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$\{ z: f^n(z) = 0 , n = 1,2,3, \ldots \}$ and $f$ is holo on an open, connected region. $f^n$ is the ($n$-th) derivative of $f$ .

How do you show this is an open set by Taylor theorem?

It's closed as $x$ limit point implies for a given $n$ there exists $k(n)$ s.t. $k(n)$ is in the region where

$f(x+k(n)) = 0$ and $|k(n)| < 1/n$.

Holomorphic functions are infinitely differentiable, and derivatives are continuous, therefore $f^n(x) =0$ as we can take the limit inside.

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Does $f^n$ represent the $n$-th derivative of $f$? I think you want to say "$f$ is holomorphic on a open connected region" instead of "$f$ is holomorphic and a open connected region", is that correct? – Paul Dec 16 '11 at 8:17
Please revise the first sentence to make more sense. As for your question, the set where a holomorphic function is $0$ is not generally open; the only case where that can happen (given that the domain is connected) is when the function is identically $0$. So, you seem to have left out some hypotheses. – Jonas Meyer Dec 16 '11 at 8:17
I think the whole question needs to be revised. What does the title "Holomorphic function open, closed zeroes" really mean? Also, what's the meaning of the last sentence: "therefore derivatives are continuous and $f(x)=0$"? – Paul Dec 16 '11 at 8:21
Changed, sorry I was in a hurry! Hope this is alright. – Adam Dec 16 '11 at 8:32
Are you asking about (the proof of) the identity theorem for holomorphic functions (applied to $f$ and $g = 0$ in Wikipedia's notation)? – t.b. Dec 16 '11 at 8:45
up vote 4 down vote accepted

Although the question is not very clear (as the comments show), this is perhaps what the OP intended to ask:

Let $\Omega \subseteq \mathbb C$ be an open, connected region and let $f : \Omega \to \mathbb C$ be a holomorphic, nonzero function. Then there does not exist any point $z \in \Omega$ where the function and all its derivatives vanish.

Here's a proof. Let $A$ be the set of all points $z \in \Omega$ such that $f^{(n)}(z) = 0$ for all $n \geqslant 0$. (I use $f^{(0)}$ to denote $f$.) We will show that $A$ is both open and closed, which implies (by connectedness of $\Omega$) that either $A = \Omega$ or $A$ is empty. Finally, if $f$ is nonzero, then $A$ cannot be $\Omega$; therefore, $A = \emptyset$.

$A$ is closed. For any $n$, the set $B_n := \{ z \in \Omega \mid f^{(n)}(z) = 0 \}$ is closed since $f^{(n)}$ is continuous. Therefore, $A = \bigcap \limits_{n \geqslant 0} B_n$ is also closed.

A is open. Suppose $z \in A$. Then since $f$ is analytic at $z$, there exists some $\varepsilon > 0$ such that for $w \in B(z, \varepsilon)$, we have $$ f(w) = \sum_{n=0}^{\infty} \frac{f^{(n)}(z)}{n!} \cdot (z-w)^n. $$ But since $f^{(n)}(z) = 0$ by definition of $A$, $f$ is identically zero inside $B(z, \varepsilon)$. Therefore, all its derivatives also vanish inside the same ball; therefore, $B(z,\varepsilon) \subseteq A$. That is, $z$ is an interior point of $A$, and we are done.

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Thankyou, my professor used shorthand as the question was written on his board near the end of the tutorial - i.e. he was rushing. I think you correctly deciphered what was going on - cheers! – Adam Dec 17 '11 at 14:28

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