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Consider a region $\omega \subset \mathbb{C}$, and a holomorphic function $f : \omega \rightarrow \mathbb{C}$. I wish to prove that if $D$ is an open disk with center $z_{0}$ whose closure is contained in $w$, then there exists $(a_{n})_{n \in \mathbb{N}}$ such that

$f(z) = \displaystyle\sum_{n = 0}^{\infty} a_{n}(z - z_{0})^{n}$, for $z \in D$.

To show this, we set $a_{n} = \displaystyle\frac{f^{(n)}(z_{0})}{n!}$. Then, plugging in to the above sum gives

$\displaystyle\sum_{n = 0}^{\infty} \frac{(z - z_{0})^{n}}{2\pi i} \int_{C} \frac{f(w)}{(w - z_{0})^{n + 1}}\ dw$, after applying Cauchy's formula for $f^{(n)}(z_{0})$ (where $C$ is the boundary of $D$).

If I could justify changing the order of the summation and integral signs, it would be pretty easy to continue and show that this equals $f(z)$. The book (Stein's complex analysis) follows a different argument to prove this, but it also switches order once, saying only that it's valid because of uniform convergence. So does this converge uniformly?

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Yes, you can prove the uniformly convergence by estimate the norm of $b_n:=\frac{1}{2\pi i}\int_C\frac{f(w)}{(w-z_0)^{n+1}}dw$.

Suppose the circle $C$ has radius $r$, and let M=\sup_{z\in C}{f(z)}<\infty. Then we have

$ |b_n|\leq\frac{1}{2\pi}\int_C\frac{|f(w)|}{|w-z_0|^{n+1}}|dw| \leq \frac{1}{2\pi}\int_C\frac{M}{r^{n+1}}|dw|=\frac{M}{2\pi r^{n+1}}\cdot2\pi r=\frac{M}{r^n}. $

Therefore, by criterion of convergence of power series, the convergent radius $R$ of $\sum_{n=0}^\infty b_n(z-z_0)^n$ is no less than $r$. By choose $D_\epsilon$ slightly larger than $D$, we have $R>r$, hence the series $\sum_{n=0}^\infty b_n(z-z_0)^n$ is uniformly convergent in $D$.

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If your series converges in D whose closure is contained in w, then indeed this series will converge uniformly in compact subsets of D (and also in D as you could choose a slightly bigger D' containing D and satisfying the same properties).