Property of analytic functions?

Question

For an open set $\Omega$, a function $f$ is analytic $z_0 \in \Omega$ if there exists a power series $\sum a_n(z-z_0)^n$ centered at $z_0$ with positive radius of convergence, such that:

$$f(z) = \sum_{n=0}^{\infty}a_n(z-z_0)^n$$

for all $z$ in a neighborhood of $z_0$.

There also exists a theorem that says if $f$ is holomorphic in a connected open set $\Omega$, has a zero at $z_0 \in \Omega$, and does not vanish identically in $\Omega$, then there exists a neighborhood $U \subset \Omega$ of $z_0$, a non-vanishing holomorphic function $g$ on $u$, and a unique positive integer $n$ such that

$$f(z) = (z-z_0)^ng(z)$$

for all $z \in U$.

However, does this second property hold for all $z_0 \in \Omega$ for any holomorphic $f$? Why or why not?

Answer 1

Well, let's suppose it held for all $z_0\in\Omega.$ Then, take an arbitrary such $z_0.$ By assumption, there is a neighborhood $U$ of $z_0$ with $U\subseteq\Omega,$ and there is a function $g$ holomorphic on $U$ and a positive integer $n$ such that $f(z)=(z-z_0)^n\cdot g(z)$ for all $z\in U.$ Given that, what is $f(z_0)$? But since that holds for all $z_0\in\Omega$ by assumption, what can we say about $f$ in $\Omega$? Do you see the contradiction?