Conjecture: Let $f:{\mathbb C}^n\rightarrow{\mathbb C}$ be an entire function in $n$ complex variables. Assume that for every $x\in{\mathbb R}^n$ there exists a $y_x\in{\mathbb R}^n$ such that $f(x+iy_x)\equiv f(x_1+iy_{x,1},\ldots,x_n+iy_{x,n})=0$. Then $f\equiv 0$. (We don't assume continuity of $x\mapsto y_x$.)

Remarks:

The conjecture is true for $n=1$: If $f$ satisfies the hypothesis, it has uncountably many different zeros, namely $\{ x+iy_x\ | \ x\in{\mathbb R}\}$. Since ${\mathbb C}$ is $\sigma$-compact, there exists a compact $K\subset{\mathbb C}$ containing uncountably many zeros of $f$. Thus the set of zeros of $f$ has an accumulation point in $K$, and by a well-known result $f$ vanishes identically.

The argument for $n=1$ does not adapt trivially to higher $n$ since holomorphic functions of severable variables have no isolated zeros.

In the conjecture it is important that $x$ consists of the real and $y$ of the imaginary parts of the variables. For example, the following is not true: Assuming that

$f:{\mathbb C}^2\rightarrow{\mathbb C}$ is entire and for every $z_1=x_1+iy_1$ there is a $z_2=x_2+iy_2$ such that $f(z_1,z_2)=0$, we have $f\equiv 0$. A trivial counterexample is $f(z_1,z_2)=z_1-z_2$.Most books on holomorphic functions in several variables devote some attention to the set of zeros of such a function. Usually this leads to a proof of Weierstrass' preparation theorem, where the discussion ends. This is of no immediate help to me, since I would actually be happy with a proof of the conjecture for

*polynomials*in $n$ complex variables! In this case, Weierstrass' theorem does not give any new information.