Does sum of real and imaginary part being bounded imply constant

Question

Let f be a entire function with sum of real and imaginary parts bounded. Is f constant? (I know Liouville's theorem but can't apply in this situation)

Answer 1

Yes, $f$ is constant.

With Liouville’s theorem, you can prove that if $f$ is not constant, $f(\mathbb{C})$ is dense in $\mathbb{C}$. Suppose at the contrary that there exists $c \in \mathbb{C}$, such that there’s no sequence in $ f(\mathbb{C})$ converging to $c$. Then, there exists $\epsilon \in \mathbb{R}$, such that $B(c, \epsilon) \bigcap f(\mathbb{C}) = \emptyset$. But then the entire function $g(z) = \frac{1}{f(z) - c}$ is bounded, which is not possible by Liouville’s theorem. Therefore, for all $n \in \mathbb{N}$, there’s a sequence in $f(\mathbb{C})$ converging to $n + i*n$.. Thus, the sum of the real and the imaginary parts cannot be bounded. In fact, we can generalize this and state that for any continuous unbouded function $g : \mathbb{C} \to \mathbb{C}$, and any non-constant entire function $f$, $g \circ f$ is not bounded.

Answer 2

If the sum of real part and imaginary part of $f$ is bounded above, say $$ \operatorname{Re} f(z) + \operatorname{Im} f(z) \le M $$ then $$ \operatorname{Re} \bigl( (1-i)f(z)\bigr) \le M \\ \implies | (1-i)f(z) - (M+1)| \ge 1 $$ so that $$ \frac{1}{(1-i)f(z) - (M+1)} $$ is entire and bounded. Now you can apply Liouville's theorem.