TITLE: Is there any holomorphic version of the tubular neighborhood theorem?
QUESTION [30 upvotes]: This question arised when I was studying Beauville's book 'Complex Algebraic Surfaces'. 
Castelnuovo's theorem says that a smooth rational curve $E$ on an algebraic surface $S$ is an exceptional curve iff $E^2=-1$. The proof in Beauville's book is to find a very ample divisor $H$ satisfying $H^1(S,\mathcal{O}_S(H))=0$ first, and then set $H'=H+kE$ where $k=H\cdot E$. The linear system of $H'$ gives a projective morphism from $S$ to $\mathbb{P}^n$ which contracts $E$, and then some topological arguments implies that the image of $S$ is actually smooth.
Although this proof is not difficult to understand, I still want a proof based on complex manifolds but not algebraic geometry.
Question: Is there any holomorphic version of the tubular neighborhood theorem?
I have several reasons to raise this question:

If we have some holomorphic tubular neighborhood theorem, we can identify some neighborhood $U$ of $E$ in $S$ with neighborhood $V$ of the zero section in $N_E$. Here $N_E$ is the holomorphic normal bundle of $E$. Then $E^2=-1$ easily implies $N_E\cong\mathcal{O}_{E}(-1)$, so $E$ can be contracted in $U$ directly. Thus we not only prove Castelnuovo's theorem but also generalize it to non-algebraic surfaces.
There exists a symplectic version of the tubular neighborhood theorem, so I guess the holomorphic case is also true.

Any answers or comments are welcome. I'll really appreciate your help.

REPLY [6 votes]: Perhaps it is worth saying that while in general a complex submanifold of a complex manifold does not admit a holomorphic tubular neighborhood, it does admit a special smooth tubular neighborhood; the leaves that get contracted are complex manifolds, but they do not vary holomorphically.