# Laurent series around different points and on annulus

1. Jun 29, 2011

### Gulli

I have to find the Laurent series for the following cases:

$$f(z)=\frac{1}{(z-a)(z-b)}$$ for $$0 < |a| < |b|$$ around $$z=0$$ $$z=a$$ $$z=\infty$$ and on the annulus $$|a| < |z| < |b|$$
I know I can do a partial fraction thingy: $$f(z)=-\frac{1}{a(b-a)(\frac{z}{a}-1)}+\frac{1}{b(b-a)(\frac{z}{b}-1)}$$ which can be expressed as $$\frac{1}{a(b-a)}\sum(\frac{z}{a})^k - \frac{1}{b(b-a)}\sum(\frac{z}{b})^k$$ for $$\frac{z}{a}, \frac{z}{b} < 1$$

The condition $$\frac{z}{j} < 1$$ is met for z = 0, but not for the other cases. I know I must use the substitution $$w=\frac{1}{z}$$ for the case where z must be taken at infinity (and thus w at 0). But I need to know how to really solve all of these cases.

Last edited: Jun 29, 2011
2. Jun 29, 2011

### HallsofIvy

Staff Emeritus
Yes, the function is analytic at z=0 so you just need to find the Taylor's series at z= 0.

The function has a "pole of order 0" at z= a because it is of the form 1/(z- b), which is analytic at z= a, divided by z- a. Find the Taylor's series for 1/(z- b) at z= a, and divide each term by z- a. (One way to do that is to write 1/(z- b) as
$$\frac{-1}{b- z}= \frac{-1}{(b- a)- (z- a)}= -\frac{1}{1- \frac{z-a}{b-a}}$$
and think about the formula for the sum of a geometric sequence.)

For z= infinity, yes, replace z by 1/z. That will give you
$$\frac{1}{(\frac{1}{z}- a)(\frac{1}{z}- b)}= \frac{z^2}{(1- az)(1- bz)}$$
That is analytic at z= 0 so you can find a Taylor's series at z= 0.

3. Jun 29, 2011

### micromass

Staff Emeritus
Hi Gulli!

It seems that you already got the case z=0. So, let's study the next case: z=a.

The thing is again to put the thing into partial fractions. Like you calculate, this gives us

$$\frac{1}{(z-a)(z-b)}=-\frac{1}{b-a}\frac{1}{z-a}+\frac{1}{b-a}\frac{1}{z-b}$$

The first part is already of the form $\frac{1}{z-a}$, so this is good. The second part is not good yet. Try to write the second part in the form

$$\frac{1}{m}\frac{1}{\frac{(z-a)}{n}-1}$$

for certain m and n. Then you can write this as a series again.

Edit: maybe Halls approach is slightly easier...

4. Jun 29, 2011

### Gulli

That's a very sneaky bit of trickery. You're saying I can just find a series for 1/(z-b) and then divide that series by (z-a)? Like this: $$-\sum(\frac{z-a}{z-b})^{k}\frac{1}{(z-a)(b-a)}$$ Won't that blow up for k<1?

Last edited: Jun 29, 2011
5. Jun 29, 2011

### micromass

Staff Emeritus
Don't you have a mistake in your series? It should be
$$-\sum(\frac{z-a}{b-a})^{k}\frac{1}{z-a}$$

This is the same as

$$-\sum_{k=0}^{+\infty}{\frac{1}{(b-a)^k}(z-a)^{k-1}}$$

This is your Laurent series around a. I don't quite see what you mean with "blow up for k<1". The k is our summation index.

By definition, the Laurent series around a is of the form

$$\sum_{k=-\infty}^{+\infty}{c_k(z-a)^k}$$

so to find the Laurent series, you'll need to write everything in terms of (z-a).

6. Jun 29, 2011

### Gulli

Actually it should be $$-\sum(\frac{z-a}{b-a})^{k}\frac{1}{(z-a)(b-a)}$$ or $$-\sum_{k=0}^{+\infty}{\frac{1}{(b-a)^{k+1}}(z-a)^{k-1}}$$

The k=0 term of this series is a singularity (for z=a), if I'm not mistaken. And yes, z-z0 is in the definition of the Laurent series, stupid of me...

