I am working on following Question;
f is analytic on $D(0,1)\setminus\{0\}$ with an essential singularity at 0, show f is not one-to-one.
I am a bit stuck. All I have been able to figure out is the following;
- $\lim_{z \rightarrow 0} f(z)$ does not exist since the singularity is not removable.
- $f(D(0,1)\setminus\{0\})$ is open by the Open Mapping Theorem
- $f(D(0,1)\setminus\{0\})$ is dense in $\mathbb{C}$, by Casorati-Weierstrass.
Now I suspect that we should try the proof by contradiction route and suppose that $f$ is one-to-one and show that then the singularity is maybe not essential, potentially removable. Or maybe we should construct two sequence converging to different limits while the images of the sequences converge to the same thing.
Cheers for the help in advance; If possible please do not give the whole answer but rather give hints and when I feel I have a solution I will post it as an answer (Of course I will accept the most useful hint as the answer to this question)
Some useful ideas are:
As you mentioned, the Cassorati-Weierstrass theorem states that $f(D(0,1)\backslash\{0\})$ is dense in $\mathbb{C},$ but also that the same holds for any disk $D(0,\varepsilon)\backslash\{0\}.$
If $f$ is one-to-one, then, for some $a \in D(0,1)\backslash\{0\}$, if we consider the pre-images $f^{-1}(D(f(a),\delta)),$ they should be, by the (holomorphic) inverse function theorem, contained in a tiny neighborhood $D(a,\eta)$ of $a.$
By 'shrinking' the original disk to something like $D(0,\varepsilon)\backslash\{0\},$ for $\varepsilon$ small, what does the 'lack of a disk' in the image of $f$ say, in conjunction with the first point presented?