TITLE: An entire function whose real part is bounded must be constant.
QUESTION [28 upvotes]: Greets
This is exercise 15.d chapter 3 of Stein & Shakarchi's "Complex Analysis", they hint: "Use the maximum modulus principle", but I didn't see how to do the exercise with this hint rightaway, instead I knew how to do it with the Casorati-Weiestrass Theorem, here is my answer:
Define $g(z)=f(1/z)$ for $z\neq{0}$,then by the hypothesis we must have that for any $\epsilon>0$ $g(D_{\epsilon>0}(0)-\{0\})$ is not dense in $\mathbb{C}$, then the singularity at $0$ of $g$ is not essential, this implies $f$ must be a polynomial, but if $f$ is a non-constant polynomial, it is easy to see that its real part must be unbounded, so $f$ must be constant.
I would like to know an answer with the the maximum modulus principle.
Thanks

REPLY [14 votes]: Let $f(z) $ be an entire  function with real part bounded,then
$f(z)$ is entire$\implies \phi(z)=e^{f(z)}$ entire
$\implies \vert\phi(z)\vert=\vert e^{f(z)}\vert= e^{Re(f(z))} \le e^M$ ,Where $Re(z) \le M$ for some fixed $M \in \mathbb R$
$\implies \phi(z)$  is constant,by Liouville's theorem
$\implies\phi'(z)=e^{f(z)}f'(z)=0\forall z\in \mathbb C$
$\implies f'(z)=0 $ in $\mathbb C$,since $e^{f(z)}\neq 0$ in $\mathbb C$ 
$\implies f(z)$ is constant