Let $f \colon M \rightarrow N$ be a holomorphism of complex manifolds. Let $p \in M$. Let $(U,\phi)$ and $(V,\psi)$ be coordinate charts on $M,N$, respectively, satisfying $U \ni p$ and $V \ni f(p)$.
I don't know what you call it, but as $f$ is holomorphic, the function $F \colon \phi(U) \rightarrow \psi(V)$ given by $F = \psi \circ f \circ \phi^{-1}$ is a holomorphism (in the regular $\mathbb{C}^n$ sense).
Define the rank of $f$ at the point $p \in M$, denoted rk$_pf$, as the rank of the Jacobi matrix of $F$ at the point $\phi(p)$. (I'm guessing readers already know this.) Here is my question:
Is there an open neighbourhood of $p$ in which the rank of $f$ is constant, or at least $\geq$ or $\leq$ rk$_pf$?
Just as in the case of smooth manifolds, rank can degenerate down but not up. That is, there's always a neighborhood of $p$ on which $\text{rk} f\geq \text{rk}_p f$. This is just because the rank is witnessed by a certain minor of the Jacobian having nonzero determinant, and this determinant is a continuous function of $p$. For an example when rank degenerates down, look no further than $z^2:\mathbf{C}\to\mathbf{C}$ at zero.