Solution: Consider the function \[ g(z) = e^{f(z)}. \] Note that \begin{align*} \vert g(z) \vert & = \left\vert e^{f(z)} \right\vert = \left\vert e^{u + \iota v} \right\vert = \left\vert e^u e^{\iota v} \right\vert = \left\vert e^u \right\vert . \end{align*}S Since $u$ is bounded, implies $e^u$ is bounded and consequently, $g$ is bounded. As $f$ is an entire function, so is $e^{f(z)}$. Now using Liouville's theorem, we conclude that $g(z)$ is constant and hence $f(z)$ is constant.