Okay, so let's start with complex analysis. Louder, lighter. Okay, so last time, there's a quick summary.

Last time, let's switch off. We do consequences of a formula. So it's valid for holomorphic F defined on this ball,

dBr of B, and Z must be in the interior of the open ball, dR of B. And we have two consequences, namely first, that holomorphic maps are not only once differentiable,

but infinitely often differentiable. So holomorphic maps smooth, infinitely often differentiable,

and they are, and even better, they can be developed into power series, so they are analytic.

Okay, so today, we will actually work a little more in this direction, and also our, we will prove one of the famous theorems, Heuvel's Theorem, hopefully, if I get to it. So let me first start with a result

which is, well, basically preliminary, or intermediate, I should say, on removable singularities. So what are these, these are Heber, the singularity in German

what does it mean? A singularity is a point where a function is not defined, and it's an isolated point. So isolated point where, say, a function,

where a function is not defined. So think of, think, for instance, of f of z, one over z, being not defined at z equals zero,

but it may also be that it's sort of an artificial undefinedness, like, I don't know, z over z, the constant function one, written this way, will not be defined at zero. However, this is not a problem, as you all see.

Okay, so we want to distinguish between the two cases, and for this case, we will say that the removable, that the singularity at zero, so here is, these examples is singularity at z equals zero,

but this singularity at z equals zero is removable here by setting the function equal to one, but it's not removable here since there is, well, we will call it a pole later on. Okay, so this, actually there's no need to define this

since I can write the theorem right away and say that I have a function defined on a ball without a single point. So this is the theorem on removable singularities. Let f from a ball be R of b without its midpoint.

So it's a punctured ball, or punctured disc, since we're in dimension two, punctured disc, to c, b holomorphic.

Okay, and now, if you add one more assumption, which won't be satisfied here, but is satisfied there, namely that the function is bounded, then we are in the second type, then we can simply define it

and conceal the function analytically. Okay, holomorphic and bounded. Then we have an analytic continuation into the singular point b,

and we can say then there exists a continuation. Duration, say, capital F from the entire disc,

b R of d to c of f, which is holomorphic. And continuation means, continuation here means that f restricted

to the previous domain without the b agrees with a given f. So it's really a continuation. So this is one real situation

where I hope you know many functions, like, I don't know, x maps to mod x, or worse, even if you map x to sine x. So minus one for negative real x, zero for zero,

and one for plus one, which are differentiable outside the origin. However, there's no continuation, in this case, not a continuous one, and in this case, not a differentiable one to zero. So this is very different here.

And again, it comes from the Cauchy integral formula, or actually from the validity of the Cauchy-Riemann equations, if you trace it back. Okay, so how does the proof go?

Well, we will actually write out a power series representation for capital F. The way we do this is, okay, to make it easier, without loss of generality, we assume b equals zero, and so we consider an auxiliary function,

namely this function H from b R of zero to C, defined by H of z equals,

and now we multiply our given F by z squared. For z non-zero, and we want it to be zero for z equals zero, since then F wouldn't be defined. We have to give a separate definition.

Okay, so now we claim that this given H, so our goal is actually to derive a power series representation for H, which in turn will give one for F, and the power series for F

will give the power series for the continuation. So in order to derive a power series, we will appeal to the theorem on analyticity, I proved last time, and in order to do so, let's prove that H is holomorphic in both cases.

So in this case, this is immediate, since F is holomorphic, z squared is holomorphic, and products preserve holomorphicity. So I want to prove that H is complex differentiable.

So for z non-zero, this is obvious. So clearly the point z equals zero is the point above her.

So let's just compute H prime of zero by a difference quotient. What's the difference quotient here? Well, the limit of z tends to zero, H of z minus H of zero over z minus zero. Plugging in the definitions,

this gives z squared F of z minus zero over z. And so this limit is the limit of z times F of z.

And now due to the fact, now I must use my assumption, due to the fact that this is bounded. So using that F is bounded, well, a bounded sequence times zero sequence is zero.

Okay, so I have that H is differentiable at zero, which is good. Let's see, okay.

And now, well, if it's complex differentiable at zero, at any point different from zero, it's holomorphic on the entire plane. So F is holomorphic, and that means by the theorem last time, which I quoted above,

which was, theorem, which number, 20 something, 26. 26 means that F is analytic, has a power series. And so let's just consider the power series.

