1:24

Well, what if f is not differentiable on this entire disk?

What if we have a situation where

f is differentiable only in a portion of the disk?

For example, the function z over z squared + 4, where is that analytic?

Well, z squared + 4 = (z- 2i) (z + 2i),

and that's equal to 0 when z is plus or minus 2i.

So this function is clearly not analytic at 2i or

at negative 2i, because at those two points, we'd be dividing by 0.

So what if I wanted a series representation in the neighborhood of 2i,

however?

What if I was really interested in values of the function close to this point 2i?

So what if I wanted a series representation in this disk and

the function is not analytic in this sector?

That's what the Laurent series are there to help us with.

Another example of a function that's not differentiable everywhere is the logarithm

function.

Remember, the principle branch of the logarithm is analytic in the entire

complex plane with the exception of the negative real axis and the origin.

So here, there are many disks in which the function is analytic.

So I can put this disk right here, and I can find the Taylor series representation

there, or I can find the Taylor series in this disk.

But the function is not analytic on the negative real axis.

So here, I can't even find an annulus of the type I just showed you in the previous

example that kind of surrounds places where the function is not analytic.

Thus I couldn't draw the whole annulus, it's going to be crossing through

the portion of the function that's not analytic.

So here's the Laurent series theorem.

Suppose f is analytic in an open set U, maybe there's a hole here, and

there's some stuff that's in there, and maybe here.

As long as I can find an annulus that fits into U,

in which f is analytic and that annulus is centered at some point z0,

and that z0 may not be part of your domain.

So this is the point z0.

going to have this inner radius r, and the outer radius uppercase R.

And all I need is that the space between these two

boundary components is entirely contained in U.

So inside I have stuff that's not contained in U, and

on the outside I have stuff that's not contained in U.

But between the two boundary components, the function is analytic everywhere.

So in this situation and

in that annulus, in that blue region, f has a series representation.

It looks a little bit different from a Taylor series.

And here's the difference.

Everything looks just like a Taylor series, but the one difference is that

the summation starts at negative infinity and goes all the way to infinity.

So we have a series that's infinite in both directions.

It's called a doubly infinite series.

So what that means is I sort of need dots indicating that

this series goes on and on on both ends of it.

I have an a0.

And then a1(z-z0), a2(z-z0) squared, just as in a Taylor series.

But I also have negative exponents k here that I need to take into consideration.

So when k is equal to -1, (z-z0) to the -1,

that's 1 over (z-z0), that gives rise to this term right here.

And then there's an a-2 over (z-z0) squared term, and so forth.

So this series could be doubly infinite in both directions, and

we'll see a bunch of the examples.

5:06

So if I have a function that's analytic in an annulus,

just like the blue region I drew below, then in that annulus the function has

a series representation that's called a Laurent series for the function.

And not only is the series convergent in that annulus, but

if I in fact make my annulus a little bit smaller.

And so those two radii are now called s and t.

So the smaller radius is s and the larger one is t.

So in this situation, we're staying away from the boundary.

In that new region that's a little bit smaller than the original annulus,

in that new region, that sub annulus that converges is in fact absolute and uniform.

And we'll be using those terms a lot.

So if you don't really remember what those means, that's quite all right.

But for those of you who are interested, this is very similar to the theorem

on Taylor series that we have in which we had locally uniform convergence

in the entire disk in which the function was analytic, and as soon as we strayed

away from the boundary the convergence was absolute and uniform.

6:47

So the function is analytic everywhere except at 1 and 2.

Now I need to find an annulus that fits into the domain where

the function actually is analytic.

So for example, I could choose the annulus z between 1 and 2 in absolute value.

This is centered at the origin.

And between these two boundary components, indeed our function is analytic,

because as long as I stay between those two boundary components and

not on the boundary itself, then I'm never dividing by 0,

because z never has the value 1 or 2.

And therefore I know it must have a Laurent series expansion.

The question's how do I find it?

So let me show you a trick for finding the Laurent series expansion.

7:32

The trick is to do a partial fraction decomposition first, and

we don't have to make it that complicated.

So instead of actually doing a real partial fraction decomposition,

I'll show you a little shortcut.

We replace this 1 in a smart way by (z-1)-(z-2).

