Solution
= Solution
The composition $g(z)=e^{-f(z)}$ is an <entire function>, and
$$
|g(z)|=e^{-\operatorname{Re}f(z)}\leq1.
$$
By <Liouville's theorem>, $g$ is constant. Since the <exponential function> never vanishes,
$$
0=g'(z)=-f'(z)e^{-f(z)}\quad\Longrightarrow\quad f'(z)=0.
$$
Thus \b[$f$ is constant]. Notice that constancy of an exponential alone need not be inverted by choosing a global logarithm; differentiating supplies the conclusion directly.