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.
 A: 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.
A: 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.
A: 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 $P = 0$.