So what is it? H, so say it has a power series, power series, equals zero, so we can write H of z equals H of zero,

which is zero, so let's erase it. But again, since A one is given by H prime of zero,

this also cancels. And so in fact, the power series starts with the second order term. And now it's A two, A two z squared plus A three zero cube and so forth, okay?

Okay, so if you have that, now we're in good shape. As you see, since we can actually write this power series really easily as a product of z squared, which means we have a power series of F, yep. So we set capital F to be H of z divided by z squared,

which means A two plus A three z and so forth. A two plus A three z plus A four z squared

and so forth, which is a nice power series. And also, this converges on any ball you like, yes? Or on any ball, sorry, what's the condition in contained in B R, little B R? So this is for z in B R, so.

Okay, so this is valid for all, so since this was valid for all z in B R, so this also is valid for all z in B R.

Okay, and so that means we found the definition of, a definition of capital F, and since this is a convergent power series on B R, we have a valid definition of F,

and this must coincide with F just by a definition of my original H. Yeah, so I can say that F of z equals little F, just by the definition, since this is H z

over z squared, and so this is F on for z non-zero in the ball, and continues it analytically with a power series to the point B equals zero. And is holomorphic on the entire ball, as I said here.

Yeah, for the reason that power series are always holomorphic functions wherever they converge.

Okay, so this shows that when we move singularities in case the function is bounded. And we should also ask ourselves why does this proof not give that these functions are differentiable at zero? Well, do you see where it occurs?

What is the argument which can't be made in the real case? Exactly, yeah, so this is really the strong property. Holomorphic functions are analytic. Yeah, the other stuff up to here is fine

in the real case.

Okay, so we will look at such removable singularities in the next statement also.

This is the identity theorem. Identity theorem for power series, or of power series.

So basically I want to sort of say everything I want to say in this class about power series now. So we want to consider what defines actually on how large a set must be in the complex plane

in order to determine the power series of a function. And the crucial term here is discrete sets. Those are sets which are not big enough in order to determine power series. So to start, let me recall what an accumulation point

Holfung's point is. So A in some set A, set in C is a collection point actually in collection in set A.

Well, in case I can close my point A by points different from little a in the set A. So this set X, A are different to my point A,

such that X, K converge to A. This is called an accumulation point here. So the sets think of sequences which converge,

but these sequences consist of points different to the limit point. Here, for instance, the accumulation points of the set Q are given by the real inverse. Okay, this you should know. Let me define discrete set.

Perhaps you know that as well, but let me define it and actually it works in the same way in the Euclidean space and the complex plane. But I want this open.

Then the set N contained in U is called discrete.

Okay, if N has no accumulation points. If N no accumulation points. Okay, what are examples?

No accumulation points. I should be careful in U. Perhaps it has something C, but I want to look at subsets here. And so I really, I'm really, accumulation points only in U.

Okay, so that means, or perhaps let me add one thing. Yeah, that's good. Yeah, what does it mean? Well, this means each such point in N

in the neighborhood such that it needs only finitely many other points of N. Cannot be that each neighborhood needs infinitely many since then there would be such a sequence. And in fact, I can make the small enough

such that this neighborhood of a given point doesn't need any other point of N just by shrinking it since initially many. So I take the maximal radius which contains only the, which contains no, none of these points.

Okay, so that means for all points in U, this in R, positive R such that little ball about this point doesn't need the set N

except possibly in the point Z which may or may not belong to N. So let me remove it and then say that little ball needs the set N or possibly the point Z

in an empty set, so. Okay, so that's the discrete set and examples are, for instance, in the integers in R or in C,

or a more sophisticated example is take numbers

one over a natural number, one over N, okay? This set as a subset of C has the accumulation point zero.

However, as a subset of C without zero, so I contain in C with zero is D. Since in fact this set, zero is the only accumulation

point in C, so if I remove it from my set U, then this, I can say that N is a discrete subset of C without zero, yep, but it's non-discrete in C.

So it really may depend on the set U, what will depend on U if the set is discrete or not. Okay, and a kind of set which is always discrete

no matter in how I choose my set U is a finite set. So, and the simplest finite set is the empty set. Okay, the empty set and the finite set is three.

Simply about the property that this here is satisfied. Why are we interested in discrete sets? The fact is that the zero set of a holomorphic function

is a discrete set, all the functions are identical to zero that's the next state name, so let you be a domain.

