I know that the zeros of a continuous function from $\mathbb{R}^n$ to $\mathbb{R}$ will from $(n-1)$-dimensional surfaces in $\mathbb{R}^n$ (unless the function itself is zero). Is this also true for a continuous function from $\mathbb{R}^n$ to $\mathbb{C}$? If I decompose a complex function $f$ as
$$f(\textbf{x}) = u(\textbf{x}) + v(\textbf{x})i,$$
for some functions $u$ and $v$ from $\mathbb{R}^n$ to $\mathbb{R}$, then it seems that the zeros of $f$ will be the intersections of the surfaces formed by the zeros of $u$ and $v$, which will be $(n-2)$-dimensional surfaces in $\mathbb{R}^n$, except for the special cases of $u = v$, in which case the zeros will still form $(n-1)$-dimensional surfaces, and the case of $f=0$.
However, if we decompose the complex function as
$$f(\textbf{x}) = \rho(\textbf{x})e^{i\theta(\textbf{x})},$$
for some functions $\rho$ and $\theta$ from $\mathbb{R}^n$ to $\mathbb{R}$, then the zeros of $f$ are simply the zeros of $\rho$, which must form $(n-1)$-dimensional surfaces in $\mathbb{R}^n$.
Which argument above is right? Why is the other wrong? Is there some subtlety with the continuity of a complex function that I'm missing?