# A trivial fact about analytic functions

In the book that I'm reading they said that this fact is trivial , but I'm not completely sure about something and I prefer to confirm it. It's about a lot of equivalences, but I have question about two.

Let $f$ be a function defined on a domain $D \subset {\Bbb C}$. Assume that $f$ has a power series expansion at each point of $D$, and let $\zeta \in D$. Then the following are equivalent:

$\quad$ i) $f^{(n)} (\zeta) = 0,\;\text{ for }\;n = 0,1,2\ldots$

$\quad$ ii) $f \equiv 0$ in a neighborhood of $\zeta$

I know how to prove that i) implies ii); I want to see the other side of the proof. I think that I must use the fact that I can write $f$ locally in the form $$f(z) = \sum_{n = 0}^\infty a_n (z - \zeta)^n = \sum_{n = 0}^\infty \frac{f^{(n)}(\zeta)} {n!} (z - \zeta)^n$$ and then by the assumption we have : $$\sum_{n = 0}^\infty \frac{f^{(n)}(\zeta)}{n!}(z - \zeta)^n = 0$$ But now I don't know what can I do, because the sum could be zero, because the sum has some positive terms, and other negative terms, and because this the sum it's zero. I don't know how to finish

Remark: I know that there are other equivalences, even one of them asserts that $f$ vanishes not only on a neighborhood,but in the whole domain $D$, but I want to see for the moment only this two equivalence. Thanks =)!

-
But ii) to i) is trivial, since $f \equiv 0$ means $f(z) = 0, \forall z \in D$, thus $f^{(n)}(z) =0, \forall z\in D$. – Sasha Nov 27 '11 at 22:06
@Sasha: May I ask why you wanted to preserve the OP's grammar? Are msh210 and I misinterpreting what was intended? I generally feel that improving grammar in posts is both helpful to the OP and also improves readability for future readers. – Zev Chonoles Nov 27 '11 at 22:10
@ZevChonoles It may be because grammar is not my strongest point, sorry. – Sasha Nov 27 '11 at 22:12
@Zev: Given how close in time the edits of Sasha and msh210 were (within a second), a reasonable guess is that Sasha was working with the original question and merely did not change the grammar, but since the edit was posted (barely) afterward it undid msh210's improvements. – Jonas Meyer Nov 27 '11 at 22:12
@Jonas: Ah, I agree that is most likely. Sorry for the confusion, Sasha :) – Zev Chonoles Nov 27 '11 at 22:15

Suppose that $f\equiv 0$ on a neighborhood $U$ of $\zeta$. The derivative of the zero function (or any other constant function) is the zero function, so $f'\equiv 0$ on $U$. Similarly, $f''\equiv 0$ on $U$, $f'''\equiv 0$ on $U$, etc. If $f^{(k)}\equiv 0$ on $U$, then in particular $f^{(k)}(\zeta)=0$.