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?
Best Answer
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 $\Omega$.