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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.

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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
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).

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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.

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