Expanding Laurent Series around Singularities

[KleZMeR]

Homework Statement

Determine Laurent Series around z = -1, z = 2, z=0
Determine pole and residue in each case, and solve series in each separate region of C.

f_1(z) = \frac {z}{(z+1)(z-2)}

Homework Equations

The Attempt at a Solution

I've determined my partial fractions as
\frac{1}{3} \frac{1}{z+1} + \frac{2}{3} \frac{1}{z-2}

and I think I know where to head now, but my confusion is the points I am expanding around; these are all singularities. I am also confused about what my regions are, given that these are singularities? Also, is Taylor series to be used on my outer radius and Laurent used on my inner, I don't know when to apply which? Any help is appreciated, thanks.

 

[vela]

Homework Statement

Determine Laurent Series around z = -1, z = 2, z=0
Determine pole and residue in each case, and solve series in each separate region of C.

f_1(z) = \frac {z}{(z+1)(z-2)}

Homework Equations

The Attempt at a Solution

I've determined my partial fractions as
\frac{1}{3} \frac{1}{z+1} + \frac{2}{3} \frac{1}{z-2}

and I think I know where to head now, but my confusion is the points I am expanding around; these are all singularities. I am also confused about what my regions are, given that these are singularities? Any help is appreciated, thanks.

What exactly is so confusing about expanding around one of the singularities?

z=0 isn't a singularity; it's a zero of the function.

The answers to some your questions, like what the regions are, are surely covered in your textbook. Try rereading the text now that you have some questions in mind.

 

[KleZMeR]

We were told to choose our own textbook, I have 1985 Arfken from library, it has 5 pages of derivations, no examples, it is honestly no help. I have been around the internet but in all examples the regions are given in the problems and they don't match the two values which I can obviously pull from the denominators.

|z|<1 , 1<|z|<2 , 2<|z|

So is this a region between R=1 and R=2, centered around -1

 

[HallsofIvy]

The only time you are going to have a Laurent series is centered on a pole! If you center a power series on point where the function is analytic, there will be NO negative powers. About an essential discontinuity, an infinite number. (Well, you can call any of those a "Laurent series" but normally the point of Laurent series is to have a non-zero, finite, number of negative power terms.

To find the Laurent series for \frac{z}{(z+ 1)(z- 2)}, about 0, use the usual MacLaurin series formula for this function or, equivalently, find the Taylor series for \frac{1}{(z+ 1)(z- 2)} and multiply each term by z.

To find the Laurent series about z= -1, find the Taylor's series for \frac{z}{z- 2} about z= -1 and multiply each term by 1/(z+ 1).

To find the Laurent series about z= 2, Find the Taylor's series for \frac{z}{z+ 1} about z= 2 and multiply each term by 1/(z- 2).

OR, since you have already determined that the function can be written as \frac{1}{3}\frac{1}{z+ 1}+ \frac{2}{3}\frac{1}{z- 2}

1) Write \frac{1}{z+ 1}= \frac{1}{(z- 2)+ 2+ 1}= \frac{1}{(z- 2)+ 3}= \frac{\frac{1}{3}}{1- (-\frac{z- 2}{3})} and use the formula for a geometric series to expand that.

2) Write \frac{1}{z- 2}= \frac{1}{(z+ 1)- 1- 2}= \frac{1}{(z+ 1)-- 3}= -frac{\frac{1}{3}}{1- /frac{z+1}{3}} and use the formula for a geometric series to expand that.

 

[vela]

Look at picture 6.12 on page 376. See how the circles divide the complex plane up into three regions? There's a different series for each region.

 

[KleZMeR]

Look at picture 6.12 on page 376. See how the circles divide the complex plane up into three regions? There's a different series for each region.

Ok, that is extremely helpful, so to clarify, there are three regions, and I am given three different z_0, and given my f(z), I will have three different Laurent series depending on my z_0, and my three regions for each z_0 are important for my plot, and also when I determine my series I will see the convergence in the denominator should reflect my region?

So another general case, given some f(z)=\frac{1}{z-1} for this I will have only two regions with some z_1 as my nearest point on the Argand diagram for which f(z) is non analytic?

Also, how do I get my TEX and regular text to not have a carriage return every time? Sorry about that.

