Let $f$ be an analytic function on a domain $\Omega\subset \mathbb{C}^n$. I am trying to prove that the dimension of $Z_f=f^{-1}(0)$ is $n-1$. This is stated in this answer, but I do not understand the proof of this fact in Chirka's "Complex Analytic Sets" (section 2.6).
Here is how the proof goes: by the identity theorem for analytic functions, we may find an open set $U$ of some generic point $a$ and $g$, some derivative of $f$ that vanishes identically on $Z_f\cap U$ but such that $dg$ is non-zero at $b$. He claims that this implies that $Z_f$ is a manifold of dimension $n-1$ in a neighborhood $b$, but I do not see how he concludes.
Clearly one may apply the implicit function theorem to see that $Z_g\cap V$ is a manifold of dimension $n-1$ containing $Z_f\cap V$ for $V$ a neighborhood of $b$. But how do we see from this that $Z_f\cap V$ is smooth? I suspect that in fact these sets are equal, but I have not been able to show it.
Furthermore, it seems to me that a detail is missing because everything written pertains to real analytic functions (for which the result is untrue). Any insight or alternative reference would be much appreciated.