One-sided product bound for an entire function

If an entire function $f=u+iv$ has $uv$ bounded above, then $f$ is constant. Indeed $F=\exp (-i{f}^2)$ is entire and $|F|=\exp (2uv)$ is bounded, so the Liouville theorem makes $F$ constant. Differentiating gives $f{f}^{\prime }=0$. The derivative of $f^2$ is therefore zero, and continuity on the connected complex plane makes $f$ itself constant. No separate boundedness of $u$ or $v$ is required.