Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $P(x,y)$ be a polynomial in two variables that is not identically $0$. Let $f:U\to\mathbb{C}$ be a holomorphic function defined on a region $U\subset \mathbb{C}$ such that $P(\Re(f(z)),\Im(f(z)))=0$ for all $z\in U$. Show that $f$ is constant.

share|improve this question
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. If you don't do this, people are less likely to answer your later questions. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, How does accept rate work?. –  Julian Kuelshammer Oct 16 '12 at 18:01
Also: Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. –  Julian Kuelshammer Oct 16 '12 at 18:03
add comment

3 Answers

Suppose $f$ is not constant. The open mapping theorem tells us that $\{f(z) : z \in U\}$ is an open set in $\mathbb C$. Therefore $\{(\Re f(z), \Im f(z)) : z \in U\}$ is an open set in $\mathbb R^2$. But we assume $P(x,y)$ vanishes on this set, therefore $P$ vanishes identically.

share|improve this answer
Very nice. This is an amazingly concise formalization of the long, informal argument I just posted. –  fgp Oct 8 '12 at 13:11
Thanks, but how to prove that if polynomial vanishes in open set than it is zero? We just know that it is continuous which means $P^{-1}(0,0)$ is closed and it contains open set $(\Re(f(U)),\Im(f(U))$ –  StudentMath Oct 8 '12 at 14:35
Oh, sorry, it is on the next reply, I got it, thanks :) –  StudentMath Oct 8 '12 at 15:42
$+1$ Nice Indeed, @StudentMath Identity Theorem forces that $P$ must vanish identically –  Bunuelian Trick May 6 '13 at 14:44
add comment

Take partials with respect to $x$ and $y$. Then you have

$P_u u_x +P_v v_x=0$

$P_u u_y+P_v v_y=0$.

Use Cauchy-Riemman equations to get

$P_u u_x-P_v u_y=0$

$P_v u_x +P_u u_y=0$.

The assumptions on $P$ force $u_x=u_y=0$. Then $v$ is also constant.

share|improve this answer
I also got it, but I could not get from last equations that $u_{x}$ and $u_{y}$ is identically zero, because there might such values $(x,y)$ such that $P_{u}=P_{v}=0$. On this values we can not say $u_{x}=u_{y}=0$. Is it right? –  StudentMath Oct 8 '12 at 14:17
@StudentMath The determinant of the system is $P_u^2+P_v^2$. If it was zero, then $P_u=P_v=0$. Therefore $P$ would have to be a constant. If the constant is zero, then $P$ is identically zero. If the constant is not zero then the condition cannot hold. Therefore the coefficient matrix of the system is invertible and the system has the unique solution $u_x=u_y=0$. –  PAD Oct 8 '12 at 18:04
add comment

Here's an informal geometric argument. From the Cauchy-Riemann differential equations it follows that a holomorphic function is locally a scaled rotation. I.e, $f(z+e) \approx f(z) + es$ for some $s \in \mathbb{C}$ (Note that $s = f'(z)$)

Assume that $f'(z) \neq 0$. $f$ is then locally invertible, since it's a scaled rotation with non-zero scaling factor. Thus, if you start out at some point $f(z)=a$, you can find a $\tilde{z}$ such that $f(\tilde{z})=a+x+iy$ ($x,y \in \mathbb{R}$), provided that $x,y$ are "small enough" (but otherwise arbitrary!). (Essentially $\tilde{z} = z + (x+iy)/s$). $P(Re(a)+x,Im(a)+y)$ then needs to be zero also, for all valid (small enough) $x,y$. This forces $P$ to be zero on some non-empty open set, which is impossible if $P \neq 0$. It follows that the assumption was wrong, i.e. that $f'(z) = 0$.

To turn this into a formal proof, you need to formalize the concept of "locally invertible". You could, for example, show that if $f$ is holomorphic at $z$, then $f(U) \supset \{f(z) + r: r\in\mathbb{C}, |r| = \epsilon\}$ for some $\epsilon > 0$.

You'll also need argue why $P$ can't be zero on a non-empty open set, but that is rather straight forward. Say $(x,y)$ lies within such a set. Then there are infinitly many $\tilde{y}$ such that $(x,\tilde{y})$ lies also within that set. But for any fixed $x$, a polynomial $P(x,y)$ can only have finitely many $y$ with $P(x,y) = 0$, unless $P = 0$.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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