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Welcome to our fifth lecture in the fifth week of our course,

Analysis of a Complex Kind.

Today we'll learn about consequences of Cauchy's theorem and integral formula.

We'll be able to pick the sum fruit of our labor.

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Let me remind you of Cauchy's theorem for simply connected domains and

also of Cauchy's integral formula that we studied during the last lectures.

Cauchy's theorem says the following.

Suppose D is a simply connected domain, so again,

that simply means it's an open, connected set.

So it's in one piece and it has no holes.

And let F be an analytic function as defined in D.

Suppose gamma is piecewise smooth closed curve in D.

So, gamma is some kind of curve and it starts and

ends at the same point, that's what makes it closed.

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By integrating over the curve gamma, the function f(z)/( z-w) to the k

+ first power.

Again, this is a very fascinating theorem, because when we evaluate this integral,

we're only taking account of values of f along the curve gamma.

However, those values are able to recover for us,

values of f and its derivative inside the domain, so inside gamma.

It's quite a remarkable theorem.

A first application of the Cauchy integral formula is Cauchy's estimate.

And this in turn will be very helpful in proving further results.

So suppose that f is analytic in an open set again.

And that open set contains a ball of radius r, including its boundary.

Okay, here is this open set, we can call U if you want to.

And there is this ball of radius r that fits into U, including its boundary.

So the boundary of the ball doesn't touch really the boundary of U.

So here's this point U0, and the ball has radius r.

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And suppose furthermore we know that the absolute value of f is

bounded above by some constant m on this boundary curve.

Then Cauchy's estimate says that for all k greater than or

equal to 0 the kth derivative of f at the center

point z0 is bounded above by k factorials.

So that's the same k as the derivative that we took here, times m, the constant

that bounds f on the circle of radius r, divided by r to the kth power.

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And the proof is actually quite simple and simply uses the Cauchy integral formula,

which I wrote down for you up here so you would remember what it looks like.

So, what we're going to do for the proof,

we're going to plug in z0, For this point w, right here.

And therefore find that fk of z0 is k factorial over 2 pi i,

times the interval over gamma f(z) divided by z- z0 to the k + 4's power dz.

So that's what we have right here, and there are absolute value signs so

we put those absolute value signs there as well.

The curve gamma, in our case, is the circle of radius r,

or counterclockwise, so I just wrote z- z0 = r right there.

But now we remember that we had some sort of triangle inequality for integrals.

In other words, we can pull the absolute value signs inside,

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at the expense of an inequality sign.

So, the absolute value of this integral is bounded above by the integral

over the absolute value of f(z) divided by the absolute value of z-

z0 to the k + first power, absolute value of dz.

And remember what the absolute value of dz means.

It just means that when we plug in the parametrization, we take the absolute

value of the derivative, gamma prime of t dt.

We don't even have to plug in any kind of parametrization.

What we observe next is, along the entire curve gamma,

no matter what t value I plug in to my parametrization,

this term z- z0 is always equal to r in absolute value.

Because my parametrization would put these z values all the way around to 0 on

the circle of radius r, but the absolute value,

the distance between 0 will always remain r.

Therefore, from this integral,

I can pull an r to the k + first power out of the denominator.

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See that right here?

It's no longer in the integral.

I keep the 2pi that I had over here and the k factorial right here.

Furthermore, in the numerator I know that f(z), by my assumption,

is bounded above by this constant m on the curve gamma.

And that's where this m comes from right there.

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And what I'm left with after doing all

that would be k factorial divided by 2pi times m.

That's my estimate for f over r to the k + 1 times the integral.

Over z- z0 = r, absolute value of dz.

I've pulled the rest out of the integral.

However, the integral that I look at right here,

that's the length of the curve gamma.

The curve gamma is the circle of radius r centered at z0,

the length of that's 2pir, and that's what you see right here.

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One of them is Liouville's theorem.

So here, I'm just reminding you of the estimate we just proved.

We proved that the kth of derivative (z0) is bounded

above by k factorial times m over r to the k.

Provided that we had the circle of radius r centered at z0,

that fits inside the domain in which f is analytic, and

that f(z) is bounded above by m,

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The consequence is Liouville's Theorem.

Suppose f is an analytic function in the entire complex plane.

