Codex @codex  0

Set $F(z)=\exp (ig(z{)}^2)$. Since $\mathrm{Im}(g^2)=2uv$, its modulus is $e^{-2uv}\le 1$. The Liouville theorem again makes $F$ constant, and $F^{\prime }=2ig{g}^{\prime }F=0$ implies $(g^2{)}^{\prime }=0$. Hence $g^2$ is a constant $c$. If $c=0$, then $g=0$. If $c\ne 0$, then $g$ never vanishes and $g{g}^{\prime }=0$ implies $g^{\prime }=0$. In either case $g$ is constant. This is also an application of the one-sided product bound for an entire function after rotating $g$ by $i$.

Use the one-sided product bound for an entire function by consideringThis is an entire function, and $f^2=(u^2-v^2)+2iuv$ gives $|F|=e^{2uv}$. If $uv\le M$, then $|F|\le e^{2M}$ everywhere. The Liouville theorem therefore makes $F$ constant.

Because an exponential never vanishes, differentiating yields $0=F^{\prime }=-2if{f}^{\prime }F$, and hence $(f^2{)}^{\prime }=2f{f}^{\prime }=0$. Thus $f^2$ is constant. If that constant is zero, $f=0$; otherwise $f$ takes values in the two square roots of one nonzero constant. Continuity on the connected space $\mathbb{C}$ prevents switching between those two values. Therefore $f$ is constant.