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I need to show, given $g : \mathbb{C} \rightarrow \mathbb{C}$ such that $g$ has an essential singularity at $0$, that for each $\delta > 0$, $\exists\; z \in \mathbb{C}$ such that $|z| < \delta$ and $|g(z)| = 1$. The question specifies that I use the Casorati-Weierstrass Theorem (and more importantly, not Picard's Great Theorem, which I think would make this trivial).

My understanding of Casorati-Weierstrass is that it says $g(U \setminus \lbrace 0 \rbrace)$ is dense in $\mathbb{C}$, for any neighborhood $U$ of $0$; that is, every point in $\mathbb{C}$ is either in this or is the limit of some sequence of points in the image. So in particular, any $w$ on the unit circle is either in the image (so we're done) or it is the limit of some points in the image. But I don't see why this cannot be the case. Why can we not have sequences of points that simply approach each point on the unit circle but never actually intersect it?

Any help would be appreciated here, with reference to my understanding above if relevant or not if not. Thanks in advance.

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    $\begingroup$ If you take $U$ as a connected neighbourhood, then $U\setminus \{0\}$ is connected, and hence $g(U\setminus \{ 0\})$ is connected. $\endgroup$ – Daniel Fischer May 10 '17 at 10:09
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As per Daniel Fischer's comment above, letting $U = D(0,\varepsilon)$ be a region (connected neighborhood) containing $0$, we then have that $G = U \setminus \lbrace 0 \rbrace$ is also connected, and as $g$ is holomorphic on $U$ (by definition of a singularity), it is continuous there and so $G$ is also connected. We have by the Casorati-Weierstrass Theorem that either $G$ contains $C(0,1)$ or you can approach $C(0,1)$ using points from it. Thus $G$ can be expressed as the union of two open sets, one "inside" $C(0,1)$ and one "outside" it, which are disjoint, violating the fact that $G$ is connected. Thus $C(0,1) \subset G$; in particular, $\exists\; z \in \mathbb{C}$ such that $|z| < \varepsilon$ and $|g(z)| = 1$.

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