# Laurent series expansion

1. Oct 12, 2013

### usn7564

1. The problem statement, all variables and given/known data
Find the Laurent expansion for

$$\frac{1}{z^2-1}$$

in the annulus $$1 < |z-2| < 3$$

3. The attempt at a solution
I've gotten to the last parts but getting stuck there.

First I expanded the denominator and did a partial fraction decomposition and arrived at
$$\frac{1}{2}(\frac{1}{z-1} - \frac{1}{z+2})$$

Then I got a series for the two terms using geometric series (disregarding the 1/2 factor for now)

$$\frac{1}{z-1} = \frac{1}{z-2+1} = \frac{1}{1+\frac{1}{z-2}} = \sum_{0}^{\infty}(-1)^j(z-2)^{-j}$$

similarly for the second term where I get

$$\frac{1}{z+1} = \frac{1}{z-2+3} = \frac{1}{3} \frac{1}{1+\frac{z-2}{3}} = \sum_{0}^{\infty}(-1)^j\frac{1}{3^{j+1}}(z-2)^j$$

The problem I have is I can't see how I can put them into a single sum (with the Laurent coefficient different for n<0 and n≤0. It would have been convenient if the j = 0 term in the "principal part" of the Laurent series was zero, but it's 1. Other than that 1 I have the exact answer I should have.

2. Oct 12, 2013

### fzero

It is not necessary that the coefficients of a Laurent series all have the same closed form expression. This is usually taken into account by using different labels for the negative power coefficients:

$$f(z) = \sum_{i=0}^\infty a_i (z-c)^i + \sum_{i=1}^\infty \frac{b_i}{ (z-c)^i}.$$

As a simple illustration, we can just note that if the function $f(z)$ is analytic (has no poles) inside the contour $C$ used to define the expansion (anywhere on the plane inside $C$, not just on the annulus), then all of the coefficients $b_i =0$, the principal part vanishes, and we have an ordinary Taylor series. So there is no general requirement that $b_i = a_{-i}$.

3. Oct 12, 2013

### usn7564

Oh, should have made myself clearer. I have the right coefficients, the answer made one integral with a generic c_j defined under it with different values depending the j. In fact it had my exact coefficients except for the negative j's it only went to -1, whereas mine goes to 0 so I cannot combine the integrals at all because I have one term extra (the j = 0 in the principal part). The c_j was defined as:

$$(-1)^j, j < 0$$
$$\frac{(-1)^j}{3^{j+1}}, j \ge 0$$

And unless I'm missing something obvious something's wrong with my sums because I can't get it to that form without an extra 1 outside the sum.

4. Oct 12, 2013

### vela

Staff Emeritus
You dropped a factor for 1/(z-2) here.
$$\frac{1}{z-2+1} = \frac{1}{z-2}\times\frac{1}{1+\frac{1}{z-2}}$$

5. Oct 12, 2013

### brmath

I guess I am confused right from the beginning because the denominator factors as (z-1)(z+1) so how can you compute a partial fraction with (z-1) and (z +2)?

I am also confused because for 1 < |z -2| <3 the function 1/$(z^2 -1)$ has no poles and therefore no Laurent coefficients. So I would say that in this annulus the function has Taylor's coefficients only.

Can someone enlighten me?

6. Oct 12, 2013

### fzero

That's a typo. If you look at the terms he actually expands, he has $1/(z+1)$.

For the given annulus, the pole at $z=1$ lies within the contour (but is not on the annulus itself). So the negative Laurent coefficients are nonzero. The function is holomorphic on the defined annulus, so both parts are convergent. If we instead consider the disc $|z-2|<1$, the function is also holomorphic and has a Taylor series, since there is a contour that excludes both poles. If we considered a unit disc centered on the pole at $z=1$, the principal part should just be one term identifying the pole. The corresponding annulus is obtained by deleting the pole from this disc.

The Laurent series always depends on the choice of contour (which may or may not include some poles) and will be convergent on an annulus that is bounded by the locations of the poles.

7. Oct 13, 2013

### brmath

Thank you for your reply. I'm glad the z+2 is a typo because it did confuse me.

Since f has just simple poles at 1 and -1 I believe he needs just the Laurent expansion from $a_{-1}$ . Looking at the pole z = 1, the expression $\frac{1}{z-1} + \frac{1}{z+1}$ looks to me like $a_1$ = 1 and the rest of the terms are Taylors series terms for $\frac{1}{z+1}$; and similarly for the pole at z=-1. Is that correct?

I think also that each expansion will converge in a circle of radius 2 -- that is until it hits the next pole. So neither of them covers the entire annulus. Are they to be combined, or do we just say that each covers part of the annulus and let it go at that?

