non-constant entire functions (Liouville) [closed]

I have to show by using Liouville's theorem, whether there are non-constant entire functions such that: $$f( \mathbb{C})$$ is in an half-plane. I found out, that this is not possible for non-constant functions, because this property must hold: $$\forall \epsilon>0 \ \forall w \in \mathbb{C} \ \exists \ \mathbb{C} : |f(z)-w|< \epsilon$$

Am I right?

closed as unclear what you're asking by Nosrati, Michael Hoppe, Arnaud D., Chris Custer, DidDec 12 '18 at 21:01

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What theorem are you using to say that the range must be dense? The present question is an elementary application of Louiville's Theorem. Suppose your half plane is the upper half plane. Just consider $$g(z)=e^{if(z)}$$ and note that $$|g(z)|=e^{- Im (f(z))} <1$$ for all $$z$$. Hence, by Louiville's Theorem $$g$$ is a constant from which it follows that $$f$$ is also a constant. [ For this note that $$\frac d {dz} e^{if(z)}=0$$ implies $$f'(z)=0$$ for all $$z$$ and hence $$f$$ is a constant]. This contradicts the hypothesis. For the lower half plane consider $$e^{-if(z)}$$, for the right half plane consider $$e^{-f(z)}$$ and for the left half plane consider $$e^{f(z)}$$. Conclusion: you cannot have a non-constant entire function whose range is a half plane.
• I dont get your argument, You define $g(z)= e^{i f(z)}$, but this is a special function. How can you conclude that this works for all functions? – SvenMath Dec 12 '18 at 8:06
• @SvenMath Assume that there is a non-constant entire function whose range is the upper half plane and get a contradiction. For this purpose I apply Louville's Theorem to $e^{if(z)}$ not to $f$ itself. – Kavi Rama Murthy Dec 12 '18 at 8:09