Let $f: \mathbb C \to \mathbb C$ be an analytic function, such that for all $z\in\mathbb C$, $$| \operatorname{Re}(f(z))\operatorname{Im}(f(z)) | \le 1.$$ I have to show that $f$ is constant.
I don't know how to apply Liouville's theorem to the question, but I can tell it is bounded, thus the theorem should apply. Any help would be appreciated.
As $\operatorname{Im}(f^2)= 2 \operatorname{Re}(f)\operatorname{Im}(f)$, hence $\operatorname{Re}(i\cdot f^2)$ is bounded, and $\exp(i\cdot f^2)$ is holomorphic, bounded, constant.
Let us define $f = u+iv$ for $u,v : \mathbb{R}^2 \mapsto \mathbb{R}$. $f$ is analytic by our assumption.
We now define the auxiliary function $g = f^2 = u^2-v^2+2iuv$ which is also analytic, since it's a composition of two analytic functions ($f$ and $z^2$). Finally, we define the second auxiliary function $h = e^{-if^2} = e^{2uv}e^{-i(u^2-v^2)}$
Now trivially, we have that $h$ is analytic, since it's a composition of two analytic functions $f, e^z$, and also, $|h| \leq e^2$ by construction. So $h$ is both analytic and bounded, which means that $h$ is constant by Liouville's theorem. But from this we must have that $f^2$ is constant, and therefore $f$ is constant, once again by construction.
Which was to be shown.
