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?

  • $\begingroup$ It works just for the zeroes, take into account that when you say the function $f$ has a zero of order $n$ around $z_0$ you mean that the function can be written as $f(z) = \sum_{k=n+1}^{\infty}a_k(z-z_0)^k$ around $z_0$, so when you factorize the $(z-z_0)^n$ you just get $f(z) = (z-z_0)^n \sum_{k=0}^{\infty}a_{k+n+1}(z-z_0)^k$ where $g(z)$ would be $\sum_{k=0}^{\infty}a_{k+n+1}(z-z_0)^k$ $\endgroup$
    – aram
    Oct 7, 2014 at 23:08
  • $\begingroup$ At all $z_0 \in \Omega$ such that $f(z_0) =0$ (assuming that $f \neq 0$ in $\Omega$. $\endgroup$
    – Tom
    Oct 7, 2014 at 23:09
  • $\begingroup$ It is not clear what you are asking. Can you please phrase the question in a self contained manner? $\endgroup$
    – timur
    Oct 8, 2014 at 1:35

1 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?


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