# Sweet homes london

### El Libro De La Madera: Una Vida En Los Bosques (Spanish Edition) Lars Mytting ((BETTER)) 💖

El Libro De La Madera: Una Vida En Los Bosques (Spanish Edition) Lars Mytting ((BETTER)) 💖 El Libro De La Madera: Una Vida En Los Bosques (Spanish Edition) Lars Mytting > https://urluso.com/2sHZ8V

El Libro De La Madera: Una Vida En Los Bosques (Spanish Edition) Lars Mytting

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Proof of the power series expansion of an analytic function

I’m doing some complex analysis, and I just need a (preferably non-rigorous) proof of the following:
If $f$ is analytic at $a \in \mathbb{C}$ and has the power series expansion $f(z) = \sum_{n=0}^{\infty} c_n(z-a)^n$ for $z \in \mathbb{C}$, then $c_n \equiv 0$ for all but a finite number of $n$.
I’ve never done proofs, so my attempt to do this may be completely wrong.
Let $r_n = \frac{1}{n!}(d^n f)(0)$, where $d^n = \underbrace{d\cdots d}_n$. We have that $r_n \equiv 0$ for $n > m$ since $f$ is analytic, and also for $n = m$ since $c_m \equiv 0$.
Now, the Cauchy Integral Formula tells us that $\int_{\gamma} \sum_{n=0}^{\infty} c_n(z-a)^n \, dz = 0$ (here, $\gamma$ represents a circle centered at $a$ of sufficiently small radius).

Is this enough?

A:

Yes, just need to remark that $\{c_n\}$ is a Cauchy sequence in $\mathbb{C}$.

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