It's my standing assumption, but in those cases where it's really important, please allow me to quote it nevertheless. So if you have a domain and a holomorphic function is domain, we have the following alternative.

Either F is identical to zero, so F of Z is zero

for all Z and U, or the set of zeros is zero is discrete, so the set, let me introduce notation of the zero set, zero set, let's call it N of F, the null set, which is the both z and U of z.

Zero set is discrete. Okay, so what do we know then?

We know for instance if we have this example in C and we have a function which vanishes at all these points and is defined on the entire complex plane C, then it has a non-discrete zero set, so then actually a holomorphic function

having all these zeros defined on all of Z of C must vanish identically, okay?

And so why is it important if we have the domain? Well, if we have a disconnected set where F is defined, then we could define it as say identical to one on one component and zero on the other,

and the statement will be false.

Okay, so let me prove this, and it's a little more, takes a little while, so let me start with the claim. The claim is that the set of accumulation point

of a zero set, z and U, such that z is accumulation point of N of F, okay?

And the claim is that this set is open, okay? It's a little indirectly defined, I know, but nevertheless, we want that the set of accumulation points is open,

so that if I have one accumulation point such as zero would be the case for the example which I mentioned, then I have little neighborhood of zero where F, which I also can approach with other zeros,

unlike the example, so that means that actually, F would be identically zero in the neighborhood of such a point in N. Okay, let me prove this, so take a point. Yeah, it may, I mean, this may also be empty. I don't care right now.

All I want to show is it's open. Okay, so I take a point in M, so that's an accumulation point. So this is an accumulation point of, well, of the zeros, right?

So that means there exists a sequence of zeros of F, so there exists a BK, the sequence BK of points in N of F, but they should not agree with B such that, so N of F means that F of BK vanishes

and also this sequence converges to B. Okay, so this is what I have if I start

with the point in the set capital M. And now the strategy is I want to give you a power series, a power series development of my function F at B. And this goes as follows, so certainly, okay,

certainly we can write F, F is a homomorph function in the neighborhood of B, we can write it as a power series, so F of Z equals, say, A naught plus A1 Z plus A2 Z squared, and so forth,

at B, I want to give you this at B, I'm sorry, and so forth, okay? And this holds for Z in a little ball, B R of B,

which is fundamental in B. Okay, well, and this in the middle of the proof, yeah? So now let's look, what do we want to show? We want to show that this power series actually vanishes,

well, if I actually have such a point, then this power series vanishes. What can I use? Well, I can use that at the BKs, I get zero. So let's do this inductively, working from one A K

to the next. So what we do is we first look at A naught. Okay, here, we're following that A naught, well, by definition, this is just F of, F of the point that I developed, F of B.

Well, F of B, now I want to invoke the BKs. Well, of course, this is F of limit BK. And now I want to use continuity, yep. And say this is limit F of BK in order to get to my zeros.

So this is F is continuous. So then I have limit of F of BK, well, each of them is zero. So this tells me that A naught is zero. Okay, so now I know this is not there.

What can I do in order to show that A one, A two and so forth vanish? Well, I divide by z minus B and use the same argument. Okay, so now, and then I keep on going. Now I consider F of z over z minus B.

Which is using the power series, A one plus A two z minus B plus A three z minus B squared and so forth. And this is valid for a start for BR of B

minus the point, remove the point B. However, the right hand side is defined also in B and it is a nice continuous function there. So that means that this function on the left hand side

has a removable singularity at z equals B. And I can actually remove it just by the value here at z equals B, which is A one. Okay, so I can say that this has a removable singularity

at z equals B and this comes from fact that this is a nice continuous function. So I'm gonna say in particular, it's continuous at z equals B.

So therefore I get that this function has a removable singularity. Well, now what is, okay, it can be removed with value A one obviously. And what is A one, so value at singularity, at singularity z equals B is A one.

And now I apply the same kind of argument as here. So I write A one is the limit. Okay, this time I must however consider the F of z.

Well, let's write one more step. It's F of z over z minus B, where B or Bk approaches, z approaches B and z is not equal to B.

But for the points Bk, I know that they give me such a sequence, so I can simply write, this is the limit of the F of Bk over whatever there is, Bk minus B and now these are zero up here.

And so this is a zero sequence and so this is zero. Okay, so showing me that A one is zero. Now I divide my function F of z over z minus B. Once again by factor z minus B and repeat the argument.

And so induction gives, following exactly the same argument gives me that AN equals zero for all N and N, actually N naught.

