 Solution

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