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I think my problem below can be solved by finding the Laurent series of $f(z) = \ln(1+\exp(z))$ about its points of singularities. (Any better suggestion is more than welcome!) How might I find such a series?

I think that the function $f(z) = \ln(1+\exp(z))$ has poles at $z=i\pi (2n-1)$ for integer $n$ but I cannot prove that it is a pole because I cannot find an $m$ such that $(z-i\pi (2n-1))^mf(z)$ is finite. Could someone please help me out?

Added: For this question we are taking the principal branch of the function.

By the way, the available options are

A) removable singularities, (Don't think so...)

B) poles, (Possible, but I can't find a suitable $m$, as mentioned)

C) essential singularities, (Possible, but I don't know how to show this either...)

D) non-isolated singularities (Not this one)

Some thoughts: I have thought of maybe obtaining the Laurent series about these $z$ values, but I don't know how to do that...

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    Or perhaps it is not a pole?2012-02-20
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    It has singularities there but not poles. Note that log has a singularity at 0 that is also not a pole, but a branch point. Your function f is therefore also multi-valued with branch points as indicated.2012-02-20
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    @WimC: Thanks, Wim! Is there a name for such singularities? (Maybe one of removable or essential?)2012-02-20
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    see [branch point](http://en.wikipedia.org/wiki/Branch_point) on wikipedia.2012-02-20
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    @WimC: Thanks, I have read this article before and I know what a branch point is. However, I am wondering what type of singularity the aforementioned ones belong to... I am sorry if I have overlooked something, but I think I need a bit more help?2012-02-20
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    @WimC: The types I can choose from are 1) removable singularities, 2) poles 3)essential singularities 4) non-isolated singularities2012-02-20
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    Then you'd have to go with non-isolated singularities. Hint: What do you mean when you write $\ln$ applied to a complex number? The principal branch? Where is this function defined?2012-02-20
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    @mrf: Thanks for replying, yes, I am taking the principal branch. In light of this new info, does the type of singularity change?2012-02-20
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    Where there is a branch point, there must also be a branch cut. That means the singularity is not isolated. Writing $z - (2n-1) i \pi = w$, $1 + \exp(z) = - w (1 + w/2 + w^2/6 + \ldots)$ so $\ln(1 + \exp(z)) = \ln(-w) + \ln(1 + w/2 + w^2/6 + \ldots) = \ln(-w) + w/2 + w^2/24 + \ldots$. It's not a Laurent series because of the $\ln(-w)$ term.2012-02-20
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    @RobertIsrael: Thank you! This is what I got but I thought I did it wrong because all the examples I have seen don't have this horrible constant term. Just for a quick clarification: I did a quick search on non-isolated singularities (mainly Wiki). Is this kind of singularity more specifically called a "natural boundary not isolated singularity"?2012-02-20
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    It's a branch cut. It's a branch cut. It's a branch cut. It's a branch cut. It is not a "natural boundary". It's a branch cut.2012-02-20
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    @GerryMyerson: ok, thanks, it's a branch cut :)2012-02-20

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