We call those entire functions because it's analytic, not just in a subset,

but in the entire complex plane.

If this function f is bounded, then it must be constant.

Now this is a totally amazing theorem.

It says that there are no non-constant

analytic functions that are analytic in the entire complex plane.

As soon as you're analytic in the entire complex plane, you can't be bounded.

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That is very different from what we know to be true on the real line.

For example, we have the function sine of x on the real line, it's very bounded.

It just goes up and down and up and down, but

it never gets bigger than 1 and never gets smaller than -1.

So it's very much bounded on the real line.

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Because this function's sine of z extends to the complex plane as

an analytic function, we know that extension can no longer be bounded.

That means, there must be some direction in which the sine function

goes off to infinity in absolute value.

So that is quite remarkable.

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Since f is n of again the entire complex plane,

we can put any ball of any radius r around that point z0.

And that ball will be one of those balls that is allowable to apply Cauchy's

estimate, because the function f is analytic in the entire complex plane.

So no matter how large a ball I choose,

it'll always fit into the domain where f is analytic.

So now we can apply Cauchy's estimate,

2f in that ball, and we're going to apply it with k = 1.

When k=1, we get the first derivative of f at z0

is bounded above by 1 factorial times m over r to the first power.

In other words, we get f prime of z0 is bounded above by m over r.

But this is true for any r greater than 0, so

we can choose a really really large value for r.

As large as we want to, so we can in fact let r go to infinity.

And if we do that, this number on the right hand side here,

gets smaller and smaller and smaller, because m is a fixed number.

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We just divide this fixed number m by a really really large number,

this number will go to 0.

In other words, the derivative of f at z0 has to be equal to 0,

because it is smaller in absolute value than any number I can come up with.

Since z0 was an arbitrary point, we can repeat this at any other point in

the complex plane and find that the derivative of f is equal to 0 everywhere.

But an analytic function whose derivative is equal to 0, is a constant function.

So, if I have a function that is analytic in the entire complex plane and

bounded, then it must be constant.

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Let's look at an example.

Suppose f is an entire function, so we could write it as u + iv,

where u is the real part, v is the imaginary part.

And suppose that the real part is less than or equal to 0 for all z and C.

That already implies that f must be a constant function.

So not even the real part can be negative or a non-positive for

all z in the complex plane.

An entire function whose real part is non positive must be constant.

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How do you prove that?

The idea is to look at the function e to the f(z) instead.

And we're going to show this function g which is e to the f(z) is constant and

then conclude that f is also constant.

So let's look at this function g.

First of all, we know this function g is also an entire function because

f is an entire function.

And now, let's find its absolute value, like the absolute value of e to the f(z).

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So this exponent here is less than or equal to 0.

It's a real number that's negative or non-positive.

And therefore, that's less than or equal to e to the 0, which is 1.

So now we know this function g is bounded.

It's an entire and bounded function, and

therefore by Liouville's theorem, g must be constant.

From there, we can conclude that f is constant by looking at

the derivative of g.

So, let's do that.

What is the derivative of g?

We'll look at this equation right here.

g'(z) is equal to, now we need to find the derivative of e to the f(z).

We need to apply the chain rule.

The derivative is e to the f(z), times f'(z).

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And we know that g is a constant function, so g prime is equal to 0.

How can the right inset be equal to 0?

So, e to the something is never equal to 0.

In other words, it must be the case that f prime of z is equal to 0 for all z.

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Another consequence of Liouville's theorem, which in turn was a consequence

of the Cauchy integral formula, is a proof of the Fundamental Theorem of Algebra.

There are many proves out there for the Fundamental Theorem of Algebra, but

this one is an amazing proof because it comes from complex analysis.

Any polynomial, p(z) of the form a0+a1z, all the way through

anz to the n where all these coefficients, a0, a1 through an,

are complex numbers, and is at least 1 and the nth coefficient is non-zero.

Any such polynomial we claim has a 0 in C, so

there exists a point z0 such as when I plug that into p, I get 0.

So no matter what polynomial I give you,

for example, p(z) = 14z to the 5th,- 3z squared,

plus iz,- (14- 2i).

So this is a polynomial, and

in this case we have n is equal to 5, and an is equal to 14.

