Solution
= Solution
Set $F(z)=\exp[-(1-i)g(z)]$. It is entire, and $\operatorname{Re}[(1-i)g]=u+v$ makes $|F|=e^{-(u+v)}\le1$. By the just-proved <Liouville theorem>, $F$ is constant. Differentiation gives $0=F'=-(1-i)g'F$. The exponential never vanishes, so $g'=0$ everywhere and \b[$g$ is constant]. This is the <entire function confined to a half-plane is constant> principle.