Flow of a vanishing Vector Field. Let us consider a complete vector field $V:\Bbb C^n\to\Bbb C^n$.
Let us call $\phi_s:\Bbb C^n\to\Bbb C^n$ its flow at time $s\in\Bbb C$, that is
$$
\dot{\phi}_s(x)=V(\phi_s(x))\;\;\forall x\in\Bbb C^n, s\in\Bbb C
$$
$$
\phi_0(x)=x\;\;\forall x\in\Bbb C^n
$$
How can I prove that $V\equiv0$ on $A$ implies $\phi_s(x)=x \;\;\forall x\in A$ and $s\in\Bbb C$?
Set $A$ as the common zero-set of a finite number of holomorphic polynomials (but maybe my statement holds for more general sets $A$, like closed sets).
 A: Assuming that $V$ is Lipschitz continuous in $x$, so that we have existence and uniqueness of solutions of
$\dot \phi_s(x) = V(\phi_x(x)), \tag 1$
then we need merely observe that
$\phi_s(x_0) = x_0 \tag 2$
satisfies
$\dot \phi_s(x_0) = 0 = V(\phi_s(x_0)), \; \forall x_0 \in A, \; s \in \Bbb C; \tag 3$
that is, constant function (2) obeys (1) on $A$; therefore, uniqueness implies it is the only such solution; $V = 0$ on $A$ implies (2) for all $s$.
Though the standard existence and uniqueness result cited and linked here is usually stated in terms of a real independent variable $s$, its extension to complex $s$ is straightforward; see this question for further details.  
A: If $V_x=0$ for some $x\in\mathbb{C}^n$, then the curve $\gamma:s\mapsto x$ is solution of $\dot\gamma(s)=V_{\gamma(s)}$ (since both sides are $0$) and $\gamma(0)=x$. Since two integral curves who coincides at a point are the same, you can conclude that $\varphi_s(x)=\gamma(s)=x$ for all $s\in\mathbb{R}$. Since it is correct for all such $x\in\mathbb{C}^n$, you get what you want.  
