I was studying several complex variables and am little bit confused about analytic continuation here. My questions are as follows:
(i) why we need connected set here? I understand a connected set cannot be a union of two open sets. So if $D$ is a proper subset of $E$, then $D = E \cup (D/E)$, which is contradiction to the connectedness of $D$. So it implies what? I'm so confused about it.
(ii) Clearly $E$ is not empty set. So $E = D$ means $f$ vanishes the whole domain $D$?
(iii) If I want to use Cauchy integral formula instead of Taylor series expansion in this proof, can I say directly, if $z \in E$, then $f = 0$ as the formula says on any $\overline{\mathbb{D}_r^n(\boldsymbol{z^{\ast}})} \subseteq D$, we have that $$ \partial^{\alpha} f(\boldsymbol{z}) = \frac{1}{(2\pi i)^n} \int \ldots \int_{\partial_0 \mathbb{D}^n_r (\boldsymbol{z^\ast})} \frac{f(\boldsymbol{w})}{(\boldsymbol{w} - \boldsymbol{z})^{\alpha+\boldsymbol{1}}} d \boldsymbol{w} $$ for all $\boldsymbol{z} \in \mathbb{D}_r^n(\boldsymbol{z^{\ast}})$?
Any help will be appreciated!
(i) If $D$ is not connected, say $D = D_1 \cup D_2$ where $D_1$ and $D_2$ are non-empty disjoint open sets, then the function $$ f(z) = \begin{cases} 0,& z \in D_1 \\ 1, & z \in D_2 \end{cases} $$ is holomorphic on $D$, vanishes on a non-empty open subset, and is not identically $0$.
(ii) Yes
(iii) Not really. That argument only shows that $f = 0$ on the union of all polydiscs in $D$ whose distinguished boundary lie in $E$. Compare the situation in one variable, if you try the same argument (it doesn't work there either).