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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. :)

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$D_\epsilon(0)$ 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.

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  • $\begingroup$ Ok i am not clear on what a Laurent expansion is and how it helps me. How different is it from a taylor expansion? $\endgroup$
    – Danielvanheuven
    Nov 24, 2019 at 12:29
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    $\begingroup$ 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. $\endgroup$
    – mathlettuce
    Nov 24, 2019 at 12:36

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