I've seen many questions already on Laurent Expansions and how to find examples like my question, however, most of them turn to geometric series and partial fractions, which I understand can be faster but when under time pressure, sometimes I just don't see the trick they want me to use and I would like to know that I always have a backup in a definition that should always work.

However, using this definition I keep getting stuck and most of the time my answer is just not correct. This particular problem is about $f(z)=\frac{1}{z(z-1)}$ where we want to express $f$ as a Laurent series around $0$ for $|z|>1$.

This is how I'd go about this problem:

We know that $f(z)=\frac{1}{z(z-1)}=\sum_{-\infty}^\infty a_nz^n$ for some coefficients $a_n$, where we use the Laurent series around $0$. Both $0$ and $1$ are simple poles, and for any $r>1$, both poles lie in the interior of de circle around $0$ of radius $r$.

We look at the open annulus of $1<r<|z|<R$ where $f$ is holomorphic. Then we know that the $a_n$ from above are given by:

$$a_n=\frac{1}{2\pi i}\int_{|z|=r_0}\frac{f(z)}{z^{n+1}}dz= \frac{1}{2\pi i}\int_{|z|=r_0}\frac{1}{z^{n+2}(z-1)}=: \frac{1}{2\pi i}\int_{|z|=r_0}g(z),$$ where $|z|=r_0$ is any circle in the open annulus ($r<r_0<R)$, on which we integrate in a positive direction.

Both poles $0$ and $1$ are inside this circle of radius $r_0$, and $g$ is holomorphic on the interior of this circle except at these poles. Therefore, we can use the residue sum theorem for $g$ to calculate the above integral, where $0$ is a pole of order $n+2$ and $1$ is a simple pole:

$$Res(g,0)=\lim_{z\to 0}\frac{d^{n+1}}{dz^{n+1}}\Big(\frac{z^{n+2}}{z^{n+2}(z-1)(n+1)!}\Big)= \frac{1}{(n+1)!}\lim_{z\to 0}\Big(\frac{(-1)^{n+1}(n+1)!}{(z-1)^{n+2}}\Big)=-1$$

$$Res(g,1) = \lim_{z\to 1}(z-1)\frac{1}{z^{n+2}(z-1)}=1$$

By the Residue Sum Theorem, the integral is equal to $2\pi i*(Res(g,0) +Res(g,1)) = 0$.

This however would lead to the $a_n$ being all 0, which is not the case for $n<-1$.

I saw the answer model which said that with a trick using geometric series you will get the answer $\frac{1}{z^2}+\frac{1}{z^3}+\frac{1}{z^4}...$. So clearly the $a_n$ are in fact $0$ for all $n>-2$, but this method does not provide me with $a_n$ for $n<-1$.

The first part of this question was about $|z|<1$ where you could use the geometric series directly; I did use this method as well when solving the first question. In that case, only the pole $0$ was inside the region of integration, and the residue, and thus the $a_n$, was equal to $-1$. This answer was correct, but now for all $n>-2$, and wrong for all $n<-1$ (as it's about a simple pole, the $a_n$ should logically be $0$ anyway for $n<-1$).

Somehow I seem to calculate the $a_n$ only for one half of the series, whereas the coefficients are incorrect for the other half. Where am I doing anything wrong in the steps above?

Edit: Finally I have maybe found what could be the problem; the calculation I used for residues uses $n+1$'th derivatives. As soon as $n<-1$, this calculation is no longer valid and the formula used for a general derivative of $1/z$ is no longer what I say it is. How to proceed now?

  • $\begingroup$ What's wrong with just writing down the geometric series? One has the expansion $\sum_{k=2}^\infty z^{-k}$, so $a_n=1$ for $n\le-2$ and $a_n=0$ for $n\ge-1$ $\endgroup$ Jun 21 '20 at 17:44
  • $\begingroup$ @AnginaSeng Please read my first paragraph; in this case I could figure it out, I just want the other described method to always work as it's not using any 'tricks'. I'm using this simple example so I can understand the concept, on an exam they might ask something a bit harder in which case I might not see how I can find an already know power series, and I would like to have a backup plan. $\endgroup$
    – Marc
    Jun 21 '20 at 17:47
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    $\begingroup$ You are belittling the geometric series by dismissing it as a "trick" (your quotation marks). I find that when I have a simple method to solve a problem, it is best to use it, rather than try a more complex method which is both time-consuming and error-prone. Here the simple method is just to write down the series from one's knowledge of the geometric series. The more complex method is to use contour integration; it is easy to fall into error thereby. $\endgroup$ Jun 21 '20 at 18:01
  • $\begingroup$ @AnginaSeng I might have phrased it wrongly; by no means am I trying to say the the method I'm attempting is better than using the geometric series. I promise, if I see how to use it in an exam I will definitely take the shorter, easier and cleaner route. However, as I said, it's happened to me multiple times that I just don't see the creative or clever way in which a problem is asking me to think, and the last resort is using definition and build it from the ground up to see where you get. $\endgroup$
    – Marc
    Jun 21 '20 at 18:15

When $n\le -2$, $-n\ge2$, $$\int_C \frac{f(z)}{z^{n+1}}\,dz=\int_C\frac{z^{-n-2}\,dz}{z-1}$$ and the integrand has no pole at $z=0$, so only the pole at $z=1$ counts, and there the residue is $1$.


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