How to prove that if $f=u+iv $ and for any $ z$, $u>0 $ or $v>0$, then $ f $ is constant Assume $ f $ is entire such that $f=u+iv $ and for any $ z $, at least one of $u(z),v(z) $ is positive.
Prove that $ f $ constant.
Well, I've been thinking about it for a while and all I could say is that $ f $ maps all of the plane into the union of $Re(z)>0 $ and $Im(z)>0 $ and its image is an open set if $ f $ is not constant (by the open mapping theorem).
It feels like this should be a contradiction, but cant tell explicitly why.
Thanks in advance.
 A: The entire function $f$ omits an open set from its values, e.g. the disk about $z_0 = -1-i$ with radius 1. Now find a 1-1 mapping $g$ from the outside of this disk to the open unit disk and consider the entire bounded function $g \circ f$. Take it from there.
Look up any proof of the Little Picard Theorem (see @TonyK's comment). It uses the same idea.
A: You don't even need Little Picard. The well-known theorems characterizing the types of singularities are enough.
If $f$ is entire, then it is either a polynomial or transcendental (meaning that its power series expansion is infinite). If it is transcendental, then the map $g(z):=f\left(\frac 1z\right)$ has an essential singularity at $0$, so its image is dense in $\mathbb C$. And since the image of $g$ is either equal to that of $f$ or has only a single element missing ($f(0)$), the image of $f$ is also dense in $\mathbb C$. A contradiction to the given constraint, which says that a whole quarter plane is missing from the image. So $f$ must be a polynomial. But if it is a polynomial, it must be constant, since all non-constant polynomials are surjective according to the fundamental theorem of algebra.
