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I was trying to use the following theoerem to prove analytic continuation. Here it is:

Let $f: \Omega \to \mathbb{C}^m$ be holomorphic where $\Omega \subseteq \mathbb{C}^n$ is open. If $\Omega$ is connected and there exists $\boldsymbol{z^\ast} \in \Omega$ such that $\partial^\alpha f(\boldsymbol{z^\ast}) = 0$ for all $\alpha \in \mathbb{N}^n$, then $f = 0$ on $\Omega$.

By the corollary of Cauchy's integral formula in polydiscs, we have $$ \partial^{\alpha} f(\boldsymbol{z}) = \frac{1}{(2\pi i)^n} \int \ldots \int_{\partial_0 \mathbb{D}^n_r (\boldsymbol{z^\ast})} \frac{f(\boldsymbol{w})}{(\boldsymbol{w} - \boldsymbol{z})^{\alpha+\boldsymbol{1}}} d \boldsymbol{w} $$ for all $\boldsymbol{z} \in \mathbb{D}_r^n(\boldsymbol{z^{\ast}})$. Then I consider the set defined by $A = \{ \boldsymbol{z} \in \Omega : \partial^{\alpha} f(\boldsymbol{z}) = 0, \forall \alpha \in \mathbb{N}^n \}$ and want to show that $A$ is both closed and open.

I was wondering if I am on the wright way to prove it? Any help will be appreciated.

Candlelight
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  • The polydisc Cauchy integral formula gives that $f$ is analytic, so if all its derivatives vanish at some point then it is identically zero, around that point and everywhere – reuns Mar 07 '23 at 22:41
  • May I ask why we need the connected set here? – Candlelight Mar 14 '23 at 00:37
  • If it is two disjoint open sets then $f$ can be $=0$ on one and $=1$ on the other – reuns Mar 14 '23 at 00:54
  • Many thanks! Could you please have a look on this question: https://math.stackexchange.com/questions/4658515/princples-of-analytic-continuation-in-several-complex-variables ? Since I am confused about how to apply connected set in a proof. – Candlelight Mar 14 '23 at 14:10

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