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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? –  Greg D Feb 20 '12 at 19:05
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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. –  WimC Feb 20 '12 at 19:06
    
@WimC: Thanks, Wim! Is there a name for such singularities? (Maybe one of removable or essential?) –  Greg D Feb 20 '12 at 19:08
    
see branch point on wikipedia. –  WimC Feb 20 '12 at 19:12
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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. –  Robert Israel Feb 20 '12 at 20:29

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