I chose those so that I could then pull the fraction apart,

make it into two fractions, and cancel out parts of the denominator in each step.

So 1 is the same thing as (z- 1)- (z- 2), because this z cancels out the fifth z.

And I have left -1- -2, which is 2- 1, which is really 1.

When I pull this fraction apart,

I find myself with (z- 1) over (z- 1)( z- 2) and

in that fraction, the (z- 1) cancels out.

And I'm left with 1 over (z- 2).

For the second fraction, I have (z- 2) divided by (z- 1) (z- 2) and

in that fraction, the (z- 2) cancels out, and I'm left with 1 over (z- 1).

So all together, 1 over (z- 1)(z- 2)

is the same thing as 1 over z- 2- 1 over z- 1.

I want to make use of the geometric series.

Let me remind you, that the sum,

n from 0 to infinity of q to the n,

is equal to 1 over 1- q, as long as

the absolute value of q is less than 1.

So how do we use that for 1 over z- 2?

9:16

What is less than 1 here?

So, it has a little bit the look of the 1 over 1- q, but not quite.

We need to get our hands on something that is less than 1 here.

Well, we know that z in absolute value is bigger than 1 and less than 2.

So, how can I rewrite that so I get terms that are less than 1?

Because z is greater than 1, I get that 1 over the absolute value of z, is less than

1 so, if we can get 1 over the absolute value of z in there then I'm good.

I also know that z is less than 2 in absolute value,

which means that z over 2 is less than 1.

So these two facts are going to be crucial for us.

So let's get back to 1 over z- 2.

If I factor out a 2, like I did right here, if I factor out a 2

from the denominator, then my factor becomes 1 over z over 2- 1.

And that's exact in the form 1 over 1- q, except the terms are reversed, so

I'm going to factor up a -1 in addition, and I'm left with 1 over 1- z over 2.

Since z over 2 is less than 1, as we showed over here,

I can now re-write the 1 over 1- z over 2

as a geometric series, where my q is equal to z over 2.

So this first part here becomes minus one half times the sum z over 2 to the k.

You still need to look at the second part, mainly the 1 over z times 1- 1 over z.

That came from 1 over z- 1 and that factored out a z.

The factoring out of the z gave me this 1- 1 over z term of

the denominator and we know that 1 over z is less than 1.

So we're going to again use the geometric series idea.

This z right here stays on the outside and

the 1 over 1- 1 over z becomes the geometric series,

sum k from 0 to infinity 1 over z to the k.

Now that I have all that, my last goal is to write this so

I really recognize it as a Laurent series.

The first thing I want to do is,

I want to bring this factor in front of the series into the series.

So that means this 2 adds another 2 to the denominator,

so the denominator becomes 2 to the k + 1.

The -1 comes inside right here and the z to the k is still there.

And for the second series, I want to bring -1 over z into the series,

so I get the -1 right here and the denominator becomes z to the (k + 1).

Well I only wrote z to the k here and the reason for

that is that I've simply started counting that k = 1.

Since the series goes to infinity,

the powers of z that I will be looking at in this series -1

over z to the k + 1, when k runs from 0 to infinity.

The exponents k + 1 run from 1 through infinity.

And so instead of saying k goes from 0 to infinity and using the exponent k + 1,

I can simply say k goes from 1 to infinity and I use the exponent k.

And my last step is to write the second series in terms of the negative powers

of k because that's really how Laurent series are written.

So, instead of calling this 1 over z to the k,

I recognize that that is equal to z to the -k.

And so instead of k going from 1 to infinity, I'm going to have k go

from -1 to negative infinity, because all these powers here are negative.

And instead, I can just write z to the k, but I use negative indices for my series.

This is what we just showed.

I can now re-off the coefficients of this Laurent series.

The first sum captures all of the positive powers of z.

So for example, a0, which is the coefficient of z to the 0, so

that's what I get k is equal to 0 is simply -1 over 2 to the 0 + 1, so 2.

a1, which is the coefficient I get when k is equal to 1,

it's -1 over 2 squared which is 4, and so forth.

But I also have negative powers of z, and those I find in the second sum here.

13:59

So I find a -1, which is the coefficient that I get when k is equal to -1,

simply 2- 1, a -2 is equal to -1 and so forth.

