0:32

And if you can find a disk centered at z0 such that

f is analytic in the whole disk with the exception of z0.

If such a disk exists, then z0 is called an isolated singularity of f.

Let's look at some examples.

The function 1 over z is analytic everywhere except at 0 because

we can't divide by 0 and therefore, this function has an isolated singularity at 0.

The function 1 over sin z has isolated singularities

whenever we're dividing at 0.

So whenever the sine function is 0,

and that's at pi,- pi, 2 pi,

-2 pi, and so forth.

1:17

And so this function has lots of isolated singularities.

It has one at 0 and I can draw a disk around 0 for example, for

example this one, so that 0 is a singularity but

the function is analytic everywhere else in that disk.

Or I can draw a disk around 2 pi.

1:39

So this function 1 over sine z has isolated singularities at 0 +

-pi + -2pi and so forth.

For the functions square root of z and logarithm of z and functions like that,

the situation is a little bit different.

Recall that these functions are analytic in their complex plane, with the exception

of the negative real access which we had to take out in order to make things work.

So if I look at the origin

2:05

right here clearly the function's not analytic at the origin.

But it still cannot draw a disk around the origin, such that everywhere in

the disk with the exception of the origin of, the function is analytic.

Because I'm still having trouble with this portion of the negative real axis that's

also sitting in that disk.

No matter how small I draw this disk, things aren't going to work out.

These functions do not have isolated singularities at the origin.

The problem is much bigger at the origin.

The function 1 over z- 2 again has an isolated singularity and the isolated

singularity this time is where the denominator is 0 and that's at z0 = 2.

So now I can draw any disk centered at 2 no matter how large and

the function is going to be analytic with the exception of 2.

2:57

Now if I have a function with an isolated singularity at z0, so

f is analytic in an annulus, the inside component is very small as radius 0 and

the outside component is r.

Then by Laurent's theorem, f must have a Laurent series expansion in that annulus.

3:16

And the Laurent series expansion, we learned about last class, is of the form.

It's an infinite sum, I have coefficients ak and then power (z- z0) to the k.

And the sum goes from negative infinity to positive infinity.

And so I have a part of this series that looks just like a Taylor series.

Only positive powers of z- z0.

But I also have negative powers of z- z0.

And that's called the principal part of this Laurent series.

3:47

Now basically three fundamentally different things can happen that

influence how f behaves near z0.

And we're going to look at the examples first that demonstrate these three

fundamentally different things that can happen.

So here again you see the Laurent series expansion of f.

Remember that f has an isolated singularity at a point z0.

And so we have this disk of radius r such that f is analytic in the punctured disk.

Here's some examples.

Consider the function f (z) is cos(z)- 1 over z squared.

Remember that cosine z has a Taylor series expansion,

1- z squared over 2 factorial + z to the 4th over 4

factorial- z to the 6 over 6 factorial and so forth.

If I subtract 1 from this function, then this 1 goes away.

4:40

And if I divide that by z squared, then I find myself with 1 over z squared

times exactly this series expansion I found up here.

Now if I actually bring the z squared inside, and

this z squared cancels with that z squared leaving us with -1 over 2 factorial.

For the z to the 4th this becomes a z squared, so

I get +z squared over 4 factorial and so forth.

5:12

So my Laurent series only has that Taylor series looking part to it.

The principal part doesn't even exist in this particular example even though

the function f clearly has an isolated singularity at the origin,

because I can't divide by 0.

Let's look at another example.

The function cosine z over z to the 4th.

So, you cannot have my my series for cos z.

And we're going to divide that by z to the 4th.

I'm going to distribute this z to the 4rth into the parenthesis list, so

I find 1 over z to the 4th- z squared over z to the 4th.

So -1 over 2 factorial z squared.

And then z to the 4th over z to the 4th, those cancel out, so

I have a +1 over 4 factorial and so forth.

5:57

In this example, I have some negative powers of z and

then lots of positive powers of z.

So this one is clearly a negative power of z, z to the negative 4th.

This one is also a negative power of z.

But the rest, they're all non negative powers of z.

6:23

The function cos 1/z.

How do we get the Laurent series for that?

I start with a Taylor series for cosine.

And I replace every occurring with z with 1/z.

That definitely gives me a series expansion for cos 1/z.

And by the theorem that the Laurent series is unique,

it must be the Laurent series of cos 1/z.

So all I have to do is look at my series for cos z.

6:54

And simply plug in 1 over z for every z.

So I have 1, and then instead of z squared over 2 factorial,

I have the 1/2 factorial and the z squared becomes 1/z squared.

