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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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    oh! I forgot about the open mapping theorem, that was my blind spot2012-08-05

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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)| 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.