And so that means that F has a power series representation which vanishes. So that means F vanishes on the open disk. Since this is really equality on the entire disk, F vanishes on B R, on the disk B R.

So F is identical to zero on B R of B. Okay, in particular, if I have such a point B and M, then I have a neighborhood in that, so it's open.

So this means claim. Okay, so this is the difficult bit of the proof. Let me now come to the easy bit,

which is a connectedness argument. I will show that not only M is open, but also M is closed. And there are different ways to show it. One would be a diagonal sequence,

argument, so if you have a sequence of accumulation,

a sequence in M, so each point in the sequence is an accumulation point, so each point is approached by a sequence, well then you take a diagonal sequence and show that the limit is also in the set as an approaching sequence.

Another way to show this is the following. So I want to show that M, no U, so M is closed or U without M is also open.

It is open. Okay, how do I do this? Well, if I take a point in U in the complement of M, that means in the neighborhood of this point, I have only finitely many zeros of F.

So if Z in the set, or if Z is not an accumulation point of N of N,

of N of F, of the zero set of F, then according to the property I erased, there exists a neighborhood of this point of Z

containing only finitely many zeros of N of F and only finitely many points of N of F.

So I choose a point. It's not an accumulation point of N of F. That means there is a neighborhood where I have only finitely many zeros.

Well, that means that in fact, I can take this entire neighborhood and it must belong to the set U without it. Okay, so this neighborhood,

this neighborhood, should have given it a name, neighborhood belongs to, is also in U set minus and U minus M, right? So all these points in this neighborhood

can also not be approached by points in the zero set since they have a finite distance to the boundary. Okay, so that means U without M, the complement of M is open and that means M itself,

so M itself is open and closed, open and closed, okay, and now you know that in a domain, M sits in U domain. Well, the only open and closed sets

are either the empty set or the entire domain. So that means either M equals U. In this case, the zero set is all of U

and the function is identically zero or M is the empty set. Well, in which case, the zero set of F has no accumulation points, so it's discrete.

N of F is discrete and this is precisely, this alternative here is precisely the claim. So we've proved what we wanted to and perhaps, is it fair if I just write the consequence which there's nothing to do about it,

so let me just write it on the board and then have a break. So now I want to apply the lemma, apply the lemma to a difference of functions, of two holomorphic functions, what do I get? Well, either they agree at most on a discrete set

or they are identical and this is the identity theorem. The identity theorem for power series that says if F, G are two holomorphic functions,

holomorphic functions, then on a domain, this is again important, on a domain U. Okay, then, well, this is the alternative of a lemma

for F minus G means either, then either, let's preserve the order, it's not there. Okay, so then either F is identical to G or,

or the coincidence set, the coincidence set is discrete. Coincidence set is, so what's the coincidence set? That's the set of sets where F of Z equals G of Z

is discrete in U. So it may still have accumulation points of the boundary of U. And so discrete set is actually the notion of finite sets but adapted to open sets, in closed sets,

closed or in compact sets, discrete sets are necessarily finite. Okay, so this means we know very well what power series are and I will draw consequences after the break. Okay, I would like to continue.

So there's one more formal statement I would like to make and then I will discuss examples. So the formal statement is just a little one but perhaps worth pointing out for power series,

which is that if I have a power series, say P of Z equals some AKZ minus B to the N.

If I have a power series and it has radius of convergence has and suppose radius

of convergence is positive, yeah. Then I claim that this power is no larger value with no epsilon say, no positive epsilon

such that I can continue this function P to a ball of slightly larger radius, R plus epsilon with a holomorphic function. So then, how do I say this?

For no positive epsilon, there is a holomorphic function defined on say a slightly larger disc, the R plus epsilon of B to C.

There's F defined on here continuing holomorphic and continuing holomorphic and such that F restricted to the ball B R coincides.

The capital R, yeah, capital R coincides with P. Okay, so that means if you have radius of convergence of the power series, it's not possible to define the function on any larger, to define a holomorphic function

on any larger disc and this is helpful in discussing power series and singularities and it's very straightforward to prove. So assume you have such an F, okay?

Then, okay, by analyticity versus power series of F. F, so this is again theorem 26. Analyticity, F has power series, converging power series on this entire ball of B.

