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What type of singularity is that at $z=\pi k+\pi/2$ for any integer $k$ for the function $\phi(z)=e^{\tan z}$? I can see that it is not removable, but I am not sure how to narrow down further. Thank you.

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    It's a bit hard to give the proper hints without giving the whole answer away, especially not knowing what you already know about the classification of singularities. Briefly, though: What do you already know about the singularities of $\tan z$ at these points, and are you familiar with the singularity of $e^{-1/z}$ at $z=0$?2012-02-19
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    @HaraldHanche-Olsen: I am guessing that they are simple poles (so order 1)? But I am not familiar with that of $e^{-1/z}$ at $z=0$. Though if I had to guess, I think it is essential?2012-02-19
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    You're right on both counts. If $e^{-1/z}$ had a pole at the origin, there should be some natural number $n$ so that $z^ne^{-1/z}$ had a removable singularity there. Can you see how that cannot possibly be true? Does it help with your original question?2012-02-19
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    @HaraldHanche-Olsen: Because the Taylor series of $e^{-1/z}$ has terms in $z^n$ for all natural numbers $n$?2012-02-19
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    Yes, that is one way to see it. Another is to note that $z^ne^{-1/z}$ is unbounded in every neighbourhood of the origin.2012-02-20
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    @HaraldHanche-Olsen: So am I supposed to argue that there isn't some $n$ such that $(z-\pi k+\pi/2)^n\exp(\tan z)$ has a removable singularity hence it is essential? Or am I missing something?2012-02-20
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    That is one way. Here is another: If $f(z)$ has a pole at $w$ then $1/f(z)$ is bounded in a neighbourhood of $w$. (If you haven't seen this one before, ask yourself why?)2012-02-20

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