Solution
= Solution
Set $F(z)=\exp(ig(z)^2)$. Since $\operatorname{Im}(g^2)=2uv$, its modulus is $e^{-2uv}\le1$. The <Liouville theorem> again makes $F$ constant, and $F'=2igg'F=0$ implies $(g^2)'=0$. Hence $g^2$ is a constant $c$. If $c=0$, then $g=0$. If $c\ne0$, then $g$ never vanishes and $gg'=0$ implies $g'=0$. In either case \b[$g$ is constant]. This is also an application of the <one-sided product bound for an entire function> after rotating $g$ by $i$.