Well, by the identity theorem then, the two must, since the two agree on the ball B R,

by the identity theorem, this is actually also the power series given up there. So identity theorem tells me that F equals P on B R

and so that means P is also the power series of F but F would be defined on this larger radius. Well, then the radius of convergence would also be R plus epsilon and this is A.

So F or P, P has radius of convergence by the theorem on power series representation on analyticity, we know that the radius of convergence is R plus epsilon, contradicting the assumption of our statement

of the proposition. Okay, so if I have radius of convergence, I cannot extend the function analytically to my entire neighborhood of the disk. Of course, it may be possible to extend it to certain points or certain larger domains

and so in order to discuss examples, let me, okay, let me make one more definition. Let F from R to C, say, be given.

Okay, then a function F tilde, say, from C to C is called analytic continuation, is the analytic continuation

of little f if and only if, well, first of all, F tilde is analytic and second, it continues. Yeah, this is what it says.

So F tilde is analytic on C and F tilde restricted to R is F. Okay, this is a situation we're given in many cases, namely, we have many examples for this,

namely think of the exponential function, the sine function or so. Consider first as a real function. Well, there is a, then we have analytic continuations by just replacing X, real X by complex Z and this gives us analytic functions

and this is also a unique way of, well, this analytic continuation is unique. I should write this out. So I say, if the analytic continuation exists, continuation exists, then it's unique.

Then it is unique. Well, why is this? Well, by the lemma, which is gone now, a function is defined, an analytic function is already

defined on any non-discrete set. R is non-discrete, yeah? So it determines the power series of a, the power series of F is already completely determined on R and so that means the result of,

well, the power series is unique and so the analytic continuation is unique if it exists. So by the lemma, or what was this? By the lemma telling me that F is,

or was it by the lemma that F is already determined on a non-discrete set. Okay, it's not necessary that we have such a continuation but since there are so many examples, it's, this is worth considering and also remember

our initial discussion of the exponential function in analysis one. What did we do? Well, we derived, basically heuristically, we derived from the differential equation,

F prime equals F, we derived the exponential series on R, heuristically, and then I said something like, well, there's just only one way to make this a power series also on the complex domain,

just by taking, replacing X by Z. Well, it's by this very simple principle I mentioned now that there's a unique analytic continuation from R to C. So there's sort of no choice, yeah, for the, if you want to extend the exponential function from the real numbers to the complex numbers,

by this principle there's a unique way of doing this. So if you want the functional equation or the differential equation of work for the exponential function, then you end up with the same exponential function on R, on C. Okay, there will be an example. So this sort of explains this to write a little note,

explains X from C to C, yeah, with X. From R to C, or R to R. Okay, the other thing is,

this is a special case of analytic continuation. Actually, I could replace, I could replace here the real numbers by any non-discrete set U, yeah, and say, well, actually I want another letter,

I guess I want a non-discrete, I need two letters. Let's call this M, M or M. N was discrete, so let's call it M. Yeah, let's, sorry, yellow choice.

Let's use yellow M here to replace R with. Okay, and let's replace then the continuation set on U, which is a subset of M, as was C, a subset of R. A superset of M, sorry. Yeah, and then I call it an analytic continuation if the continuation is analytic

and restricted to the set M, it's F, and all I need to do is I require M is non-discrete. Doesn't have to be open, just non-discrete is sufficient in order to, for a reason

that when the power series is well defined. Any error? Could be, if I'm mistaken, let me see. Okay, it's actually, it's actually the consequence. It's, yeah, should be theorem 30.

Yeah, sure. Since theorem 30 tells me two such continuations must agree, they do agree on a non-discrete set. It used to be R and now it's my non-discrete set M and then by, yeah, and also it's there,

by the theorem up there, they must agree, thank you. Okay, so that means I can talk about, I can talk about analytic continuations in general. Now there is a little problem about it, but let me first, which I want to show you on an example,

but let me first show, give you a nice example where everything works out nicely. So first example, consider the function on the reals, X maps to one over X squared plus one. Okay, so this is a simple function defined on all of R

which has, is one at zero and decays to zero at plus minus infinity. Okay, what about the power series at X equals zero? Well, in terms of the geometric series,

it's easy to write it out, but I don't want to do this. All I'm saying is there is a power series, has power series developed at zero and what is the radius of convergence? Well, you use the geometric series,

it's obvious, well, you can compute or convince yourself that the radius of convergence is one. Convergence equals one. Okay, so what is particular about.

this value 1 here, or minus 1, on the reals, nothing, right? Why should, what's the reason for the power series of this function to converge exactly on this interval here? It's hard to tell. However, if you look at the complex version, then it's obvious.

