Let $f$ : $\mathbb C \to \mathbb C$ be analytic such that $|Re(f(z)) Im(f(z))|$ $\le$ $1$ for every $z \in \mathbb C$. Show that $f$ is constant.
I know the set is bounded hence I should be able to apply Liouville Thm.
Other than this information I do not know how to approach this question.
Any help will be appreciated.
In the linked question it is shown that $\;e^{if^2}\;$ is a constant, and thus also is $\;if^2\;$ and, finally, then also $\;f\;$ is...
The above is obtained by first observing that, if we write $\;f=u+iv\;$ , then
$$\text{Im}\,(f^2)=\text{Im}(u^2-v^2+2uvi)=2uv\;\;\;\text{is bounded by the given data}$$
and thus we get that
$$\text{Re}(if^2)=\text{Re}\,(-2uv+(u^2-v^2)i)=-2uv\;\;\text{is also bounded}$$
Let $f(z)=u(z)+iv(z)$, with $u$ and $v$ real valued functions. We will write $f^2(z)=[f(z)]^2$ Note that $$f^2(z)=[u(z)+iv(z)]^2=u^2(z)-v^2(z)+2iu(z)v(z)$$
So your condition is saying that $Im \ f^2$ is bounded. That is equivalent to say that $Re \ if^2$ is bounded. Now, consider $h(z)=e^{if^2(z)}$. We have
$$|h(z)|=|e^{Re(if^2(z))}\cdot e^{i\cdot Im(if^2(z))}|\leq e^{Re(if^2(z))} < M$$
(using that $|e^{i\cdot Im(if^2(z))}|<1$) so $h(z)$ is an entire and bounded function. So it's a constant. Now it's easy to deduce that $f$ is also constant.
