# Does sum of real and imaginary part being bounded imply constant

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)

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.
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.