Yeah, now we extend to z maps to 1 over z squared plus 1. In the complex plane. And what do we discover? Well, by the standard polynomial trick, this is 1 over z plus i times z minus i.

So that means we have two singularities at plus i and at minus i, right? So, this function is actually defined on, let's see, where I remove the points plus

minus i. Well, by the previous theory, it's now clear that if I have the development point, b equals 0, the maximum radius of a disk for which this function here is holomorphic has radius 1, right? Okay. Pretty, pretty poor picture of a disk of radius 1, but I

hope you need, so here's 1 and here's minus 1, yeah? Okay. So, you see right away where this, what this radius of convergence is. So, maximal, maximal disk, about 0, about

0 on which f is defined, on which 1 over z squared plus 1 is defined, is defined, is

the disk of radius 1. And so, and actually, there exists no, we cannot continue this function holomorphically anything beyond i plus minus i since this is a pole.

Where the function tends to infinity, so there's no continuous function including these two points, which extends it. So that means, and so formally, this is actually the last statement proposition 31, yeah? We don't have a holomorphic extension to anything larger

due to the fact that these two points are poles, so that means I know right away that the radius of convergence is 1. So, taking proposition, so this is one fact, it's at least, so this shows me r is at least 1 and on the other hand, proposition 31 tells

me r is at most 1 since there's no analytic continuation, so that means r is equal to 1 without any computation, right? Of the, how is it called, the 1 over limit, superior,

nth root of mod a n formula, this is the Cauchy-Adama formula, right? I forgot that, yeah? Okay, so no need to appeal to the, no need to appeal to Cauchy-Adama, yes?

So that means if I tell you that I have a holomorphic function and here are points that's not defined

and here's a point b, then I know right away the radius of convergence of the power series, it's the maximal radius of the disk contained in the domain where the function is holomorphic, so it really is the radius of the closest singularity if it's a function of this kind, yeah? So this is a very neat feature, question, sorry, yes, else it would

be contradiction, thank you, sure. Okay, so this is very neat, now there's a little complication about analytic continuation which I do not want to hide and in fact it leads

to some modern mathematics, Riemann surfaces and we will look at, and in order to discuss

it we will look at another example, namely we'll look at the logarithm, so the next

example I want to discuss is the logarithm, so remember that the complex logarithm is

given in the form logarithm of the modulus of z plus i times argument of z and we usually define it on a slit domain, so we find on the complex plane without the, without

the negative real axis and without zero, yeah, on this set s. Okay, now I'm going to, now here's a nice problem about analytic continuations, look at the complex plane,

look here, say at the point one, this function here is for instance defined on a disk of radius one, oh dear, on a disk of radius one about the point one, yeah, it doesn't

hit the zero which is dangerous, okay, so here we have log z defined, so this is okay, I will never get it defined at zero, sure, but I can now extend it, say to a set,

I can extend it to a set like this, yeah, why not, okay, so analytic, analytic extension continuation, continuation defined and actually by this formula we precisely know what happens

is the logarithm of the modulus of z and the argument would go from zero to pi over to, I don't know, pi plus something, okay, I can look at another analytic continuation

and, which would say go from the disk, but this time let me just go like this, right, okay, another analytic continuation, okay, so that means that it may depend on the way

I continue my function what the value at a given point is, right, so this is the here what's the difference, well, you know what the difference is, it's just two pi i

in this case, yeah, the argument here would be, here the argument is zero, pi over two, if I come from here it's pi, so here I have the yellow version, I have pi plus something for the argument, on the other hand, the orange version, I get something like almost minus pi for the imaginary part here, so the difference, so in here I have

a difference of two pi i, right, and so it turns out that in fact analytic continuation

to the complex plane or to a maximal subset in the complex plane is not really the best idea, so what you need to do is sort of keep on going by analytic continuation, last

term I showed you, I showed you the helicoid surface, yeah, which is a graph, well, it's a straight line and apply a screw motion to it such that it goes up and up like a helicoidal

set of stairs, so that would be the actual, actually a nice domain for this logarithm function to be defined, you wanted to define on infinite coverings and that's the so-called Riemann surface of a function, in the case of z squared you would just need two copies

going around twice to have an analytic function well defined, I will not develop this any further, but it leads to the notion of a manifold and since it's Riemann surfaces, this was already discovered in the 1850s or so, 50 years before manifold

were actually invented, so complex analysis was a forerunner in this respect, yeah, I think