 

[vela]

Here's a simple example. Let f(z) = 1/(z-1). Suppose you want series expansions about $z_0 = 0$. There's a singularity at $z=1$. This splits the complex plane into two regions centered about $z_0=0$. The first is $\lvert z \rvert < 1$, and the other is $\lvert z \rvert > 1$.

You should know that you can expand f(z) as
$$\frac{1}{z-1} = -\frac{1}{1-z} = -(1+z+z^2+z^3+\cdots) = -1 - z - z^2 - \cdots,$$ but the thing is, this only converges if $\lvert z \rvert < 1$. This is the series representing f(z) in the first region.

To get the series for the second region, you expand in powers of $1/z$ because if $\lvert z \rvert >1$, then $\lvert 1/z \rvert <1$. You do it like this
$$\frac{1}{z-1} = \frac{1}{z}\frac{1}{1-\frac 1z} = \frac{1}{z}\left[1+\frac 1z + \left(\frac 1z\right)^2 + \cdots\right] = \frac 1z + \left(\frac 1z\right)^2 + \cdots.$$

 

[KleZMeR]

The only time you are going to have a Laurent series is centered on a pole! If you center a power series on point where the function is analytic, there will be NO negative powers. About an essential discontinuity, an infinite number. (Well, you can call any of those a "Laurent series" but normally the point of Laurent series is to have a non-zero, finite, number of negative power terms.

To find the Laurent series for \frac{z}{(z+ 1)(z- 2)}, about 0, use the usual MacLaurin series formula for this function or, equivalently, find the Taylor series for \frac{1}{(z+ 1)(z- 2)} and multiply each term by z.

To find the Laurent series about z= -1, find the Taylor's series for \frac{z}{z- 2} about z= -1 and multiply each term by 1/(z+ 1).

To find the Laurent series about z= 2, Find the Taylor's series for \frac{z}{z+ 1} about z= 2 and multiply each term by 1/(z- 2).

OR, since you have already determined that the function can be written as \frac{1}{3}\frac{1}{z+ 1}+ \frac{2}{3}\frac{1}{z- 2}

1) Write \frac{1}{z+ 1}= \frac{1}{(z- 2)+ 2+ 1}= \frac{1}{(z- 2)+ 3}= \frac{\frac{1}{3}}{1- (-\frac{z- 2}{3})} and use the formula for a geometric series to expand that.

2) Write \frac{1}{z- 2}= \frac{1}{(z+ 1)- 1- 2}= \frac{1}{(z+ 1)-- 3}= -frac{\frac{1}{3}}{1- /frac{z+1}{3}} and use the formula for a geometric series to expand that.

Thank you! So, does my choice of series I use to find my Laurent series only depend on the form of my function or partial fraction here?

 

[vela]

Ok, that is extremely helpful, so to clarify, there are three regions, and I am given three different z_0, and given my f(z), I will have three different Laurent series depending on my z_0, and my three regions for each z_0 are important for my plot, and also when I determine my series I will see the convergence in the denominator should reflect my region?

So another general case, given some f(z)=\frac{1}{z-1} for this I will have only two regions with some z_1 as my nearest point on the Argand diagram for which f(z) is non analytic?

Also, how do I get my TEX and regular text to not have a carriage return every time? Sorry about that.

Actually, when you're centered on one of the singularities, you'll only have two regions.

Use ITEX instead of just TEX tags for inline math.

 

[KleZMeR]

Here's a simple example. Let f(z) = 1/(z-1). Suppose you want series expansions about $z_0 = 0$. There's a singularity at $z=1$. This splits the complex plane into two regions centered about $z_0=0$. The first is $\lvert z \rvert < 1$, and the other is $\lvert z \rvert > 1$.

You should know that you can expand f(z) as
$$\frac{1}{z-1} = -\frac{1}{1-z} = -(1+z+z^2+z^3+\cdots),$$ but the thing is, this only converges if $\lvert z \rvert < 1$. This is the series representing f(z) in the first region.

