For Which of the following conditions entire function $f$ is constant?

Question

Let $f:\mathbb{C} \rightarrow \mathbb{C}$ be an entire function Suppose that $f=u+iv$ where $u,v$ are real and imaginary parts of $f$ respectively. Then f is constant if

1. $ u(x,y):z=x+\iota y \in \mathbb{C} $ is bounded.

2. $ v(x,y):z=x+\iota y \in \mathbb{C} $ is bounded.

3. $ u(x,y)+v(x,y):z=x+\iota y \in \mathbb{C} $ is bounded.

4. $ u^2(x,y)+v^2(x,y):z=x+\iota y \in \mathbb{C} $ is bounded.

if either of real part or imaginary part of a entire function is bounded then $f$ is definitely constant.

Fourth option is also correct as if modulus of $f$ is bounded then $f$ is constant.

I am not sure about option (3) It seems it may have some counter example.

Answer 1

All four follow easily from Liouville's theorem:

Theorem If $f$ is an entire function and $f(\Bbb C)$ is not dense in the plane then $f$ is constant.

Proof: Since $f(\Bbb C)$ is not dense there exist $p\in \Bbb C$ and $r>0$ so $$f(\Bbb C)\cap D(p,r)=\emptyset.$$This says that $|f(z)-p|\ge r$ for every $z$. So the function $g=1/(f-p)$ is bounded; Liouville says $g$ is constant, and hence $f$ is constant.

Answer 2

For 3., note that $f(z) - if(z)$ is constant by 1. It follows that $f'(0) = if'(0), f''(0) = if''(0), \dots.$ Thus all derivatives of $f$ at $0$ vanish, which implies $f$ is constant.