In the next term, the z to the 4th becomes 1/z to the 4th.

7:16

And so here, it looks like the powers of z get more and more negative as I go on.

So I here I have infinitely many negative powers of z.

So here you see three fundamental differences and

the behavior of these three functions at the origin.

The first function had no negative powers of z.

The second one had finitely many negative powers of z.

And the last one had infinitely many negative powers of z.

This very much influences how the function behaves near the isolated singularity.

And so that's why we use this distinction to classify isolated

singularities into three types.

Here's the classification.

Suppose z0 is an isolated singularity of an analytic function f.

And so that means,

we have a Laurent series expansion in a punctured disk around z0.

We see this singularity is removable,

if all the coefficients belonging to negative powers of z

minus z0 are actually equal to 0, so ak = 0 for all k < 0.

And those are the k's that give rise to negative powers of z- z0.

8:30

We say the isolated singularity is a pole,

if I have finitely many negative powers of z- z0.

So in other words, there exists a largest negative power.

So if there exists an N, such that a -N is non-zero, so

that negative power occurs, so I have an a- N term in my Laurent series expansion.

And then, we know for the powers to the right of them, but nothing to the left.

That's where things stopped.

So ak = 0 for k less < -N.

The index N is also called the order of the pole.

And finally, we say the singularity is essential if we have

infinitely many negative powers, fz minus z0.

9:21

So here again, a table that illustrates the definition.

We say that z0 is removable if I have no negative powers of z, nothing.

It's a pole of order N If the most negative power of

(z- z0) is (z- z0) to the -N.

We say it's a simple pole,

as this is a special case of a pole, if the order of the pole was actually 1.

So, the term for this to the left is the a-1 term.

And the singularity is essential if I have infinitely

many terms that have negative powers of z- z0.

Let's study removable singularities in a little bit more detail.

So remember, a singularity is called removable if,

once I form the Laurent series of the function centered at z0,

I actually have no negative powers of z minus z0.

So, all the aks are equal to 0 for k < 0.

Here's another example.

The function sine z over z.

Remember the Taylor series for sine z was z-

z cubed over 3 factorial + z to the fifth,

over 5 factorial and so forth.

If I divide that by z, I find myself with z/z which

is 1- z squared over 3 factorial + z to the fourth over 5 factorial and so forth.

10:50

Since this is a series of extension of sine z/z, and since the series expansion

is unique, this must be the Laurent series of sine z / z centered at the origin.

But this Laurent series look like a Taylor series.

It has no negative powers of z in it.

11:11

I can in fact plug in, z = 0 into this series, and

when I plug in z = 0 into this series, all these terms go away.

What I'm left with is 1.

So if I define f(z) to have the value of 1,

f(z) = 0, then I have a function that's defined, and

has a Taylor series in a whole disk, and it becomes, therefore, analytic.

So the function that is defined as sine z over z when z is non-zero and

1 when z = 0, we make that into an analytic function.

We have removed the singularity that sine z over z had at the origin.

11:51

Here's the theorem.

Supposed this 0 is an isolated singularity of f,

then z0 is a removable, if and only if f is bounded near z0.

We just saw that, once we've removed the singularity,

we're left with an analytic function.

An analytic function is clearly bounded near z0 but

the theorem says that this goes both ways.

Once you have determined that you have an isolated singularity and

the function happens to be bounded nearby,

you know automatically it happens to be a removable singularity.

Next, let's look at poles.

Remember that z0 is a pole of order N of a function f,

if the Laurent series of f that's centered at z0,

satisfies that the a-N term is non-zero, but

any other ak's for k < -N are equal to 0.

Here's another example, sine z over z to the 5th.

12:59

1 over 3 factorial times 1 over z squared + 1 over 5 factorial and so forth.

The most negative power of z is 1 over z to the 4th.

There's no 1 over z to the 5th or 1 over z to the 6th or anything like that.

And so, we have a pole of order for N 0, for this function.

13:23

Isn't there a how to recognize poles?

Suppose z0 is an isolated singularity of a function f, then z0 is a pole, if and only

if, the absolute value of the function goes to infinity as z approaches z0.

We can look at that in our example.

13:41

Sine z over z to the 5th, we see its Laurent series expansion right here.

As z goes to 0, 1 over z to the 4th goes to infinity.

And it's the term that dominates the entire series.

All these parts here behave nicely as z goes to 0, and

these two terms go to infinity, but 1 over z to the 4th goes to infinity much

much faster than 1 over z squared, and so it dominates this series.

The 1 / z squared can't make up for the what the 1 / z to the fourth does.

And so this function goes to infinity as z goes to z0.

