TITLE: zeros of holomorphic function in n variables
QUESTION [5 upvotes]: 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.

REPLY [8 votes]: The conjecture is obviously false even for $n=2$. Check $f(z,w)=(w-z^2)(z-(w+1)^2)$. Write $z=x+iy$ and $w=u+iv$. Given $x$ and $u$, I can make first term zero unless $u>x^2$. I can make the second term zero unless $x>(u+1)^2$. Since both inequalities can not be true, we are done.