0:31

Then f has a Laurent series expansion in that punctured disc, and

the Laurent series expansion looks like this.

You have coefficients ak and powers (z-z0) to the k, for

k from negative infinity to infinity.

And the series converges locally uniformly in this punctured disk.

And it converges uniformly soon as we stay away from the boundary of the that disk.

Now notice the following, what happens when we integrate f over a circle of

radius rho centered at z0, where rho is a little bit less than r?

So this red curve that I just drew is a compact set inside the punctured disk,

and therefore the convergence of the series that represents f is uniform

on that curve.

That means when we're integrating f over this curve,

we're really integrating the sum over the curve, but because the convergence is

uniform, we can interchange the summation and the integration.

That is typically not the case, but in a uniform convergence,

you're allowed to interchange.

And therefore, we find ourselves with the sum on the outside and

the integration on the inside.

That makes it quite easy to evaluate the integral over f(dz),

because now all we have to do is find all these individual integrals that

have nothing to do with f itself.

We simply have to find out what is the integral over (z-z0) to the k over

the circle of radius rho centered at z0.

We distinguish two pieces, namely the case where k is not equal to negative 1 and

the case where k is equal to negative 1.

If k is not equal to -1, then this function (z-z0) to

the k actually has an antiderivative that is analytic

in the entire complex plane with the exception of z0.

And the antiderivative is 1 over (k + 1) (z- z0) to the k + 1,

because when I differentiate this function on the right,

the k + 1 comes down, cancels out with the 1 over k + 1,

then the exponent becomes a k, which is exactly what I wanted.

The only time when that doesn't work is when k is equal to negative 1,

because that would be dividing by 0 in that case.

So we have an antiderivitive, which means the integral,

because we're integrating over a closed curve, amounts to being 0.

When k is equal to negative 1, however, I don't have an antiderivative of that type,

and therefore the integral needs to be evaluated in a different way.

We can certainly use the Cauchy integral formula, which is complete overkill, but

we can also calculate the integral directly.

Let me remind you how to find such integrals.

What we need is, we need to parameterize the curve over which we want to integrate.

We're integrating over a circle of radius rho centered at a point z0.

One way to parameterize such a circle is with

a parameterization gamma of t equals z0 plus rho e to the it.

We'll need the derivative of this parameterization,

and it is again the f prime of t equals rho i e to the it.

Now we simply plug that in.

So instead of z, I write z0 plus rho e to the it and subtract the z0, and

now I need to multiply where the derivative of the parameterization.

And you'll notice that a lot of things cancel out.

In the denominator, the z0 goes away and I'm left with rho e to the it,

and now even the rho e to the it cancels out, so

all I'm left with is the integral from 0 to 2 pi of idt, and that equals 2 pi i.

So when k=-1, the integral that I'm interested in equals 2pi i.

Now remember what we actually were trying to find.

We're trying to find the integral over f(z)dz where f

had an isolated singularity at z0.

So f was analytic in the punctured disk punctured at z0

of some radius r with an isolated singularity at z0.

And we had looked at this Laurent series expansion, and

we were able to bring the summation to the outside and the integration to the inside.

And we're now left with this integral z-z0 = rho over z-z0 to the k dz.

We saw, that unless k = -1, this integral is not even there.

So this whole integral,

we just realized is equal to 0 as long as k is not equal to -1.

And so there's only one term left in this sum, namely the term where k = -1.

When k = -1, the integral equals 2 pi i, and the only thing that's then

left is the coefficient here, namely the a-1 coefficient, in that case.

And so we end up with 2 pi i times a -1.

In other words, if I integrate a function with an isolated singularity,

around that singularity for the curve that's small enough to fit into the disk

with a function that's otherwise analytic, then I get 2 pi i times a -1, where

a -1 is the coefficient of the 1 over z-z0 term in the Laurent series expansion.

So, therefore this a -1 gets very special attention, and that's why I've been

having you calculate a -1 in so many examples in the previous in-video quizzes.

This a -1 is so special that it even gets a name.

If f has an isolated singularity at

z0 with this Laurent series expansion right here,

then the residue of f at z0 is exactly this a -1 term.

So this a -1 term is called the residue of f at z0.

6:20

We already calculated the Laurent

series expansion of the function 1 over z-1 times z-2.

We know this function has isolated singularities at z = 1 and z = 2.

And so there's lots of different ways to find any y in which the function

is indeed analytic, in which we can find a Laurent series expansion.

If we were interested in the residue of this function at the point 1, then we

need to find a Laurent series expansion in a punctured disk centered at 1.

For example, this one right here.

So that's the disk we used, and we found that in that disk, f has a Laurent

series expansion which is -1 over z-1 times the sum that you see right here.

The sum has only positive powers of z-1.

The only negative power of z-1 is right here, and

since it's the power z-1 to the -1, you see the coefficient a -1 right up here.

In other words, a minus 1 equals negative 1,

which means the residue of f at 1 is negative 1.

If, on the other hand, we wanted to find the residue of this function

at the isolated singularity at 2, we need to find the Laurent series expansion in

the punctured disk centered at 2.

Again, we do a partial fraction decomposition,

we find that the Laurent series is 1 over z-2,

minus the sum right here which contains only positive powers of z-2.