Again, this goes both ways.

If you have a pole, the function goes to infinity, but also, if the function goes

to infinity as we approach z0, then you must be dealing with a pole.

14:29

By the way, if f has a pole at z0, then you could look at 1 over f,

and then that function will have a removal singularity at z0 and vice versa.

Why is that the case?

Well, if f is a pole,

we know that means that the absolute value of f goes to infinity, as z approaches z0.

Our Riemann's theorem, we learned about removable singularities.

If I look at 1 over f(z), because f(z) goes to infinity,

1 over f (z) must go to 0, and that's clearly a dominant function here, is z0,

in Riemann's theorem that means I'm dealing with a removable singularity.

Finally, let's look at essential singularities.

Remember that an essential singularity means that the Laurent series of f

has infinitely many negative powers of z minus z0.

Here's an example, f (z) equals e to the 1 over z.

How do I find the Laurent series?

We remember the Taylor series for the exponential function.

e to the w = 1 + w + w squared over 2 factorial

+ w cubed over 3 factorial, and this is actually true for any w.

So this is the Taylor series centered 0.

If you plug in w =1 over z,

then we find e to the 1 over z = 1 + 1 over z.

We simply replace every w with a 1 over z.

So 1 over 2 factorial z squared and so forth.

And so that's the series that you see right here.

Fully there are infinite negative powers of z in that series.

Again, because the Laurent series is unique and because it just found

one series this must be the Laurent series for e to the 1 over z.

So we see that this series has infinitely many negative powers of z,

therefore, e to the 1 over z has an essential singularity at the origin.

Let's look at what happens when z approaches 0.

What happens to e to the 1 over z?

Let's start with zs approaching the origin along the x- axis.

And in fact, why don't we start by approaching from the right?

16:37

So I'm plugging in positive values of x.

What happens to e to the 1 over x?

Well, as x goes to 0, 1 over x goes to infinity.

Because x was positive, it goes to positive infinity.

If I plug in larger and larger exponents into the exponential function,

the exponential function simply goes to infinity.

17:01

Now let's do the same thing but from the left.

Still take real values on the real axis but approach the origin from the left.

So here I have x values that are less than 0.

This time, because x is negative, 1 over x goes to negative infinity.

And, therefore, the exponential function of 1 over x goes to 0.

Remember the exponential function looks like this.

17:25

And so when I plug in very, very small negative numbers,

exponential function is practically equal to 0.

Finally let's plug in values on the imaginary axis.

So let's see what happens as I approach the origin on the imaginary axis.

So these are points of the form ix.

Then f(z) is e to the 1 over ix to e to the 1 over ix is

the same as e to the minus i over x.

The exponent is purely imaginary and

I know that all of these numbers, with purely imaginary exponent,

they all lie in the circle of radius 1, centered at the origin.

18:17

I have seen sequences for which the function goes to infinity.

Other sequences for which the function goes to 0 and yet

other sequences for which the function bounces around in the unit circle.

So, there's no unified behavior whatsoever and

therefore f does not have a limit as z approaches z0.

And that's the typical behavior for an essential singularity.

18:40

Here's the theorem.

Suppose that z0 is an essential singularity of f.

Then for every point in the complex plane there exists this sequence zn converging

to the essential singularity such that the function values converge to w0.

What does that mean?

I'm going to draw two complex planes, one for my z's and one for my f(z)'s.

Pick any point over here in the w plane, any point w0.

Here's my essential singularity.

The theorem says that you can now find points zn that converge to z0 such

19:26

And this is possible for any w0 that you choose.

In other words, not only does the function f not have a limit as z approaches z0,

it's as spread out as it it gets.

Let's look at an example.

Suppose f is the function e to the 1 over z.

We know that f has an essential singularity at the origin.

Let's pick a point w0, say 1 + root 3 i.

Here's another picture.

So by the theorem of Casorati–Weierstrass

there must exist points zn approaching z0 somehow,

such that their images approach 1 + root 3i.

Let's see if you can find such points.

In fact, we can find zn's such that e to the 1 over zn which is f(zn)

gets not only close to 1 but actually equals 1+ root 3i.

So how do we do that?

Well it's pretty simple, the logarithm comes to our help.

We simply need then 1 over zn is a logarithm of 1 + root 3i.

And remember, the logarithm is a multi valued function.

The logarithm of a point z is the natural logarithm of

the absolute value of z + i times an argument of z.

Which means four point z in the complex plane, if we wanted to find its logarithm,

we need to find its distance from the origin.

That's the absolute value of z.

Then we need to find an argument.

The argument of z is this angle plus a multiple of 2pi.

