Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $g$ be nonconstant and entire, give the relation between the sets $\operatorname{cl} \{ z \in \mathbb{C} : |g(z)| > 1 \}$ and $\{ z \in \mathbb{C} : |g(z)| = 1 \}$.

I'm probably missing something big here as all I got is that when we replace $g$ by an infinitely differentiable $f:\mathbb{R} \rightarrow \mathbb{R}$, the latter is not a subset of the former.

share|improve this question
3  
Try using the open mapping theorem. –  Julian Rosen Aug 4 '12 at 13:51
    
I take it that cl here stands for closure? (That is not very standard notation. An overbar is more common, but is a bit ambiguous in complex analysis, of course.) –  Harald Hanche-Olsen Aug 4 '12 at 14:11
    
@HaraldHanche-Olsen: I've seen it and used it outside complex analysis (including general topology), it seems pretty standard to me. It might have to do with the fact that I've been dabbling in logic recently, but an overbar usually brings in mind first a tuple, then conjugation and only then closure (and then a line). –  tomasz Aug 4 '12 at 20:19
    
oh! I forgot about the open mapping theorem, that was my blind spot –  lucas Aug 5 '12 at 0:13
add comment

2 Answers

up vote 1 down vote accepted

We have $$\{z\in\Bbb C:|g(z)|=1\}\subset\overline{\{z\in\Bbb C: |g(z)|>1\}}.$$ Indeed, let $z_0$ such that $|g(z_0)|=1$. Then $g(z_0)\in g(\Bbb C)$, which is open, by the open mapping theorem. Hence we can find $r$ such that if $|\omega-g(z_0)|<r$ then $\omega\in f(\Bbb C)$. In particular, with $\omega_n:=(1+2^{-n})g(z_0)$ for $n$ large enough, we can write $\omega_n=g(z_n)$, and $|g(z_n)|=1+2^{—n}>1$. Hence $g(z_0)=\lim_{n\to +\infty}|g(z_n)|$ is necessarily in $\overline{\{z\in\Bbb C: |g(z)|>1\}}$.

Note that the inclusion is in general strict, as the choice $g(z)=z$ shows (it also helps to have an intuition of the problem).

share|improve this answer
add comment

let $z_0$ be a point of $\mathbb{C}$ such that $|g(z_0)|=1$. Let $U$ be a neighborhood of $z_0$. Then use the mean-value property of holomorphic function to show that $U\cap \{z\in\mathbb{C}: |g(z)|>1\}$ is not empty.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.