If $P(\Re(f(z)),\Im(f(z)))=0$ where $f$ is holomorphic, than $f$ is constant.

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.

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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.

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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 – Un Chien Andalou May 6 '13 at 14:44

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.

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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

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 $P = 0$.

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