To get the series for the second region, you expand in powers of $1/z$ because if $\lvert z \rvert >1$, then $\lvert 1/z \rvert <1$. You do it like this
$$\frac{1}{z-1} = \frac{1}{z}\frac{1}{1-\frac 1z} = \frac{1}{z}\left[1+\frac 1z + \left(\frac 1z\right)^2 + \cdots\right] = \frac 1z + \left(\frac 1z\right)^2 + \cdots.$$

That is the best thing I have heard anyone say, ever! In my class notes I see that I can have no divergent points, i.e all series must converge.

 

[KleZMeR]

Actually, when you're centered on one of the singularities, you'll only have two regions.

Use ITEX instead of just TEX tags for inline math.

Wait, for my f_1(z) = \frac {z}{(z+1)(z-2)} centered on a singularity I will only have 2 regions? Or for a f(z) with only one term in the denominator?

 

[KleZMeR]

OK, so with my z_0 = -1 and my function above, I am getting two regions, I think I get it, I think.

So, region 1 is centered at -1, and my first singularity at 2 splits my region into two, and so, my inner region is |z|&lt;3, and the other is |z|&gt;3

But what I now wonder, do my series still need to be split up between |z|&lt;1 and |z|&gt;1 due to convergence?

 

[vela]

Your region isn't quite right. $\lvert z \rvert = 3$ is a circle of radius 3 centered at the origin, but you want it to be centered about $z_0 = -1$.

 

[KleZMeR]

Ok, so 2&gt;|z| and 4&gt;|z|, that doesn't make sense to me? I do understand the general case you gave me in number 7 above, but this problem is confusing me, do I need multiple series for each term when inside the inner region? A series for z&lt;|1| and |z|&lt;4 and |z|&lt;2 ? is |z| not just the magnitude of a vector?

 

[KleZMeR]

1&lt;|z-1|&lt;3

 

[KleZMeR]

Ok, so it's the magnitude from my center point, I understand that, but what has confused me, is that I guess I am supposed to use my geometric series for values |z|<1 , but my region is centered at -1 , and so I have a sub-region inside my inner circle that is |z|<1 , and then the rest of my inner circle's region, and THEN I have a region outside my circle and that needs a different series representation? Either I am seriously over thinking this or I'm just confused or both, so given a general case centered at the origin, which I understand, it seems this problem is vastly different.

 

[vela]

If the region is centered at z=-1, then the series will be in terms of powers of (z - (-1)), not powers of z, so the requirement for convergence in the inner circle will be of the form |z+1|<c where c is some constant (not necessarily 1).

 

[KleZMeR]

OK, so I believe I have found the Laurent series for my inner region by using the geometric series:

\sum_{n=0}^\infty \frac{(-1)^n (z-2)^n}{3^{n+1}} - 2 \sum_{n=0}^\infty \frac{(z+1)^n}{3^{n+2}}
and from each term I got:
|z-2|&lt;3 and |z+1|&lt;3

I hope I am expressing this right? The sum of the two series from each partial fraction is the series of my function inside the region? Or is each series representative of each singular point which are two of my three center points I expanded about? Something feels off about just adding these two series, I think they each represent my expansion about a point, but I just wonder what happens to the other partial fraction term?

As noted in comment #4 above there are multiple ways to do this, and I see that there are negative terms so this indicates that this is centered on a singularity?

Thank you everyone for helping me get thru this on a Saturday, I really needed it.

 

[vela]

No, that's not correct. If you expand about z=2, for example, everything has to be written in terms of powers of (z-2). You originally found
$$f(z) = \frac{z}{(z+1)(z-2)} = \frac 13 \frac{1}{z+1} + \frac 23 \frac{1}{z-2}.$$ The second term is already a power of (z-2). Using the series you found for the first term, you therefore have
$$f(z) = \frac 23 \frac{1}{z-2} - \sum_{n=0}^\infty \left(-\frac 13\right)^{n+1} (z-2)^n.$$ That's the series for f(z) about z=2 which converges for |z-2|<3. Similarly, the series for f(z) about z=-1 that converges for |z+1|<3 would be
$$f(z) = \frac 13 \frac{1}{z+1} - \sum_{n=0}^\infty \frac{2}{3^{n+2}} (z+1)^n.$$ (I'm assuming you found the correct expansions.)