Last edited: Jun 29, 2011
7. Jun 29, 2011

### micromass

Staff Emeritus
Why to you multiplicate this series by $\frac{1}{(z-a)(b-a)}$? Am I missing something here? Should you just multiplicate it by $\frac{1}{z-a}$

That doesn't make any sense, how can a term of a series be a singularity?
Do you mean that the k=0 term of this series is not defined for z=a (as you divide by 0)? That's certainly true.

8. Jun 29, 2011

### Gulli

The extra 1/(b-a) is there because Halls of Ivy forgot it when he rewrote -1/(b-z), but it's not really important since it's a non-zero constant.

Yes, I did mean that that term (with k=0) is "undefined" because we divide by 0, isn't that a problem?

9. Jun 29, 2011

### micromass

Staff Emeritus
Yes, indeed!

No, because the point of the Laurent series is that we won't evaluate it in a. We want to evaluate the Laurent series with points close to a, but not equal to a. Thus, the term

$$\frac{1}{z-a}$$

will always exist, since we never evaluate it in a.

The Laurent series is defined in a punctured neighborhood of a, that means that it isn't defined in a itself.

10. Jun 29, 2011

### Gulli

Ah, that last part (not being bothered by dividing by "approximately zero") makes it a whole lot easier. All I really have to do is find any series of (z-z0) and I'm done, well, except for the annulus case.

Then again, how do I get a series out of something as ugly as $$\frac{w^2}{(1- aw)(1- bw)}$$

Last edited: Jun 29, 2011
11. Jun 29, 2011

### micromass

Staff Emeritus
You need to express it as a series with terms w2. So you only need to find the series of

$$\frac{1}{(1-aw)(1-bw)}$$

and then multiplicate the result by w2. Now, split the above fraction into partial fractions and write each partial fraction as a series by again using the identity

$$\frac{1}{1-z}=\sum_{k=0}^{+\infty}{z^k}$$

12. Jun 29, 2011

### Gulli

So I should get a w^2 times a series in (w-0)? Doesn't that violate the definition of the Laurent series which says only a constant is allowed in front of the sum?

13. Jun 29, 2011

### micromass

Staff Emeritus
No, because you can bring the w2 in the sum in the end. For example, you might get

$$w^2\sum_{n=0}^{+\infty}{w^n}=\sum_{n=0}^{+\infty}{w^{n+2}}$$

14. Jun 29, 2011

### Gulli

I get:

$$\sum(-\frac{a^{k+1}}{b}+1+ab^{k-1})z^{-k-2}$$ with z=1/w.

So how do I tackle the annulus case? And btw, thanks for all your speedy replies.

15. Jun 29, 2011

### micromass

Staff Emeritus
For the annulus case, we start again from the partial fractions

$$\frac{1}{b-a}\frac{1}{z-a}-\frac{1}{b-a}\frac{1}{z-b}$$

we want to represent these fractions as series again. For the fraction $\frac{1}{z-b}$ there is no problem as $|z|<|b|$. But the $\frac{1}{z-a}$ might pose problem. For that, do the trick

$$\frac{1}{z-a}=\frac{1}{z}\frac{1}{1-\frac{a}{z}}$$

The factor $\frac{1}{z}$ is already in the good form. The other factor can easily be represented as a series now.

16. Jun 30, 2011

### Gulli

So if, say, the annulus were between a/2 and 3a/2, then the anwer for the case z=a would already be a solution?

17. Jun 30, 2011

### Gulli

For the annulus |a| < |z| < |b| I get:

$$f(z)=\frac{1}{(b-a)(z-a)}-\frac{1}{(b-a)(z-b)} = \frac{1}{(-\frac{z}{b}+1)(b-a)b}+\frac{1}{(b-a)(1-\frac{a}{z})z}$$

That would give me:

$$\frac{1}{b(b-a)}\sum(b^{k}z^{k}) + \frac{1}{b-a}\sum(a^{k}z^{-k-1})$$

Is this any good?

Last edited: Jun 30, 2011
18. Jun 30, 2011

### micromass

Staff Emeritus
Yes, but there's a small typo. You have to write b-k in the first series.

19. Jun 30, 2011

### Gulli

Yes, you're right, it should be minus k. So am I allowed to leave this solution as it is or do I have to rewrite it as a single sum in z (if that's even possible)?

20. Jun 30, 2011

### micromass

Staff Emeritus
No, what you have seems ok!