Let be $a_{1}; a_{2}$ 2 complex numbers whose modules are $> 1$. I define on $\mathbb{C}^{2} - \{0\}$ the following equivalence relation $\equiv$ : $(z_{1}; z_{2}) \equiv (z_{1}'; z_{2}') \Leftrightarrow \exists n \in \mathbb{Z} | (z_{1}; z_{2}) = (a_{1}^{n}z_{1}'; a_{2}^{n}z_{2}') $. I will wrote $Q$ the quotient (which is a compact complex manifold) and $\pi : \mathbb{C}^{2} - \{0\} \mapsto Q$ the canonical projection (which is holomorphic).
In a problem, I've shown the closure of $\{(x; y) \in \mathbb{C}^{2} -\{0\}\times Q | p(x) = y \}$ in $\mathbb{C}^{2} \times X$ is analytic$**$ and I want to infer a contradiction. I've tried a lot of things but it doesn't solve the problem.
Could you help me please?
$**$The definition of analytic is given here https://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/agbook.pdf page $91$ definition (4.1).
I wish you a good day.
Edit : I forget the hello at the beginning.
I think I found : let $x$ be in $Q$. The assumption of being analytic implies that their exists $U$ an open neighbourhood of $(0; x)$ and $\{f_j\}$ holomorphic maps on $U$ such that $(\mathbb{C}^{2}- \{0\} \times Q \cup \{0\} \times Q) \cap U = \{f_j = 0\}$.Now, let be $(z_{1}; z_{2})$ an element such that $p(z_{1}; z_{2}) = x$. Let $D$ be the line passing by $(z_{1}; z_{2})$ and $0$. For $x'$ close enough to $x$ and different from $x$, their is a clue $j$ such that $f_{j}(z_{1}; z_{2}; x') \neq 0$. We may choose $x'$ such that any represantant of $x'$ are on $D$. By the fact the $0$ of a non constant holomorphic function are isolated, their is a neighbourhood $W$ of $0$ in $\mathbb{C}^{2}$ such that $f_{j}(.; x')$ are $0$ on $D \cap W$ only in $(0; 0)$. But one get a contradiction since we can found a represantant $(z_{1}'; z_{2}')$ of $x'$ on $D$ and for $k \leq 0$ integer large enough, $0 \neq (a_{1}^{k}z_{1}'; a_{2}^{k}z_{2}') \in W$ so that $f_{j}(a_{1}^{k}z_{1}'; a_{2}^{k}z_{2}', x') = 0$.