Laurent series around different points and on annulus

[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.

 

[HallsofIvy]

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.

 

[micromass]

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...

 

[Gulli]

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.

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?

 

[micromass]

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: $$-\sum(\frac{z-a}{z-b})^{k}\frac{1}{z-a}$$ Won't that blow up for k<1?

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.

Hi Micromass. Why do I need to write everything in terms of (z-a)?

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).

 

[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...

 

[micromass]

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

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

The k=0 term of this series is a singularity (for z=a), if I'm not mistaken.

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.

 

[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?

 

[micromass]

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, indeed!

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

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.

 

[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 $$\frac{w^2}{(1- aw)(1- bw)}$$

 

[micromass]

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

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

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

 

[Gulli]

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

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

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?

 

[micromass]

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?

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

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

 

[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.

 

[micromass]

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.

 

[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?

 

[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?

 

[micromass]

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

 

[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)?

 

[micromass]

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)?

No, what you have seems ok!

 

[Gulli]

Great, thanks a lot, I'm going to try more examples now.