0
$\begingroup$

I'm a little confused about a one part of a specific example of a Laurent series that was given by a prof. It seems like it should be pretty straightforward. I want to expand $f(z)=\frac{1}{z^2-z}$ as a Laurent series about $z=i$. I did it one way, and he did it another.

It's analytic at this point, so I think the Laurent series should just be in the form of a regular power series expansion. We know that $f(z)=\frac{1}{z-1}+\frac{1}{z}$. I agree with the example that $\frac{1}{z-1}=\frac{1}{i-1}\cdot\frac{1}{1-(-\frac{z-i}{i-1})}=\sum_{n\geq 0} \frac{(-1)^n}{(i-1)^n}(z-i)^n$. Here is where I'm confused. I would have written $\frac{1}{z}$ as: $$ \frac{1}{z}=\frac{1}{i+(z-i)}=\frac{1}{i}\cdot\frac{1}{1-(\frac{z-i}{-i})}=\sum_{n\geq0}\frac{(-1)^n}{i^{n+1}}(z-i)^n $$

No principal part, which makes sense to me.

The prof instead did: $$ \frac{1}{(z-i)+i}=\frac{1}{z-i}\cdot\frac{1}{1-\frac{-i}{z-i}}=\sum_{n\geq 0}(-i)^n(z-i)^{-(n+1)} $$

This seems to be giving us a principal part for the Laurent expansion and a non-zero residue. Can someone explain why he did this?

Thanks!

EDIT: There was a nice comment pointing out that the series I wrote and the series the prof wrote have two different domains of convergence. I'm a little confused in general then what the prof's expansion is saying about our function. Does it matter that it has an infinite principal part? What does that mean?

$\endgroup$
4
  • $\begingroup$ If you want a series expansion in some disc centered at $i$, then it seems your calculation, rather than your professor's, is correct. For one, the latter converges only outside $D(i,1)$. However, it is strange to ask for a Laurent series around a point, since it is then also a Taylor series. Usually one asks for the Laurent series in an annulus. $\endgroup$ – solstafir Jul 19 '16 at 4:20
  • $\begingroup$ @solstafir There were other parts of the example expanding around singularities. It's still an annulus, just a punctured disk. Good point about convergence though, thanks! $\endgroup$ – Dizzy Jul 19 '16 at 4:26
  • $\begingroup$ Yes, but $f$ only has singularities at $0$ and $1$, not $i$. $\endgroup$ – solstafir Jul 19 '16 at 4:27
  • $\begingroup$ @solstafir Yes, of course. My point was that we usually just declare the domain of the series to be an annulus. You're correct that the domain function is also defined at $i$ though. $\endgroup$ – Dizzy Jul 19 '16 at 4:34
1
$\begingroup$

To expand on my original comment, let's investigate the domain of convergence of each series. First, we have

$$\frac{1}{z-1} = \sum_{n=0}^\infty (-1)^n \left(\frac{z-i}{i-1}\right)^n$$

which converges for $|z-i|<\sqrt{2}$. For $\frac{1}{z}$, we have two different candidates for our series. You have

$$\frac{1}{z} = \frac{1}{i} \sum_{n=0}^\infty (-1)^n \left(\frac{z-i}{i}\right)^n$$

which converges for $|z-i|<1$. Your professor has

$$\frac{1}{z} = \frac{1}{z-i}\sum_{n=0}^\infty (-1)^n \left(\frac{i}{z-i}\right)^n$$

which converges for $|z-i|>1$.

Putting each of these together, you have a Taylor series for $f$ which converges for $|z-i|<1$ (which can indeed, if you wish, be interpreted as a series defined on the punctured disc), whereas your professor has a Laurent series for $f$ which converges in the annulus $1<|z-i|<\sqrt{2}$. Perhaps your professor intended for you to expand your function in some annulus centered at $i$, other than the punctured disc?

$\endgroup$
4
  • $\begingroup$ Yeah, this is more or less what you mentioned in your comment. I guess I'm more interested in what it means that the prof's expansion doesn't have an analytic part to the series. Any ideas? $\endgroup$ – Dizzy Jul 19 '16 at 13:23
  • $\begingroup$ Sorry, I'm not sure if I understand your question. What do you mean by "an analytic part to the series"? Your professor's series is analytic in the annulus $1<|z-i|<\sqrt{2}$. It is, however, not analytic in the disc $|z-i|<1$ (or anywhere outside of the annulus). $\endgroup$ – solstafir Jul 19 '16 at 14:27
  • $\begingroup$ It's terminology. All the negative powers of the Laurent expansion form the principal part. The rest form the analytic part. An analytic function's Laurent expansion only has an analytic part, i.e. it's power series expansion. $\frac{1}{z^2-z}$ is analytic at $z=i$, so I would expect the Laurent expansion to only have an analytic part, which is what we see in my expansion. The prof's expansion has a non-zero principal part, which is surprising to me. $\endgroup$ – Dizzy Jul 19 '16 at 16:02
  • $\begingroup$ Oh, I understand now. In that case, here is the explanation: $\frac{1}{z}$ is analytic in the disc $|z-i|<1$. It is also analytic in the region $|z-i|>1$. However, note that the circle $|z-i|=1$ includes the pole of $\frac{1}{z}$ at $0$, so it is not analytic on this circle. Therefore, you cannot represent it by a series which converges in subsets of both $|z-i|<1$ and $|z-i|>1$. Your professor's series for $\frac{1}{z}$ converges in the latter, but not the former. In particular, it does not contain any analytic part because it does not converge in any disc around $i$. $\endgroup$ – solstafir Jul 19 '16 at 16:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.