Suppose $X$ is a normal space (compact, if this would help here), and $f : X \to \mathbf{C}$ is continuous. Set $K = f^{-1}(\{0\})$ and let $E \supset K$ be some open neighborhood of $K$ such that $f$ is bounded away from $0$ on $X \backslash E$.
My question is the following. Is it possible to construct (or prove the mere existence of) a continuous map $g: X \to \mathbf{C}$ such that $g$ vanishes nowhere and it coincides with $f$ on $X \backslash E$? My intuition tells me yes. However, I have difficulties constructing such a $g$.
If $u : X \to [0,1]$ is an Urysohn function which vanishes outside $E$ and equals $1$ on $K$, setting $g := f + u$ will work when $f$ is real-valued and positive on $E$, but otherwise it might brake down. This can be avoided by writing $f(x) = |f(x)| e^{ i \arg f(x) }$, and then setting $g(x) := ( |f(x)| + u(x) ) e^{i \arg f(x) }$, however this is ill-defined whenever $f(x) = 0$, so then the question would be if $x \mapsto e^{i \arg f(x) }$ can be continuously extended from $X \backslash K$ to $X$, in such a way that this extension does not vanish. If we think of this as a map from $\overline{X \backslash K}$ (continuously extended to $\partial K$) into $\mathbf{C}$, then Tietze's extension theorem gives us an extension, but this might vanish on $K$. This seems like a dead end to me.
There are probably some other constructions that I'm overlooking. Or maybe there exists some cunning counter-example?
Any advice is welcome. :-)