Laurent Series Explained | How to Determine Laurent Series | Complex Analysis #9

- [Instructor] Hello, and welcome.

In this video we're going to talk a bit about

Laurent Series.

We'll start with the theorem, and then we will continue

by determining all possible Laurent Series

for each function here below.

So let's get going.

The theorem tell us that if a function is analytic

on some annulus domain, then we know that the function

is going to be equal to its Laurent Series on this domain,

and the Laurent Series is in fact made up of

two separate series:

The first one, which include all the positive powers

of z minus z-naught is called an analytic part,

since this part will always be analytic,

while the second one contains all the negative powers

of z minus z-naught, and is called a principal part.

And the coefficients in the Laurent Series can be determined

by using this formula here, and here c is a closed curve

inside the annulus surrounding z-naught, but you don't

really have to remember this formula,

since we never really use it.

Instead we manipulate known geometric series

to determine the coefficients, just like we did

in the case with Taylor series.

But what is an annulus domain?

An annulus domain looks something like this.

It's all the z values which lies between two circles.

The inner circle is determined by the radius R-one.

While the second circle is determined by radius R-two.

And this is a typical annulus domain.

Looks pretty much like a doughnut, right?

But our annulus domain doesn't need to look like this,

we can have one case where R-one is equal to zero,

then the domain contains all the z values inside the

circle R-two except z-naught.

But we can also have that R-two is equal to infinity.

Then the domain would be all the z-values outside

the circle R-one, but we can also have these two cases

at the same time.

And then the domain would look something like this.

Which means that the domain would include

all the z values on the whole plane

except z-naught.

And by using this annulus domain we're able to

capture, isolate the single lattice, and that is why

Laurent series, unlike Taylor series, still works

if the point you are expanding about, z-naught

is an isolated singularity.

And some other good things to know, is that

the coefficient a minus one is called the Residue

of the function f at z is equal to z-naught.

And this residue will be really useful

later on in the course when we're starting

to do contour integrations.

And note that if you find that a function is

analytic on the whole disk, then all the coefficients

for the negative powers of z minus z-naught

will be equal to zero, and that means that the

Laurent series is going to be reduced to a Taylor series,

and this comes from a fact that the only thing that

separates a Taylor series and a Laurent series

is that a Laurent series also contains this principal part.

And the last thing I would like to touch on is that

these two parts of a Laurent series

converges on different places in the complex plane.

The analytic part converges for all z values

inside the big circle R-two.

While the principal part only converges for z values

outside the smaller circle, R-one.

And that is why the sum of these two series,

which makes up our Laurent series will only be

valid for z values in between these two circles.

Let's continue by doing some examples.

So here I would like us to determine all possible

Laurent series for each function here below.

And we can do that by using geometric series,

because we know that the Laurent series for the function

f around some point z-naught should look

something like this.

And we can get something that looks pretty much exactly

like that by using these two geometric series here above.

And here I can also note that the second of these

two geometric series is in fact only a special case

of a first series, since if you insert one divided

by W as the W in the first series, you get the second one.

So if we know for some constant b, insert b divided

by z minus z-naught as our one divided by W,

in the second geometric series, we get the following.

And this thing will only be valid if the absolute value

of b is smaller than the absolute value of z minus z-naught.

And if we now for some constant c insert

c times z minus z-naught as our W in the first

geometric series, we get the following.

And this thing will only be valid if the absolute value

of the z minus z-naught is smaller than one

divided by the absolute value of c.

And here we can see that if we take the sum

of these two series, we get something that looks

pretty much exactly the same as our Laurent series.

And the sum will only be valid inside some annulus

determined by our coefficients b and c.

So in short I'm saying that if we use these two

geometric series and choose our constants b and c smart,

we will be able to create whatever Laurent series

we would like.

So let's continue with our first example.

And the easiest way to determine all possible Laurent

series is to mark all the important points

on a graph, so we know that we are expanding about

the point z is equal to zero, and we also know

that the function has a pole at z is equal to two.

And that means that we can create two separate domains,

one inside the circle and one outside of it.

And each of these two domains will have

their own Laurent series.

For our first domain, we would like to use

the first geometric series, since the series

must be valid inside the circle.

So we can start by factoring out the minus one,

to switch places on the numbers on the denominator.

And then we can continue by factoring out the half

to create a one in the denominator.

And now we can use our first geometric series

with W is equal to z, and c is equal to a half,

which will give us the following.

And by moving one divided by two from outside

to the inside of the sum, we get our final expression.

