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.