {1}{z-a}}={\frac {1+{\frac {a}{z}}+\left({\frac {a}{z}}\right)^{2}+\cdots }{z}}} it follows that holomorphic functions are analytic, i.e. they can be expanded as convergent power series.In particular f is actually infinitely differentiable, with f (n) (a) = n ! 2 π i ∮ γ f (z) (z − a) n + 1 d z. {\displaystyle f^{(n)}(a)={\frac {n!}{2\pi i}}\oint _{\gamma }{\frac {f(z)}{\left(z-a\right)^{n+1}}}\,dz.}

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

The theorem stated above can be generalized. The circle γ can be replaced by any closed rectifiable curve in U which has winding number one about a. Moreover, as for the Cauchy integral theorem, it is sufficient to require that f be holomorphic in the open region enclosed by the path and continuous on its 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 f (z) = 1 / z, defined for | 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 inversion formula to construct the holomorphic function from the real part on the boundary. For example, the function f (z) = i − iz has real part Re f (z) = Im z. On the unit circle this can be written i / z − iz / 2. Using the Möbius transformation and the Stieltjes formula we construct the function inside the circle. The i / z term makes no contribution, and we find the function − iz. This has the correct real part on the boundary, and also gives us the corresponding imaginary part, but off by a constant, namely i.