Can a holomorphic function be globally represented by a power series on an open connected set? I have a theorem in my book (Stein) which says:

Suppose $f$ is holomorphic in an open set $\Omega$. If $D$ is a disc centered at $z_0$ and whose closure is contained in $\Omega$, then $f$ has a power series expansion at $z_0$
$$f(z) = \sum_{n=0}^{\infty} a_n(z-z_0)^n$$
for all $z \in D$.

Let's say our open set $\Omega$ contains $0$; let $P(z)$ be the power series of $f$ centered at $0$. Obviously, if $\Omega$ is not connected, $P(z)$ does not have to represent $f$ at all points of $\Omega$. However, what about the case where $\Omega$ is connected? Are there any counterexamples?

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 $h\equiv 0$.
