Let $\sf C$ be an equidimensional algebraic curve of $\mathbb{C}^n$. Let $x$ be a point of $\sf C$ witch is a regular/smooth point.
I want to prove that there exists a Zariski open set $O$ such that $x\in O$ and ${\sf C}\cap O$ is diffeomorphic to the tangent space of $\sf C$ at $x$.
It seems obvious on a drawing but i don't manage to prove it correctly.
This is false:
Take the smooth hyperbola $H\subset \mathbb C^2$ given by $xy=1$.
It is algebraically isomorphic to $\mathbb C^*$ and its tangent space $\mathbb T_x H $ at any point $x\in H$ is a line, diffeomorphic to $\mathbb C$.
However any open Zariski neighbourhood $H\cap O$ of $x$ in $H$ will be isomorphic to $\mathbb C^*$ with maybe a few points deleted, so that it will not be simply connected and thus definitely not diffeomorphic to $\mathbb T_x H \simeq \mathbb C$ .