An entire function whose real part is bounded must be constant.

As other posters have commented, the standard approach here would be to invoke Liouville's Theorem. One way to do this is to consider the entire function $e^{f(z)}$.

Observe that $|e^{f(z)}| = e^{\Re f(z)}$, which is bounded by our assumption on $\Re f(z)$.

Then $e^{f(z)}$ is an entire bounded function, and hence (by Liouville's Theorem) constant.

From this, we conclude that $f(z)$ is constant as well.

A simpler way is to use Liouville's Theorem: consider $g(z) = 1/(1+b - f(z))$ where $\text{Re}(f(z)) \le b$.

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