Proving Entire Function f(z) is Constant | Complex Analysis Proof

[nickolas2730]

Q:Let f be entire and suppose that I am f(z) ≥ M for all z. Prove that f must be a constant function.

A: i suppose M is a constant. So I am f(z) is a constant which means the function is a constant.

Am i doing this right ?
but i don't think there will be such a stupid question in my assignment ...
Or M is not even a constant? M can be anything?

THanks

 

[micromass]

M is a constant alright. But why is I am f(z) a constant?? It depends on z, so I don't find it trivial that it is a constant...

Mod note: I moved the thread to calculus and beyond.

 

[jackmell]

Q:Let f be entire and suppose that I am f(z) ≥ M for all z. Prove that f must be a constant function.

A: i suppose M is a constant. So I am f(z) is a constant which means the function is a constant.

Am i doing this right ?
but i don't think there will be such a stupid question in my assignment ...
Or M is not even a constant? M can be anything?

THanks

Suppose I have the function

$$f(z)=e^{ig(z)}$$

and I know that $|f|<100$ everywhere. Then that must mean f is a constant right? But for $e^{ig}$ to be a constant, that must mean g(z) is a constant. So, suppose I have the expression:

$$e^{ig}=e^{i(u+iv)}$$

and $|e^{ig}|$ is always bounded, say less than 100. Then what must be the constraints on the function u+iv for that to work?

 

[nickolas2730]

am i doing this prove by liouville's theorem?
can i do it like this:
suppose f = 1/e^f(z)=1/e^ Im(f) < 1/e^M
so it is bounded
by liouville's theorem, it is constant