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Let $f:\Omega\longrightarrow \mathbb{C}$ be analytic, and let $z_0\in\Omega$ s.t. $$f (z)=\sum_{n\geq 0} a_n(z-z_0)^n,\quad\forall \left|z-z_0\right|<r,$$ for some $r>0$, and some complex-valued sequence $(a_n)_{n\in\mathbb{N}}$.

Now there may exist some $R>r$ s.t. the disc (except perhaps its outline) $D(z_0,R)\setminus C(z_0,R)\subset\Omega$, and so $f$ is still analytic inside it.

How can I show then that $f$ still converges to the series $\sum_{n\geq 0} a_n(z-z_0)^n,\forall r\leq\left|z-z_0\right|<R$, in the case that such $R$ exists?

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  • $\begingroup$ Isn't that a standard result of the Taylor expansion from the Cauchy Integral Formula? $\endgroup$
    – Torsten Schoeneberg
    Commented Jun 11 at 6:23
  • $\begingroup$ @TorstenSchoeneberg How exactly? Even if we show that the Cauchy Formula still evaluates to $a_n$ outside the small circle, we don't know if the series converges or diverges, no? $\endgroup$
    – virtualcode
    Commented Jun 11 at 11:29
  • $\begingroup$ Well it certainly needs work to prove it, but the sources I know do state that if you develop a function $f$, analytic on $\Omega$, into its Taylor Series around a point $z_0 \in \Omega$, then that Taylor Series converges on any open disk around $z_0$ which is fully contained in $\Omega$. (And by the identity theorem for analytic functions, that one must be the same series as the one you have on the potentially smaller disk with radius $r$.) $\endgroup$
    – Torsten Schoeneberg
    Commented Jun 11 at 20:36

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