Let $f$ be an entire function such that $\Re(f(z))*\Im(f(z))\ge0$ for all $z$ then $f$ is constant. Prove or give contradicting example. I know about Louisville's theorem and Cauchy–Riemann equations but I don't see how to use them in this situation. I've tried to bound $|f(z)|$ but I could only show that $|f(z)|\le\Re(f(z))+\Im(f(z))$. I would like a hint.
Write $f(z) = u(z) + iv(z)$, so that $u=\Re(f)$ and $v = \Im(f)$. The trick to many of these problems is to find a suitable auxiliary function: as Daniel Fischer mentioned, a good candidate would be a function whose real or imaginary part is $uv$. How about $g(z) = f(z)^2 = u(z)^2-v(z)^2 + 2i\,u(z)v(z)$? By hypothesis, $\Im(g(z))\geq 0$ for all $z$ . . .
You don't need anything as strong as Picard's Theorem. The fact that the image of a non-constant entire function must be dense will suffice. This fact is a simple consequence of Liouville's Theorem.