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The integral I want to evaluate is the integral of $\displaystyle\frac{z^n}{e^z-1}$ over the circle $|z| = 3 \pi $. I was thinking of using Cauchy's Residue Theorem, but then I got stuck on evaluating the residues at $z=0$, because I couldn't figure out how to change the form into one where I could find a Laurent series.

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    One formula you should be remembering for your test (it's probably the most useful one for calculating residues) is that $\operatorname{Res}_{z_{0}} \frac{f}{g} = \frac{f(z_0)}{g'(z_{0})}$ if $f(z_{0}) \neq 0$ and $g(z_{0}) = 0$ and $g'(z_{0}) \neq 0$. (i.e. $z_{0}$ is a simple zero of $g$). You can find other useful ones on the Wikipedia page on the [residue theorem](http://en.wikipedia.org/wiki/Residue_(complex_analysis)).2011-05-13

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OK, this is an attempt to rewrite my answer without mentioning L'Hôpital or Riemann, as requested.

The integrand $f(z) = \frac{z^n}{\mathrm e^z-1}$ is the quotient of two holomorphic functions, each of which we can write as a power series at any point $z_0$. For $z_0=0$, this yields

$$f(z)=\frac{z^n}{\left(\sum_{k=0}^\infty z^k/k!\right)-1}=\frac{z^n}{\sum_{k=1}^\infty z^k/k!}=\frac{z^n}{z\sum_{k=1}^\infty z^{k-1}/k!}=\frac{z^n}{z\left(1+\sum_{k=2}^\infty z^{k-1}/k!\right)}\;.$$

If $n$ is a positive integer, we can cancel one factor $z$ to obtain

$$f(z)=\frac{z^{n-1}}{1+\sum_{k=2}^\infty z^{k-1}/k!}\;.$$

The denominator doesn't go to zero as $z\to0$, and thus there is no pole at $z=0$.

On the other hand, if $n=0$, we have

$$f(z)=\frac{1}{z\left(1+\sum_{k=2}^\infty z^{k-1}/k!\right)}\;.$$

Thus we can determine

$$\lim_{z\to0}zf(z) = \lim_{z\to0}\frac{1}{\left(1+\sum_{k=2}^\infty z^{k-1}/k!\right)}=1\;,$$

and this is the residue of $f$ at $z=0$.

The residues at $z=\pm2\pi\mathrm i$ can be determined similarly using $\mathrm e^{z}=\mathrm e^{z\mp2\pi\mathrm i}$.

