# complex proof with weird domain term

Consider the function $$f(z)=e^{1/z}$$ defined in $$\mathbb{C}$$ \ {$$0$$}. Show that for any $$\epsilon > 0$$ and any $$w \in \mathbb{C}$$ \ {$$0$$} there is some $$z \in D_\epsilon (0)$$ \ {$$0$$} such that $$f(z) = w$$.

This is the problem i have. I am not a math student but i have to follow this math course (i am an astronomer). So it might be a dumb question. My question is what does $$D_\epsilon(0)$$ mean especially the zero because D is a domain right? So what does the $$0$$ mean? Is it a domain with only zeros or something? That is the main question but ofcourse an answer to the full thing would also be fine. :)

$$D_\epsilon(0)$$ is the disk with center $$0$$ and radius $$\epsilon$$, i.e. $$D_\epsilon(0) = \{x \in \mathbb{C} | |x| < \epsilon\}$$.
A domain is a connected open subset of a finite dimensional vector space, so indeed $$D_\epsilon(0)$$ is a domain.
Hint for your exercise: Consider the Laurent expansion of $$f$$ at $$0$$, i.e $$f(h) = \sum_{n \in Z} a_n h^n$$. Show that there are infinitely many non-zero $$a_n$$'s with $$n<0$$. Then you can conclude using the Cassorati-Weierstrass - Theorem.
• It can also terms with negative powers, i.e. a Taylor series is of the form $/sum_{n \in \mathbb{N}} a_n x^n$ and a Laurent series is of the form $/sum_{n \in \mathbb{Z}} a_n x^n$. In fact, Laurent series are just a generalization of Taylor series. Nov 24, 2019 at 12:36