Suppose $f:\mathbb{C}^d \to \mathbb{C}$ is a holomorphic function such that ${\partial f \over \partial z_k} \neq 0$ for $k = 1, \ldots, d$ whenever $f(z) = 0$. Suppose also that the zero set of $f$ intersects $\mathbb{R}^d$ nontrivially. Let $$ A = \{x \in \mathbb{R}^d: f(x) = 0\}. $$ Since $f$ is holomorphic and ${\partial f \over \partial z_k} \neq 0$ on $A$, we also have that $df \neq 0$ on $A$ where $d$ is the exterior derivative with respect to the real coordinates on $\mathbb{R}^d$.
I want to conclude that $A$ is a real-analytic manifold. My only hesitation is that $f$ is complex-valued instead of real-valued? Even though $f$ is complex-valued, would we still say that $f$ is a real-analytic function since its restriction is defined on $\mathbb{R}^d$ and analytic there? Would we still say that $A$ is a real-analytic manifold?