So, all the negative coefficients are simply -1 and

all of the positive coefficients are -1 over the power of 2.

Here again is the picture of where the functions analytics,

analytic everywhere except at the points 1 and 2.

And so, in the first example,

we chose an annulus that ran between those two points, 1 and 2.

What if I chose a different annulus?

14:42

So I could simply look at this annulus and the outside component is unbounded,

that's also considered an annulus, and f is analytic in that annulus as well.

So in that annulus, f must have a Laurent series expansion as well.

Let's find it.

The trick is again the same as before.

I do my partial fraction decomposition, so 1 over (z- 1)(z- 2) is 1 over

z- 2- 1 over z- 1, but this time I simply need to factor a little bit differently.

We want to use the same geometric series guide.

So what can we conclude from this condition?

From this condition, we can conclude only that 2 over z

is less than 1, so that's a fact we can use here.

The infinity part doesn't really help us all that much.

But, in addition, because 2 over z is less than 1 if I look at 1 over z,

which is half of 2 over z.

That's clearly also less than 1 in absolute value, so

those are the two facts we're going to use now in this range that we're looking at.

15:42

So, the idea is that 1 over z- 2, I'm going to factor out the z,

and I'm left with 1- 2 over z inside the parentheses.

And 2 over z is less than 1 in absolute value.

And so I already have a perfect term here for

A geometric series expansion and for the second term, I do the same thing,

I factor out a z and I'm left with 1- 1 over z in.

And again, 1 over z is less than is 1 and absolute value.

So again, I can simply use my geometric series for that term as well.

The first term, I have a 1 over z factored out and

then my q is 2 over z so I have the sum 2 over z to the k.

16:23

And for the second term, I have a 1 over z on the outside and

then my q is 1 over z, so the sum of 1 over z to the k.

So, it looks like I only have negative powers of z all over the place.

So again, my goal is to write this so

I recognize exactly what the negative powers of z are, so

the first step is to bring this 1 over z and this 1 over z inside of the sums.

If I do that then, in the first sum,

my terms are going to be 2 to the k over z to the k + 1,

because I get extra z in the sum goes from 0 to infinity.

But I'm concerned mostly with the powers of z.

What are those?

So, the powers of z in the denominator start at 1.

Even though k = 0 is the first k I plug in, the powers really start at 1.

So, if I want to write this in powers of z, then I want to have k run from 1 to

infinity and have a z to the k in the denominator.

17:23

How's the numerator related to that?

Well, the power of 2 in the numerator is just one less than the power of

the denominator.

So, with my new summation index, I need to call the numerator 2 to the k- 1.

And then, I do the same thing with the second sum when I pull

the z to the inside, my terms become 1 over z to the k + 1.

And I'm going to sum over those, k from 0 to infinity.

But at a closer look, I notice that the exponents for

z run from 1 through infinity.

So, I'm going to run my series from 1 through infinity.

The numerator isn't being affected by that, so 1 over z to the k.

18:48

All right, so we have seen that 1 over (z- 1) times (z- 2) has

a Laurent series in the annulus where z is between 1 and 2 in absolute value.

But has also another Laurent series expansion where z is

greater than 2 in absolute value, and those are quite different from each other.

What are the coefficients a k in this last example?

We know this, that k only takes negative values, so a0 isn't even there.

a0 = 0, and so is a1, and so is a2, and so forth.

All the nonnegative powers of z aren't there, and so,

all those corresponding coefficients are all equal to 0.

What's a -1?

19:55

What's a -2?

a- 2 is 2 to the -2- 1- 1.

So, 2 to the 2 -1, that's 2- 1, so that's 1.

a -3 is 2 to the -3- 1- 1.

So, 2 squared- 1 so that's 3 and so forth.

Those are very different from our first example.

But those are, again, not the only annula I could choose.

Again, here's my picture.

Analytic everywhere except I need to avoid the points 1 and 2.

And so far, I chose any that has centered at 0, but I don't have to do that.

I could also choose an annulus that's centered elsewhere.

21:14

As long as I keep the strict inequality, f is analytic in this annulus,

whose inside component is really, really small.

So, you must settle Laurent series there, as well.

