Proof
The argument, first given by Cauchy, hinges on Cauchy's integral formula and the power series development of the expression
- .
Suppose ƒ is differentiable everywhere within some open disk centered at a. Let z be within that open disk. Let C be a positively oriented (i.e., counterclockwise) circle centered at a, lying within that open disk but farther from a than z is. Starting with Cauchy's integral formula, we have
To justify the interchange of the sum and the integral, one must notice that in the intersection of |(z − a)/(w − a)| ≤ r < 1 and some closed domain containing C, ƒ(w)/(w − a) is holomorphic and therefore bounded by some positive number M. So we have
The Weierstrass M-test says the series converges uniformly, and thus the interchange of the sum and the integral is justified.
Since the factor (z − a)n does not depend on the variable of integration w, it can be pulled out:
And now the integral and the factor of 1/(2πi) do not depend on z, i.e., as a function of z, that whole expression is a constant cn, so we can write:
and that is the desired power series.
Read more about this topic: Analyticity Of Holomorphic Functions
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