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Is the zero set Z(f) of the holomorphic function $f$ a complex manifold? If not what could be the conditions required for it to be.

We know that for a topological space to be a complex manifold it has to be compact, have an open cover and homomorphisms from each open subset to the set of complex numbers such that they are holomorphic (analytic) and their inverses also.

user111
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It's not always a complex manifold: Severn Schraven pointed out that the zero set of $f(z_1,z_2)=z_1z_2$ has a non-manifold point $(0,0)$; specifically, a neighborhood of $(0,0)$ in $Z(f)$ is not even homeomorphic to an open subset of $\mathbb{C}$, let alone being complex-diffeomorphic to it.

A sufficient condition is that the implicit function theorem applies to $f$, which for holomorphic functions $f:\mathbb{C}^n\to\mathbb{C}$ means: at every point of the zero set $Z(f)$, at least one of the partial derivatives $\partial f/\partial z_k$ is nonzero. The implicit function theorem then provides a holomorphic parameterization of small pieces of $Z(f)$. It is discussed in Analytic implicit function theorem and Implicit function theorem for several complex variables.

(I used to think that nonvanishing of $\nabla f$ on $Z(f)$ is also necessary for $Z(f)$ to be a complex manifold, but Daniel Fischer pointed out that $f^2$ has the same zero set. )

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    The nonvanishing of $\nabla f$ isn't necessary, for example $Z(f^2) = Z(f)$. But locally that's the only way, if $Z(f)$ is a complex manifold near $\mathfrak{z}$, then there is a neighbourhood $U$ of $\mathfrak{z}$ and a holomorphic $g\colon U \to \mathbb{C}$ such that $Z(f)\cap U = Z(g)$, $\nabla g \neq 0$, and $f = g^k$ for some $k\in \mathbb{N}\setminus {0}$ on $U$. – Daniel Fischer Mar 19 '17 at 20:58