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I believe $e^{z - \frac{1}{z}}$ has essential singularities at $z = 0$ and $z = \infty$ (in both cases because of a $\frac{1}{z}$ in the exponent) but I'm having a hard time proving this. How can one show this?

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@JulienGodawatta the definition for singularities at infinity I'm using is the one given on – HJS Oct 2 '12 at 23:08
Forgive me, I've never seen this before. I'll delete my comment. – user39572 Oct 2 '12 at 23:10
No problem, any ideas on how to go about this? – HJS Oct 2 '12 at 23:12
@Cocopuffs That's what I'd like to do but it seems ugly. I wrote $\sum \frac{1}{k!}(z-1/z)^k$ then expanded out the $(z-1/z)^k$ term using the binomial theorem, but that's a big mess. – HJS Oct 2 '12 at 23:14

Well, it's clear that we've got singularities in those places, and that they're isolated (I leave confirmation to you). In any punctured neighborhood of $0$, we have $e^{z-\frac1z}=\cfrac{e^z}{e^{\frac1z}}$. Since $e^{1/z}$ has an essential singularity at $z=0$, then $1/e^{1/z}$ can't have a pole there (for then the singularity of $e^{1/z}$ at $z=0$ would be removable). Also, it can't approach zero there (for then $e^{1/z}$ would blow up, meaning its singularity at $z=0$ would be a pole), nor can it approach any non-zero value in the limit (for then the singularity of $e^{1/z}$ at $z=0$ would again be removable). Therefore, $1/e^{1/z}$ has an essential singularity at $z=0$, and since $e^z$ is analytic and non-zero at and about $z=0$, then $e^{z-\frac1z}=\cfrac{e^z}{e^{\frac1z}}$ has an essential singularity there, too.

Now, take $z\mapsto\frac1z$, so that the result is simply the reciprocal of the original function. Since the original function has an essential singularity at $z=0$, then (by arguing similarly to the above) so does the resulting function. That's equivalent to the original function having an essential singularity at $\infty$, so we're done.

Did I use any results you've not seen before?

Let me rephrase all of my statements in terms of an arbitrary function $g$ with certain properties. In particular, suppose $g$ has an isolated singularity (so it's essential, removable, or a pole, yes?) at $z=a$, and that $g$ is analytic and non-zero in the punctured disk $$P(a;r):=\{z\in\Bbb C:0<|z-a|<r\}$$ for some sufficiently small $r>0$. Hopefully, it's clear that in such a case, $1/g$ is is also defined and analytic in $P(a;r)$. Now, I claim that the following statements all hold:

(1) There is some non-zero $c\in\Bbb C$ such that $c=\lim_{z\to a}g(z)$ if and only if there is some non-zero $d\in\Bbb C$ such that $d=\lim_{z\to a}(1/g)(z)$. That is, one of $g$, $1/g$ has a removable singularity that can be repaired with a non-zero function value if and only if the other does, too.

(2) If $0=\lim_{z\to a}g(z)$, then $g$ has a removable singularity at $z=a$ and $1/g$ has a pole there. Furthermore, if $m$ is the order of the pole of $1/g$ at $z=a$, then the "repaired version" of $g$ has $z=a$ as a zero of order $m$.

(3) If $\lim_{z\to a}|g(z)|=\infty$, then $g$ has a pole at $z=a$ and $1/g$ has a zero there. As in (2), the orders match up. In fact, this is just the converse of (2).

(4) $g$ has a pole (of order $m$) at $z=a$ if and only if $1/g$ can be repaired to a function with a zero (of order $m$) there, and vice versa. Note that this just puts (2) and (3) together into one result.

(5) $g$ has an essential singularity at $z=a$ if and only if $1/g$ has an essential singularity there, too.

It's a good exercise to prove these general statements. (1) should be simple to prove, (2) and (3) shouldn't be too tricky, and (4) is pretty much immediate from (2) and (3). Using (1) and (4)--together with the fact that every isolated singularity is necessarily (a) essential, (b) removable, or (c) a pole--it should be fairly straightforward to prove (5). (Can you see how?)

Now, you should be able to see that the function $g(z)=e^{z-\frac1z}$ satisfies all our hypotheses (with $a=0$), as does $g(1/z)=(1/g)(z)$. Thus, ruling out the possibility of a removable singularity or pole by using (1) and (4), it follows that we must be dealing with an essential singularity in both cases. Since $g(z)$ has an essential singularity at $z=0$, then by (5), so does $g(1/z)=(1/g)(z)$, meaning by definition that $g(z)$ has an essential singularity at $\infty$.

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Intuitively I understand, but how do I make a statement like "$e^{1/z}$ has an essential singularity at $0$ and so $1/e^{1/z}$ cannot have a pole there since otherwise the singularity of $e^{1/z}$ at $0$ would be removable" rigorous? – HJS Oct 2 '12 at 23:20
I've added on more precise versions of the claims I made, generalized to other functions. Hopefully, that helps see how you can get the rigor you want, but let me know if you've got any other questions. – Cameron Buie Oct 3 '12 at 0:52

Denote by $f(z) := e^z e^{- \tfrac{1}{z}}$. For $z_n = \tfrac{1}{n}i$ you see $f(z_n)$ stays within the range $\lvert f(z_n) \rvert = 1$, a compact set. But $\lvert f(\tfrac{1}{n}) \rvert \to 0$. Therefore, since $z_n \to 0$, you can conclude that $f$ has neither a removable singularity in $0$ nor a pole.

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Sorry, there is a mistake in that. I'll correct it soon. – k.stm Oct 2 '12 at 23:25
So, if I have it right, you've shown that there is no way we can define a limit as $z\rightarrow 0$ for the function, so there is no way we can continuously extend it, I see how this shows that it is not removable, but not how it must not be a pole. – HJS Oct 2 '12 at 23:27
If $0$ was a pole, then $\lvert f(z_n) \rvert \to \infty$ whenever $z_n \to 0$. Depending on your definition of pole, this is easily seen. – k.stm Oct 2 '12 at 23:29
It seems simple, could you give me a quick argument for that? (My definition of pole would be finitely many but non-zero terms in the Laurent series with negative exponent). With it then here's my argument, there must be a singularity at $0$ because $|f(1/n)| \rightarrow \infty$, it cannot be a pole because $|f(z_n)|$ remains bounded, and so is an essential singularity? Is that right? – HJS Oct 2 '12 at 23:36
Write $f(z) = z^{-k}g(z)$ with $g$ holomorphic in some compact neighbourhood $K \ni 0$ without a zero of $g$ in it, then $g$ is bounded in $K$ and $\lvert g(z) \rvert > \varepsilon > 0$, so $\lvert f(z) \rvert = \lvert z^{-k} \rvert \lvert g(z)\rvert > \varepsilon \lvert z^{-k} \rvert$ and the latter term diverges for $k \geq 1$. – k.stm Oct 2 '12 at 23:42

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