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.