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The Residue Theorem

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Introduction to Complex Analysis

Universidad Wesleyana

4.8 (594 calificaciones) | 34K estudiantes inscritos

This course provides an introduction to complex analysis which is the theory of complex functions of a complex variable. We will start by introducing the complex plane, along with the algebra and geometry of complex numbers, and then we will make our way via differentiation, integration, complex dynamics, power series representation and Laurent series into territories at the edge of what is known today. Each module consists of five video lectures with embedded quizzes, followed by an electronically graded homework assignment. Additionally, modules 1, 3, and 5 also contain a peer assessment. The homework assignments will require time to think through and practice the concepts discussed in the lectures. In fact, a significant amount of your learning will happen while completing the homework assignments. These assignments are not meant to be completed quickly; rather you'll need paper and pen with you to work through the questions. In total, we expect that the course will take 6-12 hours of work per module, depending on your background.

Destrezas que aprenderás

Power Series, Complex Analysis, Mapping, Optimizing Compiler

Revisiones

4.8 (594 calificaciones)

5 stars
510 ratings
4 stars
75 ratings
3 stars
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NS

Jun 25, 2018

The prof makes it easy to understand yet fascinating. I enjoyed video checkpoints, quizzes and peer reviewed assignments. This course encourages you to think and discover new things.

DS

Nov 15, 2018

it was very challenging course , not so easy to pass the assignments but if you have gone through lectures, it will helps a lot while doing the assignments especially the final quiz

De la lección

Laurent Series and the Residue Theorem

Laurent series are a powerful tool to understand analytic functions near their singularities. Whereas power series with non-negative exponents can be used to represent analytic functions in disks, Laurent series (which can have negative exponents) serve a similar purpose in annuli. We’ll begin this module by introducing Laurent series and their relation to analytic functions and then continue on to the study and classification of isolated singularities of analytic functions. We’ll encounter some powerful and famous theorems such as the Theorem of Casorati-Weierstraß and Picard’s Theorem, both of which serve to better understand the behavior of an analytic function near an essential singularity. Finally we’ll be ready to tackle the Residue Theorem, which has many important applications. We’ll learn how to find residues and evaluate some integrals (even some real integrals on the real line!) via this important theorem.

The Residue Theorem17:12

Impartido por:

Dr. Petra Bonfert-Taylor
Former Professor of Mathematics at Wesleyan University / Professor of Engineering at Thayer School of Engineering at Dartmouth

Transcripción

Welcome to the third lecture in the seventh week of our course,

Analysis of a Complex Kind.

Today we'll learn about a really powerful theorem, namely the residue theorem.

Let's start with the motivation.

Let me remind you about isolated singularities.

Remember that if f has an isolated singularity at a point z0, that means

f is analytic in some disk centered at z0 with the exception of z0.

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.

Let's look at some examples.

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.

But we notice these are all even powers of z.

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.

Therefore, the residue of f at -i is -1 over 2i, which is one-half i.

Here finally is the residue theorem,

the powerful theorem that this lecture is all about.

Suppose D is a simply connected domain, and suppose f is an analytic function

in this simply connected domain, except for isolated singularities.

So here's my domain D,

and I denoted the isolated singularities with these little x's.

Let C be a simply closed curve in D.

And suppose z1 through zn are those isolated singularities

of f that lie inside of C.

This curve should not pass through any of the singularities.

So here's my curve C, and

it surrounds these two isolated singularities that I've denoted z1 and z2.

Then the theorem says the integral of f over this curve C = 2pi

i times the sum of the residues of f at the points zk that are inside the curve C.

In the particular example I drew here, we would be simply getting 2pi i

times the residue of f at z1 + the residue of f at z2.

We've seen something like this already.

We notice that if we integrate a function with an isolated singularity around

the circle centered at the isolated singularity and inside the region where

the function is analytic, then what we get as the integral is actually exactly

2 pi i times the residue of the function at the isolated singularity.

So the residue theorem is an extension of this fact that we just discovered.

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.

In order to be able to really take advantage of this very powerful theorem,

we'll need some more strategies and

techniques that will help us to calculate residues.

So far, we've been calculating entire Laurent series

in order to find the residue of a function at a singularity.

But there are actually other ways to find residues, and

that's what we'll learn about in the next lecture.

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