Analyticity of holomorphic functions

In complex analysis a complex-valued function ƒ of a complex variable z:

f(z)=\sum_{n=0}^\infty c_n(z-a)^n
(this implies that the radius of convergence is positive).

One of the most important theorems of complex analysis is that holomorphic functions are analytic. Among the corollaries of this theorem are

Proof

The argument, first given by Cauchy, hinges on Cauchy's integral formula and the power series development of the expression

\frac 1 {w-z} .

Let D be an open disk centered at a and suppose ƒ is differentiable everywhere within an open neighborhood containing the closure of D. Let C be the positively oriented (i.e., counterclockwise) circle which is the boundary of D and z a point in D. Starting with Cauchy's integral formula, we have

\begin{align}f(z) &{}= {1 \over 2\pi i}\int_C {f(w) \over w-z}\,\mathrm{d}w \\[10pt]

&{}= {1 \over 2\pi i}\int_C {f(w) \over (w-a)-(z-a)} \,\mathrm{d}w \\[10pt]
&{}={1 \over 2\pi i}\int_C {1 \over w-a}\cdot{1 \over 1-{z-a \over w-a}}f(w)\,\mathrm{d}w \\[10pt]
&{}={1 \over 2\pi i}\int_C {1 \over w-a}\cdot{\sum_{n=0}^\infty\left({z-a \over w-a}\right)^n} f(w)\,\mathrm{d}w \\[10pt]
&{}=\sum_{n=0}^\infty{1 \over 2\pi i}\int_C {(z-a)^n \over (w-a)^{n+1}} f(w)\,\mathrm{d}w.\end{align}

Interchange of the integral and infinite sum is justified by observing that f(w)/(w-a) is bounded on C by some positive number M, while for all w in C

\left|\frac{z-a}{w-a}\right|\leq r < 1

for some positive r as well. We therefore have

\left| {(z-a)^n \over (w-a)^{n+1} }f(w) \right| \le Mr^n,

on C, and as the Weierstrass M-test shows the series converges uniformly, the sum and the integral may be interchanged.

As the factor (z  a)n does not depend on the variable of integration w, it may be factored out to yield

f(z)=\sum_{n=0}^\infty (z-a)^n {1 \over 2\pi i}\int_C {f(w) \over (w-a)^{n+1}} \,\mathrm{d}w,

which has the desired form of a power series in z:

f(z)=\sum_{n=0}^\infty c_n(z-a)^n

with coefficients

c_n={1 \over 2\pi i}\int_C {f(w) \over (w-a)^{n+1}} \,\mathrm{d}w.

Remarks

 \frac 1 {(w-z)^{n+1}}
gives
f^{(n)}(a) = {n! \over 2\pi i} \int_C {f(w) \over (w-a)^{n+1}}\, dw.
This is a Cauchy integral formula for derivatives. Therefore the power series obtained above is the Taylor series of ƒ.

External links

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