Why can't an analytic function be injective in a neighborhood of an essential singularity? Let $D \subset \mathbb{C}$ be a domain and let $a \in D$. Suppose $f: D \smallsetminus \{a\} \to \mathbb{C}$ is analytic and that $a$ is an essential singularity of $f$. Show that $f$ cannot be univalent (= injective) in any neighborhood of $a$.
This is a trivial consequence of the Picard theorem. But I don't know if there is any elementary approach.
 A: This is a simple consequence of the Casorati–Weierstrass theorem:

If $f: D \smallsetminus \{a\} \to \mathbb{C}$ is holomorphic and $a$ is an essential singularity of $f$ then $f(D \smallsetminus \{a\})$ is dense in $\mathbb{C}$.

One proof of the Casorati–Weierstrass theorem is via Riemann's theorem on removable singularities and is given on the Wikipedia page I linked to above. It's very simple:
If $f(D \smallsetminus \{a\})$ is not dense then by definition of density we can find $w_{0} \in \mathbb{C}$ and $\varepsilon \gt 0$ such that $\{w\,:\,|w - w_0| \lt \varepsilon\} \cap f(D\smallsetminus\{a\}) = \emptyset$. But then $g(z) = \dfrac{1}{w_0 - f(z)}$ is bounded and analytic on $D \smallsetminus \{a\}$ and the straightforward Riemann theorem on removable singularities allows us to extend $g$ to an analytic function $\tilde{g}: D \to \mathbb{C}$. The function $\tilde{f}(z) = w_0 - \dfrac{1}{\tilde{g}(z)}$ coincides with $f$ on $D \smallsetminus \{a\}$ and a quick check shows that $a$ must either be a removable singularity or a pole of $f$ (depending on $\tilde{g}(a) \neq 0$ or $\tilde{g}(a) = 0$). Either way this contradicts the hypothesis that $a$ is an essential singularity of $f$.

To prove that $f$ cannot be injective on any (pointed) neighborhood $W$ of the essential singularity  $a$, let $x \neq a$ be any point of $W$. Choose $0 \lt \varepsilon \lt \frac{|x-a|}{2}$ so small that both $U = \{z\,:\,0 \lt |z-a| \lt \varepsilon\}$ and $V = \{z\,:\,0 \leq |z-x| \lt \varepsilon\}$ are contained in $W$. Observe that $U$ and $V$ are disjoint and open.


*

*By Casorati-Weierstrass, $f(U) \subset \mathbb{C}$ is dense (since $a$ is also an essential singularity of $f\,|_{U}$) and by the open mapping theorem $f(V)$ is open (since $f$ is not constant). Therefore $f(U) \cap f(V) \neq \emptyset$. 

*If $f$ were injective then $f(U) \cap f(V) = \emptyset$ by disjointness of $U$ and $V$, contradicting $1.$
This shows that  $f$ cannot be injective (univalent) on any neighborhood $W$ of the essential singularity $a$.