So that's non-zero and so the assumptions are satisfied.

The theorem says, there exists a point z0.

I can plug in a certain z0 that makes the right hand side equal to 0.

Because I have no idea what that point is.

This theorem doesn't show me how to find that point,

it just guarantees the existence of such a point.

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This is a proof by contradiction.

So suppose to the contrary,

that you can find one polynomial p as in the theorem, that has no zeros.

So no matter what you plug in there, you're not going to get 0.

Then we can look at 1 over p(z), and since p was an entire function and

it's never equal to 0, we're never dividing by 0.

Therefore, this new function f, which is 1 over p, is also an entire function.

Our goal is to apply Liouville's theorem to f.

So, we want to show that f is a bound that entire function,

thereby using Lioville's theorem making f a constant function.

And then we can conclude that p also must be constant.

So how do we do that?

We want to show that f(z) is bounded.

So we want to show f(z) is bounded above by some constant M.

We can do that by showing in turn that p(z), which is 1/f(z),

is greater than or equal to some constant bounded away from 0.

If that is the case then f(z), which is 1/p(z), is bounded above.

So that's our goal now.

Well, here's p(z) again and I rewrote it a little bit.

I just factored out a z to the n from p(z).

So here's my a n, a n-1 only had a z to the n-1, so

I have factored out too many z's.

We need to divide by z here, all the way through a0 divided by z to the n.

Now let's put absolute values on there.

The absolute value of p(z) is the absolute value of z to the n times the absolute

value of what's in parentheses.

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And my claim is that is greater than or

equal to the absolute value of z to the n times, I

put absolute values around the individual terms and negative signs in between them.

And what I'm using here is the reverse triangle inequality, which you may or

may not remember, so let me remind you.

It is true in general that the absolute value of the sum

of two complex numbers a + b is greater than or

equal to the absolute value of a minus the absolute value of b.

We also have the regular triangle inequality which said that a + b is

less than or equal to the absolute value of a + the absolute value of b.

But this one up here is the reverse triangle inequality, and

that's the one we're using over here.

We're using a generalized version with more than two terms,

but it's the same idea.

So the absolute value of a n plus a n-1 over z and so forth is greater than or

equal to the absolute value of the first term minus the absolute value of

the second term minus the absolute value of the third term and so forth.

But now the great thing is we're going to let z go to infinity.

So as z gets larger and larger and larger, this

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whole remainder goes to 0 because we're dividing by z.

So when you plug in a really large value for z, the coefficients are fixed, and

so these terms get smaller and smaller and smaller.

a n is some fixed number as well, and then we're multiplying that

fixed number with this larger and larger and larger and larger term.

So altogether that goes to infinity as z goes to infinity.

So p(z) just gets larger and larger and larger.

And therefore f(z) goes to 0 as z goes to infinity.

Well, f(z) going to 0 as z goes to infinity means it's

going to be rather small, as long as z is large enough.

So for example, you can find a radius R,

such that outside of the ball of radius R centered at 0,

f(z) is less than or equal to 1 because it has to go to 0.

So I already know that f(z) is less than or equal to 1 out there.

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But inside this closed disk of radius R,

f is a continuous function and therefore cannot go off to infinity.

So f is bounded by some constant m.

Inside here altogether is bounded by whichever is bigger, little m or 1.

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Therefore, f is bounded in C.

But by Liouville's theorem, that implies that f is a constant function.

But if f is a constant function, then 1/f, which is p,

must also be a constant function.

And that's a contradiction because we said p is a polynomial with a n

being non-zero and n is at least 1, so that is not a constant function.

So we were able to prove the Fundamental Theorem of Algebra using

basically the Cauchy Interval Formula.

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A consequence is that you can factor polynomials in C,

any polynomial of the form that we just saw before.

So p(z) was a0 + a1 z, and so

forth through a m z to the m can be rewritten as a product.

(z-z1)(z-z2) all the way through z-zn,

where these numbers, z1, z2, through zn, are the zeros of p.

So those points, when you plug them in, p=0.

They're not necessarily distinct, but there's m of them right here.

Here's an example.

The polynomial p(x) = x squared + 1 has no zeros in R.

You can graph that, x squared + 1 is just a parabola shifted up by 1.

