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 $x\mapsto y_x$.)


  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.


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.

  • $\begingroup$ Thanks. Nice answer. I'll have to see whether there are additional hypotheses in the application that I have in mind. $\endgroup$ – M Mueger Feb 2 '18 at 17:01

Not a complete answer, but perhaps in the many-variable case you could leverage Theorem 5.1 from

Chirka, E.M., Complex analytic sets. (Kompleksnye analiticheskie mnozhestva), Moskva: ”Nauka”. Glavnaya Redaktsiya Fiziko-Matematicheskoj Literatury. 272 p. R. 3.70 (1985). ZBL0586.32013.

Which states that for an analytic set $Z \subset \mathbb{C}^n$, like the zero set of $f(z) = 0$, the regular part $\operatorname{reg} Z \subset Z$ (which consists of all points around which $Z$ is a complex submanifold) consists of a union of connected components $\operatorname{reg} Z = \bigcup_j S_j$, which is locally finite. That is, for any compact $K\subset \mathbb{C}^n$, $K$ intersects at most finitely many of the $S_j$. The singular part $\operatorname{sng} Z = Z \setminus \operatorname{reg} Z$ is also an analytic set, but of lower dimension. So a similar argument applies to it.

The above reference also contains a wealth of other information about complex analytic sets that might be of use.

  • 1
    $\begingroup$ Well, the zero-set of a holomorphic function is an analytic set and therefore locally a unit of holomorphic (n-1)-manifolds. But I don't see how this helps approaching the conjecture, where real and imaginary parts of the variables are treated differently. $\endgroup$ – M Mueger Feb 2 '18 at 15:32
  • $\begingroup$ As the counter example in Oleg's answer showed, the obstruction to the conjecture is local (unlike in the $n=1$ case). Had there been non local obstruction, the local finiteness of the union of connected components might have come in useful. $\endgroup$ – Igor Khavkine Feb 3 '18 at 10:12

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.