= Solution
Use the <one-sided product bound for an entire function> by considering
$$
F(z)=\exp[-if(z)^2].
$$
This is an <entire function>, and $f^2=(u^2-v^2)+2iuv$ gives $|F|=e^{2uv}$. If $uv\leq M$, then $|F|\leq e^{2M}$ everywhere. The <Liouville theorem> therefore makes $F$ constant.
Because an exponential never vanishes, differentiating yields $0=F'=-2iff'F$, and hence $(f^2)'=2ff'=0$. 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. \b[Therefore $f$ is constant.]
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