It never crosses the x-axis no matter what x I plug in there.

It's never going to be equal to 0.

So I cannot factor that, because if I could factor that,

then it would have some zeros.

So I cannot factor this in R.

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But in C, if I allow complex numbers to be plugged in, I can factor z squared + 1.

It has the two zeroes i and -i.

Because when I plug in i for z, i squared is -1 + 1,

is equal to 0, and similarly -i.

And so in C, I can factor this as (z-i)(z+i).

You can try this out.

If you multiply through,

you get z squared plus i z minus i z minus i squared.

What is that?

That is z squared.

i z and -i z cancel out.

i squared is -1 with a minus sign here, gives me a +1.

That's the polynomial I started with.

Another consequence of the Cauchy Integral Formula is the following result.

We'll skip the proof here.

It's called the Maximum Principle.

Suppose f is an analytic function in a domain D and

suppose there exists a point somewhere in D, z0,

such that the absolute value of f(z) is largest at that point in z0.

Then f must be constant.

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So f cannot have a maximum value in absolute value

anywhere in a domain where it is analytic.

There's always a direction where it's going to get bigger.

This is very non-true in R.

You can draw lots of functions that have maxima.

There you go.

So this function has a maximum right here and a maximum right there.

Or the sine function, Has

lots of local maxima value, there and there and there.

But if you look at the sine function, it's extended to the complex plane.

If you make this three-dimensional and

look at the absolute value of sine z on the real axis, that's the axis that,

that's the absolute value at, you now go like this.

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Suppose D is a bounded domain and f is a function that's continuous

in D including its boundary, and analytic in D.

Then the absolute value of f reaches its maximum on the boundary.

So here's this D.

And f is analytic in D but still continues on the boundary.

We know since D is a bounded domain and f is continuous on the boundary and

the domain itself, that f must have a maximum somewhere.

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Let's look at an example.

Let's look at the function f(z) = z squared- 2z.

And suppose we wanted to know

what is the maximum of the absolute value of f on this square?

The square q, consisting of all x + iy, where x and y are between 0 and 1.

So I drew this square right there.

Well since this function z squared- 2z is analytic in set q, and

still continues from the boundaries, in fact an entire function.

We know that the maximum of absolute value of F must occur on the boundary, so

on these red curves.

So all we have to look at are these red curves, so let's look at them.

On gamma one, the first curve down here, the imaginary part is 0.

So Y is equal to zero, and the real part is between 0 and 1.

So X is between 0 and 1.

Y is equal to 0, so what is f(z)?

F(z) is simply f(x), because y is equal to 0.

And f(x) is x squared minus 2x.

The absolute value of that is the absolute value of x times x- 2.

That is actually 2- x times

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What does that function look like?

2- x times x is actually an upside down parabola

that intersects the x axis at 2 and 0, and so

its maximum must be at one.

We're only interested in this portion anyway.

And so the maximum is right there and

the maximum occurs at x equals one and f(1) is equal to 1.

So we know that on gamma one f(z) is bounded of f by 1.

We do the same thing for our gamma two.

Gamma two is the boundary here on the right.

On gamma 2, the imaginary part is between zero and

1 and the real part is fixed at 1.

So, f(z) = to f(1 + iy).

So we plug in 1 + iy for z.

We get 1 + iy squared, minus 2 times 1 + iy.

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And that's what you see down here.

You notice the 2iy's cancel each other out, and

we can combine the 1 and the 2 into a negative 1 minus y squared.

The absolute value is y squared plus 1.

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The maximum of that occurs, obviously, when y is the largest possible.

Which is y = 1.

The maximum occurs at y = 1, and we have f(z) is bounded by f(1 + i).

Which is equal to 1 squared plus 1, or 2 on gamma 2.

You do the same thing for gamma 3 and gamma 4,

there you observe that f(z) is bounded above by f(i).

F(i) is equal to i squared, which is -1.

Minus 2 i.

The absolute value of i squared minus 2 i is actually square root of 5.

And so on gamma 3 and gamma 4, f is bounded by the square root of 5.

That's the largest of all the values we found in the boundaries, so

this must be the maximum of f.

So we see that f(z) is bounded of f by f(i) which is square

root of 5 on all of Q.