8. Oct 13, 2013

### fzero

I am not 100% sure what you mean with your notation. If $a_i$ is a coefficient in the Laurent series, I'm not sure what (paraphrasing) "getting a Laurent expansion from $a_i$" would mean. As I said, the Laurent expansion depends on a point as well as a choice of contour around that point. The coefficients from one contour to another can differ depending on the location of the poles with respect to the chosen contours.

Only the pole at $z=1$ lies inside the contour chosen here. Correspondingly, the term $1/{z-1}$ generates the negative power terms, while $1/{z+1}$ gives the positive powers.

If we chose another contour that also enclosed the pole at $z=-1$, then that contour would necessarily lie outside of the annulus of convergence for the Laurent series. We can no longer rely on the geometric series expansions to compute the Laurent series, since the expansion does not result in a convergent series, and so we don't have a uniqueness theorem to use to identify the expansion as the Laurent series.

Since we are dealing with $f(z)=1/(z^2-1)$, we can't ignore either of $1/(z\pm 1)$, but must include both. We can consider the radius of convergence of each geometric expansion separately. The intersection of the resulting domains leads precisely to the specified annulus, $1< |z-2| <3$.

9. Oct 13, 2013

### brmath

By "from" I was referring not to $a_i$ but specifically to $a_{-1}$ and meant the expansion $a_{-1}, a_0,a_1,$etc. If there were a non-zero term $a_{-2}$ would that not imply that the pole at 1 is not simple? Yet we know it is.

So seeing that 1/(z-1) is sitting there explicitly and that the other term is holomophic in that neighborhood, I assumed $a_{-1}$ = 1 and the other $a_{-i}$ would be 0. Is that wrong?

It never occurred to me to think of a contour that would include both poles.

Finally, I'm surprised when you say the choice of contour would affect the coefficients (not that I know so much, so why shouldn't I be surprised). For example, if we took a small circle for the contour and then doubled the radius, would that affect the coefficients? I thought the Cauchy integral formula tells us that is invariant as long as f is holomorphic in the circle. Are you thinking of some other kind of contour like an ellipse or a rectangle or some jagged thing which is not even convex? I was under the impression that integral would still not change in value, as long as f stays holomorphic.

I know this is a homework section so I can move these questions to the analysis section if you think it proper. But I appreciate the answers. I don't think I ever really learned this stuff.

10. Oct 13, 2013

### jackmell

That was very confusing to read.

Allow me to summarize:

$$\frac{1}{z^2-1}=1/2\left(\sum_{n=0}^{\infty} \frac{(-1)^n}{(z-2)^{n+1}}-\sum_{n=0}^{\infty} \frac{(-1)^n (z-2)^n}{3^{n+1}}\right),\quad 1<|z-2|<3$$

and that is no different than writing it using a single summation if you just manually group the terms accordingly:

$$\frac{1}{z^2-1}=\sum_{-\infty}^{\infty} a_n (z-2)^n$$

11. Oct 13, 2013

### fzero

I think that you are being confused by the fact that the expansion is around $z=2$ rather than an expansion around the pole at $z=1$. Since $z=1$ is a pole, the expansion there should indeed have $a_{-1} \neq 0$ and $a_{i<-1} =0$.

I'm not thinking of wierdly shaped contours. I'm just thinking from the point of view where we fix the point we're expanding around and then carelessly choose a contour that might as well be a circle. There are 3 possibilities (in the sense of topological equivalence classes); namely, we include:

1. no poles
2. the pole at $z=1$
3. both poles

For cases 1 and 3, one or the other of the geometric expansions is not convergent. For choice 2, if our contour happens to stray from the domain of convergence, then we just have to deform the contour a bit while including the pole to get a sensible expansion.

I haven't actually thought about this in a while, nor probably in as much detail, so I might have been overfocusing on this issue. If one is clever enough to guess the domain of convergence in advance, then we can always choose the right contour at the beginning. Also, if we're using alternatives to the contour integrals to define the expansion, then we don't need to worry as much about the contour either.

But as I mentioned above, I think the main point of confusion we're having is not so much the contour, but the center point of the expansion.

12. Oct 13, 2013

### brmath

You are completely right. I really didn't understand how all those z-2's were getting in there. And I don't see anywhere in the problem that says explicitly that you should expand around z = 2. However, if you are doing that, you can have a plain Taylor's series in a circle of radius 1, which of course is not going to cover the annulus. So I now understand the idea is to use a contour which includes one of the poles. I'll review the problem again in that light.

That said why would one use z =2? It seems like a hard way to do something easy. But perhaps it was a teaching point (in which case I'm busy learning).

13. Oct 13, 2013

### fzero

The first thing you should do with a problem like this is actually draw the annulus and identify the poles. Then you can either pick a contour, or sketch in the domains of convergence for explicit series expansions.

It's clear that expanding around $z=2$ is a bit more work than $z=1$. You can draw interesting conclusions from using either point, though.

14. Oct 13, 2013

### brmath

I can see that the expansion around z=2 is more work, and have been thinking about it. Thanks for all your insights.

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