And therefore this term here represents the term for k equals negative 1, and

therefore the 1 right here, that's my a negative 1 for this Laurent series.

So the residue of this function f at the point 2 is 1.

So again, here's the procedure.

If you want to find the residue of a function at an isolated singularity,

what you need to do is you need to find the Laurent series expansion

centered at that singular point, and then read off the a -1 term.

Here's another example, the function f(z) = sin z over z to the 4th.

We saw that this function has a pole, x = 0.

So it's an isolated singularity.

And we found this Laurent series to be 1 over z cubed + -1 over 3

factorial times 1 over z, and so forth.

So here, we have two terms with negative powers of z,

and this term in particular is the k = -1 term.

And this is the coefficient a-1 right there,

because it's the coefficient in front of the 1 over z term.

And therefore the residue of f at 0 is -1 over 3 factorial,

which is minus one-sixth.

Here let's do another example, the functional cos(1/z).

We saw this function as an essential singularity a 0.

So the Laurent series expansion has infinitely many terms with

negative powers of z.

9:16

There is no 1 over z term.

Therefore the coefficient of the 1 over z term must be 0.

And therefore the residue of this function at 0 is equal to 0.

How about sin(1/z)?

That function also has an essential singularity at the origin.

But the Laurent series expansion has all odd powers of z.

And indeed, there is a 1/z term in the Laurent series expansion, and

therefore the residue of f at 0 is equal to 1.

Now, remember the function cosine of z-1 over z squared.

We noticed that that function has a removable singularity at the origin,

because when you look at the Laurent series expansion,

there's no negative powers of z left.

Therefore, again there is no 1 over z term, and the residue of f at 0 equals 0.

And here's a final example.

The function 1 over z squared + 1.

The denominator factors into z-i times z + i.

And again we can do a partial fraction decomposition.

I'm going to do it the similar way that I've showed you a few lectures ago.

I'm going to write the numerator is (z+i)-(z-i).

The z's cancel out with each other and i minus minus i is 2i.

So my numerator is now 2i instead of 1.

But if I divide by 2i,

the new 2i in the numerator cancels out with the denominator 2i.

The reason I did this is so that I can cancel things out again.

I can now pull the fraction apart at the negative sign,

and I notice that when I look at the first fraction, the z + i term cancels out.

And I'm left with 1 over z-i.

And if I look at the second fraction, the z-i terms cancel out.

And I'm left with 1 over z + i.

1 over 2i's in front of it, and so

here's my partial fraction decomposition of the function f.

This function has isolated singularities at z = i and z = -i.

We want to find the residue at each of those points.

So near i, I need to write this function in terms of powers of z-i.

I already have 1 over 2i times 1 over z-i.

The rest represents something that is analytic near i.

I don't even have to bother finding the exact representation.

I know an analytic function has a Taylor series expansion,

and so I can write this analytic function as a Taylor series,

okay, from 0 to infinity, ak (z-i) to the k.

And together,

these two must form the Laurent series, by uniqueness of the Laurent series.

Therefore, the term for k equals negative 1 is the one here up front.

And a-1, this 1 over 2i, which equals minus one-half i.

Similarly, when I look at f near -i,

my goal is to write f as a Laurent series with powers of z + i.

I already have one such power of z + i right here,

together with its negative sign of course, and the 1 over 2i in front of it.

So that you can see right here.

And the rest is an analytic function near -i,

which again will have a power series representation

ak (z + i) to the k, for k from 0 to infinity.

Together, the series and the first term from the Laurent series

expansion of 1 over z squared + 1 near -i, and therefore,

this must be my a -1 term for this particular Laurent series.

14:35

Here's an example.

Let's again look at the function 1 over z squared + 1,

which is analytic in the entire complex plane with the exception of

the isolated singularities at i and -i.

I drew a bunch of curves, C1, C2, C3, and C4 in here.

Suppose we first integrate over the curve C1, which is the red curve.

By the residue theorem, I need to find 2 pi i times the residue of the function at

the point that's surrounded by the curve, which in this case is just the point i.

The only point that's surrounded by the curve C1.

And we already found that residue, it's minus one-half i.

So all together, the 2's cancel out.

The i's multiply to -1, cancel out the other -1, so the residue is equal to pi.

If I use the curve C2 instead, then this curve C2 surrounds the point -i.

And therefore I need to look at 2 pi i times the residue of f at -i,

which we also found, it's one-half i.

And I find the integral equals -pi.

Now if I integrate over the curve C3,

I notice that curve surrounds two isolated singularities, namely i and -i.

So if I integrate around C3,

I need to look at 2 pi i times the sum of those two residues.

But the two residues are minus one-half i and one-half i, and

they cancel each other out.

So together I get that the integral over C3, f(z)dz = 0.

Finally, let's look at this blue curve right here.

It doesn't surround any singularities, and so the sum is empty,

and therefore I get the integral over C4 = 0.

Here again is this very powerful theorem that we just learned about.

In a simply connected domain in which f is analytic except for

isolated singularities, with a simple closed curve which we're integrating,

we can find the integral over f, over this curve, is 2 pi i

times the sum of the residues of those singularities that are surrounded by C.