We also spoke of the principal branch of the logarithm.

Which we denoted with an uppercase L, which was the natural log of the absolute

value of z + i times uppercase Argument of z.

And the uppercase Argument is an especially chosen argument,

21:30

Just as an example, the logarithm of 2i.

So that's the number right here, would be the natural log of the length.

So the distance of 2i from the origin + i times an argument of 2i.

And now we know the distance of 2i from the origin, that's 2, so

that's the natural log of 2.

And the angle that 2i forms with a positive real axis,

it's a 90 degree angle.

So in radians, that is pi over 2.

But because we're talking about lowercase argument here we have to take into

consideration all possible arguments.

So possibly 2pi n get's added to that.

On the other hand if we talk about the principal logarithm of 2i, that

is equal to ln of 2 + i over 2 because pi over 2 is between negative pi and pi.

So now we want to find the logarithm of 1 + root 3i.

So we need to find the distance from the origin.

We need to find the length of 1 + root 3i.

But we simply find that by the Pythagorean Theorem.

It's the real part squared plus the imaginary part squared.

22:52

And so that's the square root of 1 + 3,

which is the square root of 4 and that's 2.

We also need to find the angle but

this is a special triangle that we're talking about.

This length here is 1 and this length is 2 and this length was root 3.

This was one of the special triangles.

This is a triangle where this angle right here is a 60 degree angle, and

that's pi over 3.

So we know the argument of (1

+ root 3i) = pi over 3 + 2pi n.

23:34

We know that the logarithm of 1 + root 3i is

the natural logarithm of the lengths, so

the natural logarithm of 2 + i times the argument.

And that's what you see right here.

So we want 1 over zn to equal this number.

So therefore zn needs to be the reciprocal of this number so

lets make that as our zn.

What happens to zn as n goes to infinity?

Well, as n goes to infinity the denominator gets larger and larger so

zn itself goes to 0.

So indeed we just found points zn that go to 0 as n goes to infinity.

We're hoping that f of these points will approach our special point,

w0 which we chose to be 1 + root 3i.

Let's check it.

24:42

E to the ln of 2, e and ln cancel each other out.

That's simply 2.

E to the I pi over 3 is next, an E to the 2 N pi I is equal to 1.

Remember, if we have a sum in the exponent, that becomes a multiplication

once we pull the exponent's function apart at the exponent.

And what is E to I pi over 3,

e to the i pi/3 is very similar to the example that I just looked at.

Pi/3 is that 60 degree angle.

And it's simply the number that we looked at earlier but

scaled down to have a length 1.

So we had looked at this number up here, 1 + root 3i.

And now this number gets scaled down to length 1.

So the number now is one half plus i times three over two.

We multiply that by this two right here and we're back to i number double this.

So even the 1 over n is m d double the 0.

Not only did we find zn's approaching our essential singularity

as used sterile such that the f(z) is converged to w0.

But more is true.

Our f(z) answer all equal to w0.

So we're showing you something that is even stronger than Casorati-Weierstra

wanted.

We observed a much stronger result that is also true, but

it's much harder to prove for essential singularities which is Picard's Theorem.

Suppose that z0 is an essential singularity of f.

26:15

Then for every w0 in the complex plane with at most one exception

you can find those sequences that we just found.

A sequence zn converging to the essential singularity.

Such that f of c n is equal to w0 for all n.

Here's the situation.

We have our essential singularity and for any point

w zero with at most one exception, so there's one point that might not work.

But for any other w zero in the complex claim

you can find the sequence of points zn converging

to z0 such that f(zn) = w0 for all n.

So they land right on top of w0 for all n.

27:01

Let's finish with an example.

The function e to the 1 over z has an essential singularity at 0.

And we already saw a sequence zn such that f(z) and z = 1 + root 3i.

And you could see that we could've done the same kind of construction for

any other point.

With one exception, either the 1 of z is not going to

be equal to 0 because the exponential function does not take the value 0 So

for 0, that wouldn't have worked.

w0 = 0 is the one exception in Picard's theorem.

But for any other w0, you could have done the same construction that we just did and

found a sequence zn converging to 0,

such that the f of zn's equal to the chosen point.

For example, take w0 = 1.

So we want z's such as f of z is equal to 1.

We want either the one over z to be 1, how can we make either the one over

z be equal to 1 whether that means 1 over z means to be of the form two n pi i and

so therefore z n could be chosen over 2n over Pi i.

Those zn's clearly go to 0, s m goes to infinity and f(zn) = 1 for all n.

In the next lecture we'll start learning about the resident theorem and

how it relates to functions with isolated singularities.