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Let $f$ be holomorphic on the open unit disk. Is it possible to conclude that therefore it has no poles inside this disk and therefore the Taylor series expansion of $f$ at $0$ has at least convergence radius $1$?

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Yes, this is true. The general result is:

Theorem: Let $U\subset\mathbb{C}$ be a domain and $f:U\rightarrow\mathbb{C}$ holomorphic. Let $r(z)$ be the radius of convergence of the power series representing $f$ at $z\in U$. Then $$r(z)\ge \mathrm{dist}(z,\partial U)=\inf_{w\in \partial U} |z-w|$$

If $f$ cannot be analytically extended outside $U$, then that inequality becomes an equality (since, if $r$ was greater, the power series would be an analytic extension).

The theorem follows directly when looking at the proof of analyticitiy of holomorphic functions. This in turn, is essentially a consequence of the Cauchy integral formula in which you consider contour integrals along a circle around $z$. The point is then realizing that the value of the integral doesn't really depend on the radius of that circle, as long as it is contained in the domain.

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