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    I wouldn't call this l'Hôpital but rather [Riemann's theorem on removable singularities](http://en.wikipedia.org/wiki/Removable_singularity) (Riemannscher Hebbarkeitssatz), but the conclusion remains the same, of course.2011-05-12
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    @joriki Two other poles enter the picture, at $\pm2\pi\mathrm{i}$. Their contributions to the residue formula add up when $n$ is even and cancel each other when $n$ is odd.2011-05-12
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    @Didier: Yes, that's what my last sentence refers to. But the question was about the residue at $z=0$.2011-05-12
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    @joriki Hmmm, you are right, twice. Sorry, shouldn't be awake at this hour.2011-05-12
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    @Theo: I'm not sure I understand your comment. I was taking as given that the ratio of two holomorphic functions is meromorphic and is thus either holomorphic at $z=0$ or has a pole there. I was suggesting to use L'Hôpital to show that the limit is finite and hence there can't be a pole; I didn't spell out the conclusion that the function must then be holomorphic because that's the only other option for a meromorphic function. Am I wrong that in this way of looking at it I don't need Riemann's theorem?2011-05-12
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    I don't follow your explanation :(. We haven't covered how to use either L'Hopital's or Riemann's Theorem on removable singularities, though the question is on a practice final.2011-05-12
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    @I Love Cake: OK, I'll try to rewrite it.2011-05-12
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    @I Love Cake: I think the point of joriki's argument is that you can notice that $\frac{1}{e^z-1}$ has a pole of order 1 at $0, \pm 2\pi i$ (one of seeing this is by writing out the taylor expansion for $e^z$) . Like Didier said in the comments, the poles at $\pm 2\pi i$ cancel out with one another. Now, if $n \geq 1$, then the zeros of $z^n$ will cancel out with the pole at zero. If not, then you can calculate the residue of the simple pole by using the limit that he prescribed.2011-05-13
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    @user1736, I think I understand a little better. But how do you calculate a residue of a simple pole using a limit? Am I missing some essential theorem?2011-05-13
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    When you're saying "because that's the only other option for a meromorphic function", I think you are implicitly using Riemann's theorem (I don't know how to prove this without it).2011-05-13
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    I just don't see how exactly you're applying l'Hôpital because this for me is a result on functions of a real variable (isn't it?). The basic thing is of course that $e^{z} - 1$ has a simple zero at $0$, so you *know* that you can write $\frac{1}{e^{z}-1} = \frac{a_{-1}}{z} + \sum_{n \geq 0} a_{n} z^{n}$. Now to determine $a_{-1}$ you're looking at the limit $a_{-1} = \lim_{z \to 0} \frac{z}{e^{z}-1} = \lim_{z\to 0}\frac{1}{\frac{e^{z}-1}{z}}$. This *looks like* l'Hôspital but I don't see a precise relation. Now for $n \gt 0$ Riemann's theorem tells you that the singularity is removable.2011-05-13
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    @I Love Cake: Do you know about Laurent series? If $f$ only has a simple pole at $z=a$, then the only negative power in the Laurent series centered at $a$ is the -1 power. This means that we can write $f(z) = c_{-1} \frac{1}{(z-a)} + \sum_{n=0}^\infty c_n (z-a)^n$. The residue of $f$ at $a$ is going to be the coefficient $c_{-1}$, so one way of recovering this is by multiplying $f(z)$ by $z-a$ and taking the limit as $z \to a$, since the nonnegative powers in the sum will vanish. Does that make sense?2011-05-13
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    @Theo: I think you can actually derive L'hopital's rule for analytic functions. If $f$ has a zero of order $m$ and $g$ has a zero of order $n$, one way of finding the limit of the quotient is to factor out the zeros and see what cancels out. Alternatively, you can also differentiate $f$ and $g$ that many times directly, so that you eventually end up with the same expression. I guess you can make the argument that this means that you don't really ever need l'hopital's rule, but it should still be valid.2011-05-13
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    @user: yes, I agree with that (but that's essentially the proof of the Riemann theorem, and that is my whole point!), the formula is valid (and *much easier* to prove than de l'Hôpital - which is rather subtle). By the way, I didn't see your comments before I submitted mine, I didn't mean to imply yours were insufficient :) @joriki: sorry for all the pings it really is a quibble thus isn't worth making such a fuzz about.2011-05-13
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    @Theo: It's cool. Your comment actually appears above mine, so I guess I should actually be the one apologizing.2011-05-13
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    @user: Ok, very good. Actually, I think you should elaborate your comments into an answer because this very much looks like what I Love Cake is looking for (I'm too tired now and I've written too many answers, recently)2011-05-13
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    Ok, so now I think we're all happy, I guess :)2011-05-13
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    @Theo: Thanks for explaining about L'Hôpital -- I wasn't aware that it's not valid for arbitrary complex functions; it seems I had some oversimplified notion of it following directly from the fact that the derivative is a linear approximation to the function, which apparently it doesn't -- I'll have to think about that :-)2011-05-13
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    @Theo: I think I see now what I was confused about -- I didn't distinguish between the case where $f$ and $g$ (in $f/g$) are differentiable at $z_0$ and the more general case where the limit of $f'/g'$ exists, but the individual derivatives don't necessarily exist at $z_0$. So is it correct that in the first case, where $f'$ and $g'$ exist at $z_0$, the rule follows directly from the linear approximation character of the derivative and carries over to the complex case without complications?2011-05-13
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    Thanks everyone. I think I understand now.2011-05-13
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    @joriki: Yes, this is exactly what I was trying to point out (not very clearly, as I see now): The existence of the limit can be established exactly as you outlined (and is a simple and weaker version of de l'Hôpital). Note also that you're using crucially that the singularity at $0$ can be removed because this is one of the hypotheses of the residue theorem. Anyway, I think we agree now.2011-05-13