# Let $f:\mathbb{C} \rightarrow \mathbb{C}$ be entire and $\exists M \in\mathbb{R}:$Re$(f(z))\geq M$ $\forall z\in\mathbb{C}$. Prove $f(z)=$constant [duplicate]

Possible Duplicate:
Liouville's theorem problem

Let $f:\mathbb{C} \rightarrow \mathbb{C}$ be entire and suppose $\exists M \in\mathbb{R}:$Re$(f(z))\geq M$ $\forall z\in\mathbb{C}$. How would you prove the function is constant?

I am approaching it by attempting to show it is bounded then by applying Liouville's Theorem. But have not made any notable results yet, any help would be greatly appreciated!

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## marked as duplicate by lhf, Freeman, t.b., Zhen Lin, Zev ChonolesOct 29 '11 at 18:53

You should consider the function's behavior at $\infty$. – Mark Oct 29 '11 at 17:24
@lhf You are correct! We should probably close this then. – Freeman Oct 29 '11 at 17:32
@LHS, you as owner can just delete the question, can't you? – lhf Oct 29 '11 at 17:34
@lhf unfortunately it appears it needs a moderator – Freeman Oct 29 '11 at 17:35
@LHS, I've flagged the question to get a moderator's attention. – lhf Oct 29 '11 at 17:40

Consider the function $\displaystyle g(z)=e^{-f(z)}$. Note then that $\displaystyle |g(z)|=e^{-\text{Re}(f(z))}\leqslant \frac{1}{e^M}$. Since $g(z)$ is entire we may conclude that it is constant (by Liouville's theorem). Thus, $f$ must be constant.
I think "Consider the function $\displaystyle g(z)=e^{-ff(z)}$" should be replaced by "Consider the function $\displaystyle g(z)=e^{-f(z)}$." – learner May 27 '13 at 5:16
Think about what the image domain of the function will look like. Can you postcompose $f$ with a simple holomorphic function on this domain (e.g. a Möbius transformation) such that the new function $g$ you obtain is bounded?