Can a holomorphic function be globally represented by a power series on an open connected set?

The set in which a power series converges is always an open disk together with some subset of the boundary of the disk. So, if a function $f$ can be represented by a single power series centered at $0$ on an open set $\Omega$, then $\Omega$ must be contained in an open disk $D$ (possibly of infinite radius) around $0$ such that $f$ extends holomorphically to $D$ (and conversely, since if $f$ is holomorphic on such a disk $D$ then its Taylor series at $0$ converges to it on the whole disk).

This gives lots of counterexamples. For instance, if $f(z)=\frac{1}{z-1}$, then $f$ is holomorphic on $\Omega=\mathbb{C}\setminus\{1\}$ but cannot be represented by a power series centered at $0$ on this domain since otherwise $f$ would need to extend holomorphically to all of $\mathbb{C}$ since $\mathbb{C}$ is the only disk centered at $0$ that contains $\Omega$.

A power series $f(z) = \sum_{n\geq 0} a_n(z-z_0)^n $ converges on a disk centered on $z_0$. So assuming that $\Omega$ is not a disk implies that there is a function which cannot be given by a single power series.

Pick a fairly general $\Omega$ and let $\{z_j\}$ be a sequence of points in $\Omega$ acummulating on every boundary point of $\Omega$. By a Theorem of Weierstrass there is a holomorphic function $h\colon \Omega \to \mathbb{C}$ vanishing on each $z_j$ and nowhere else.

Since $\Omega$ is not a disk, the only way $h$ could be given globally by a unique power series would be $h$ being holomorphic on a disk $D$ containing $\Omega$. This would lead us to a contradiction since $D$ would contain some boundary point of $\Omega$ which is also an acummulation point for the zeros of $h$ and the identity principle would tell us that $h\equiv 0$.