Again, my tricks are exactly the same as before.

z- 1 is less than 1, so I'm going to use that fact that z- 1 is less than 1,

and try to find the geometric series expansion using that fact.

So, how do I write 1 over z- 2 using that z- 1 is less than 1?

I'm going to write that as 1 over (z- 1)- 1.

And the only thing I need to do is now flip the order of these two terms, so

it looks like a geometric series.

So I get an extra negative sign from flipping, and

I have 1 over 1- z- 1, so the q in the geometric series is z- 1.

And I find that this is equal to negative sum times the sum k = 0

to infinity (z- 1) to the k, just the geometric series.

And this is true for z- 1,

an absolute value between 0 and 1.

And so my function itself, which is 1 over z- 2- 1 over z- 1, the 1 over z- 2,

I replace with my geometric series.

And then, all I have left is 1 over z- 1 which already looks like

a term of a Laurent series.

I don't have to do anything with it, and

the last thing I want to do is combine these two into one series.

And it turns out 1 over z- 1,

that is actually the same thing as (z- 1) to the -1.

And that's exactly the same form as the terms in the sum up front, and so

I can pull this last term into the sum, have the sum starting at -1 and

instead of at 0, so I take into consideration this new term as well.

I have a negative sum in front of the new term as well as in front of the sum,

so we'll have a -1 there, all together.

And this is my Laurent series in the annulus,

where z- 1 is between 0 and 1, in absolute value, and

so in this annulus, all the aks are equal to -1.

Well, not all the aks, but this is true for

k from -1, 0, 1 to and so forth.

The other ones are equal to 0.

So, ak = 0, for k less than or equal to -2.

Here's another example.

The function sine z over z to the 4th is really analytic entire complex plane,

unless I'm dividing by 0 which happens when z is equal to 0.

So it's analytic in the complex plane minus the origin.

I can find a Laurent series in an annulus, that's arbitrarily large on the outside

component so I might just as well call that infinity, but I can't draw that.

And the inside component has radius 0.

So how do I find the Laurent series centered at 0?

Or remember, for sine z itself, so if the denominator wasn't there,

we would have Taylor series expansion.

And that is given by z- z cubed over 3 factorial + z to the 5th over 5

factorial and so forth.

And uses this summation and notation for that.

So, if I wanted to look at sine z over z to the 4th,

I need to just divide each of those terms by z to the 4th.

z divided by z to the 4th is 1 over z cube.

25:17

So I get a -3 is equal to 1.

Then, the 1/z term right here,

that belongs to a-1.

So a-1 is actually -1/3 factorial.

And there's no 1 over z squared term, so a-2 must be 0.

Similarly, there's no constant term.

So a0 is equal to 0.

And then I come to the positive powers of z.

So right here, I have a power of 1, so 1 over 5 factorial must be my a1.

There's no z squared term, so a2 is 0, a3 is

negative 1 over 7 factorial, and so forth.

So far, we have found all of the coefficients ak by

actually computing all the rosters.

26:11

Remember that for Taylor series, so

where a functional is actually analytic in an entire disk,

we found that the aks in the Taylor series expansion can be computed

as the kth derivative of f, f at that center point z0 divided by k factorial.

Wouldn't it be nice if for the Laurent series we had a similar formula?

So how about these Laurent series, when our function is now analytic only

in an annulus centered at some point z0?

Well, f may not even be defined at z0 because all we're seeing is that f is

analytic in any other region, but we know nothing about f at z0.

And so, we can't even speak of the kth derivative of f at z0.

So, we need some new approach here.

Remember Taylor series for a second.

We know that ak is the kth derivative of f, f z0 divided by k factorial.

That's what we had up here.

27:10

The Cauchy's theorem, the kth derivative of f at z0

can be found by k factorial over 2 Pi i,

times the integral of f divided by z minus z0, to the k plus 1.

And we're integrating over some radius s.

For any s less than R.

We'll get you a similar fact for Laurent series.

Here's the theorem.

If f is analytic in an annulus, and the inside radius is little r and

the outside radius is upper case R, and it's centered at this point z0,

at which we know nothing, so f is analytic in this annulus, but that's all we know.

Then we know f has a Laurent series expansion, our first theorem and

we know actually more.

The aks can be computed as 1 over 2 Pi i,