And this thing will only be valid if the absolute

value of z is less than one divided by one half,

which is the same as saying that

the absolute value of z must be less than two,

which is exactly what we wanted.

For our next domain it will act to use the second

geometric series, since the series must be

valid outside of the circle,

and to do that we have to make our function

look a little bit more like a function in the formula.

If we start by factoring out one divided by z

we get the following.

And now we can actually use the second geometric series

with W is equal to z and b is equal to two.

And this expression can be simplified to the following

by factoring in one divided by z.

But if you remember the principal part which

this thing corresponds to, should start at

N is equal to one and not N is equal to zero.

And we can easily do that by using this neat trick.

Let's say that we change the variables.

Let's say that k is equal to N plus one.

Then the sum would go from k is equal to one, to infinity,

and the expression inside would be two raised

to the power of k minus one, divided by z raised

to the power of k.

And now the trick is that k is only a name

for a variable.

And we can use whatever name we would like,

so I can rename it now to N and we get

exactly what we like to have.

And this sum will only be valid if

the absolute value of z is bigger than two,

which is exactly what we wanted.

In our next problem we will have a same function

but you are expanding about a different point.

This time we're expanding about z is equal to one,

which means that we will get two different domains

when we mark the important points.

And for first domain we like to use

the first geometric series again.

But this time we like to have that W is equal to

z minus one, since we're expanding about

point z is equal to one this time.

And we could do that by breaking up R minus two

in the denominator to minus one, minus one,

and then we can just factor out the minus one

to switch places on the terms in the denominator.

And now we can once again use our first

geometric series with W is equal to z minus one.

And the constant c is equal to one.

This gives us a series which is valid

inside the right domain.

And for our next domain we would like to write

the expression in the form of a second geometric series,

but since we still would like our W to be equal

to z minus one, we can start exactly the same

as before, and by now factoring out one divided

by z minus one, we can once again use

our geometric series.

And in this case we have that W is equal to

z minus one, and our constant b is going to be

equal to one, so that means that we get

the following expression.

And this expression can be simplified as the following.

And to get it to start with N is equal to one instead

of zero, we just have to decrease all the N

inside the sum by one since we are raising our

starting point by one.

And this sum will be valid outside of the circle,

like we wanted.

For our last example, we start by marking the important

points, which are the point of expansion,

z is equal to zero, and the poles for our function

which are one and two, and here we can see

that we can create three different regions.

One inside the smaller circle,

one between the two circles,

and one outside of a big circle.

So in all cases where you don't really see

how you can use a geometric series by the way,

I recommend you to always start by rewriting the

expression with the help of partial fraction expansion

so that you can determine the Laurent series

by observing each of these two terms,

one at a time.

For our first domain, we already determined earlier

that one divided by z minus two is equal to the following.

That since this is valid for z values inside the circle

with radius two, we know that it will also work

in our domain here.

And for our second term, one divided by z minus one,

we can rewrite it with the help of the first

geometric series by factoring out minus one

and then by using the first geometric series

we get the following expression.

And if we now add these two terms together again,

we get that the function's Laurent series

is equal to the following.

And since the sums start at the same points,

we can just add them together

and get the following results which will be valid if

the absolute value of z is less than one

since this is the only place in the complex plane

where both of these two series are valid

at the same time.

So, until now, the Laurent series we have determined

have only contained either the analytic part

or the principal part, but in this second region here

we will get them both at the same time.

And the trick is to let one of these two terms

up here be equal to the principal part

while letting the other term be equal to

the analytic part, and which one to choose to what

comes from experience.

But in most cases you will see that one way

is absolutely more easy than another.

Like in this case here we can still use that

one divided by z minus two is equal to the following.

So this term can correspond to the analytic part

we are searching for.

And if we now try to rewrite our second term,

so that corresponds to our principle part

we are all done.

So one divided by z minus one, can be rewritten

as the following, by factoring out one divided by z,

and if we now use our second geometric series

we get the following.

And by simplifying this expression a bit, we see

that they have managed to find the principal part

for a Laurent series, so if we just add these two parts

together, we get something that is valid inside

these two circles, which is exactly what we wanted.

For the last domain, we know that one divided

by z minus two from earlier is equal to the following.

And we can still use this expression for this bit here.

So the only thing we really have to do

is to determine the Laurent series for one divided

by z minus one.

That is valid in this domain, but we have actually

already done that, we did that in the last example.

So we already know that this thing

is going to be equal to the following.

And if we just add these two results together

we get that the Laurent series for the function f

is going to be equal to this thing here,

which only is valid if the absolute value of z

is bigger than two.

Thanks for watching.

(gentle acoustic folk music)