 

[KleZMeR]

Yes! There was an inversion I wasn't quite understanding, thank you! Expanding around z=-1 gave me a power of (z-2), and expanding around z=2 gave me a power of (z+1). I will read more on this, I honestly don't want to just know how to answer these, but I want to know why the answers are such.

So I will do this again for the outer region and look for convergence with your comment #7.

For my Laurent series about 0, which has 3 regions, I used Taylor series and then multiplied thru by Z from comment #4, and I found for this term:

<br /> \frac{1}{3} \frac{1}{z-2}

that

\frac{-1}{3} \sum_{n=0}^\infty \left ((\frac{f^n}{n!})\right) (\frac{z}{2})^{n+1}.

Where {f^n} is analyzed at 0.
This is only one term from my partial fraction, so how does this differ about a non-singularity?
And for expanding around z=0 with three regions, I am wondering how do I determine which series representation I should use for each region.

 

[vela]

This is only one term from my partial fraction, so how does this differ about a non-singularity?

The series doesn't contain any negative powers of $z$ that can blow up when z=0. In fact, it's just the plain old Taylor series about $z_0=0$ for that term. This is what Halls meant above. If you expand about a regular point, you just get the Taylor series about that point. Note this is only for the Laurent series that applies to a region containing the point $z_0$. For other regions, you can have negative powers, but they'll never blow up because $z=z_0$ isn't part of the region.

And for expanding around z=0 with three regions, I am wondering how do I determine which series representation I should use for each region.

It's all about convergence.

 

[KleZMeR]

The series doesn't contain any negative powers of $z$ that can blow up when z=0. In fact, it's just the plain old Taylor series about $z_0=0$ for that term. This is what Halls meant above. If you expand about a regular point, you just get the Taylor series about that point. Note this is only for the Laurent series that applies to a region containing the point $z_0$. For other regions, you can have negative powers, but they'll never blow up because $z=z_0$ isn't part of the region.It's all about convergence.

Ok, so |z|<1 and |z|>1 is crucial. So, say, expanding a series for |z|>1 and |z|>4 is redundant? Regardless of some singular point outside |z|<1 ?

 

[vela]

Can you give me an example of what you're asking about? I don't understand what you're getting at.

 

[KleZMeR]

Expanding around z=0, which is not a singularity, produces three regions, given this specific problem and the two existing singularities. There are two separate regions outside of |z|<1 due to my singular points. Therefore, do I expend looking for convergence in both regions outside |z|<1 ? Won't my results look the same for say 4>|z|>1 and |z|>4 ? (Assuming that there is say some singularity at 4)

 

[vela]

No, the series for f(z) won't look the same in the two outer regions. Each region will have its own expansion.

Find the series for one of the regions, and then check to see if it converges in the other region. You'll find it doesn't.

 

[KleZMeR]

No, that's not correct. If you expand about z=2, for example, everything has to be written in terms of powers of (z-2). You originally found
$$f(z) = \frac{z}{(z+1)(z-2)} = \frac 13 \frac{1}{z+1} + \frac 23 \frac{1}{z-2}.$$ The second term is already a power of (z-2). Using the series you found for the first term, you therefore have
$$f(z) = \frac 23 \frac{1}{z-2} - \sum_{n=0}^\infty \left(-\frac 13\right)^{n+1} (z-2)^n.$$ That's the series for f(z) about z=2 which converges for |z-2|<3. Similarly, the series for f(z) about z=-1 that converges for |z+1|<3 would be
$$f(z) = \frac 13 \frac{1}{z+1} - \sum_{n=0}^\infty \frac{2}{3^{n+2}} (z+1)^n.$$ (I'm assuming you found the correct expansions.)

f(z) = \frac 23 \frac{1}{z-2} + \sum_{n=0}^\infty \left(-\frac 13\right)^{n+2} (z-2)^n - \sum_{n=1}^\infty \frac{3^{n-2}}{(-1)^n(z-2)^n} .

So, would this be my entire series in both regions, assuming my two series are correct? (second term for inner region and third term for outer region?) Or is it sufficient to write them separately in the form you suggested above? i.e one for |z-2|<3 and |z-2|>3 ?

 

[vela]

You should be able to figure out the answer to that question on your own. If you let z=10, for instance, would the two sides of the equation be equal?