Solution: We consider the function \[ g(z) = e^{-f(z)},\ z \in \mathbb{C} . \] As $f$ is an entire function, so is $g$. Note that \begin{align*} \vert g(z) \vert = \left\vert e^{-f(z)} \right\vert & = \left\vert e^{-[\mathrm{Re}(f(z)) + \iota \mathrm{Im}(f(z))] } \right\vert \\ & = e^{- \mathrm{Re}(f(z)) } \leq e^{-2}. \end{align*} Now we will use Liouville's theorem to conclude that $g$ is a constant function as it is entire and bounded. Thus, $e^{-f(z)} = k$ (constant) for any $z \in \mathbb{C} $ and this proves that $f$ is a constant function.