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:


1. 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.
2. The argument for $n=1$ does not adapt trivially to higher $n$ since holomorphic functions of
severable variables have no isolated zeros.
3. 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
$f(z\_1,z\_2)=z\_1-z\_2$.
4. 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.



zeros of holomorphic function in n variables
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.