Riemann also talked of analytic Gebilde and there was a long theory before it became conceptually clear what to do, yeah, but it's the same thing as with the angle, where's the angle function really defined, well, if you work on the unit circle, the

angle is not continuous, there's a two pi gap, where is the angle function nicely defined, well, it's on the covering, it's on the real line, which I imagine is going around infinitely often on the unit circle and same thing here for the logarithm with

the discrete theory of power series and now let me use the last 50 minutes for famous

theorem, the theorem by Liouville, okay, again, again a consequence, what's going

on up there, hey, this is the class, okay, so, Liouville's Theorem, okay, so,

what we look at are functions defined on the entire complex plane, so these are simply

called entire functions, an entire function, an entire function is a holomorphic function

defined on all of R, holomorphic function defined on the entire plane, C, C, okay,

and there are many examples, yeah, famous examples are polynomials, polynomials, what else do we have, the exponential, sine, cos, okay, but we also have, we also have but not one over z squared plus one, so there are many more, okay, if you look at these

functions with, from a point of view of boundedness, well, then you realize polynomials are not bounded, exp is not bounded, well, sine and cosine look, in the real version,

look bounded, right, well, they are bounded functions then restricted to reals, however, if you go to the purely imaginary values, then you know that sine IT is not bounded since it's E to V, yeah, it's E to V, what is it, IT minus E to the minus IT over 2

or sinh, something like, what is it, sinh, I'm missing a 2I here, sorry, is it correct now before you ask, hmm, I think this, oh dear, let me look, E to the, oh no, here

I have the I, sorry, yeah, you're right, here's the I, so I don't need it there and now, but that means I must flip the order of the minuses, okay, I'm sorry, okay, well, don't need the plus here, okay, whatever this is, it's the sinh function

over 2I, over I, or minus, oh, it's actually plus the sinh function and whatever, in any case, this has limits plus minus infinity if you go to, with T to plus minus

infinity, so the analytic continuation of sine defined over reals or cosine defined over reals is unbounded, yeah, so all these functions are unbounded on sine and on the complex plane and the theorem is that this is always so with entire functions,

okay, so theorem, Liouville's theorem says each, okay, each, well, if it's not, if it's not unbounded, then it's bounded, well, if it's bounded, then it's only a constant.

Each bounded entire function, entire function is constant, yeah, so essentially, each interesting entire function is unbounded, that's another way to state it, yeah, okay, and so once

again, statement, certainly not true for real analysis, but in fact, it's not true for some other partial differential equations, okay, there's something to say, but it's

much more complicated, so I don't want to mention that in detail, okay, so here's the simple proof, the proof is just inspect the Cauchy, the Cauchy formula for the first derivative with the Cauchy integral formula for F, and we derived in the course

of showing that F is smooth, we derived formulas for F prime and the nth derivative, here we need the formula for F prime, okay, so I start with, say, pick a point in C

and then I have this formula by Cauchy, which was theorem 25, we have that F prime of Z is one over two pi I, and it will be an n factorial, but this is the first derivative,

so it's one, and it's the boundary, D B R of Z, or do I, well actually, I want capital R since R will be large in a minute, F of Zeta, and now unlike the original Cauchy integral formula, this comes from differentiating, so it's Zeta minus Z squared

D Zeta, okay, and this, since we have an entire, so this, originally this works for any ball, any disc contained in the domain of definition U, well here it's an entire function, it's defined on C, so this works actually for all R, yeah, this

is come F entire, we can use any disc we like, well, and all we need to do now is just use the length estimate and the boundedness, so the numerator is bounded,

well, this is a circle of length two pi R, so you see, and this is approximately one over R squared, so this will altogether give one over R by the length estimate, so let's do this in detail, F bounded means, so say, F is less than some constant C,

now we use the length estimate in order to estimate the modulus of F prime, which is one over two pi times the length of the circle, let's just write it out, D B R

of Z times, well, the modulus of the soup of soup of F of Zeta over Zeta minus C

squared, and now the soup taking over D B R of Z, and now this is two pi R, it's estimated by, well, the numerator by C, and the denominator is a constant, it's just

R squared, so what do we get altogether, we get a one over R squared, oh, one over R, so actually equal, but here must be a, this way, better this way, okay, this is one, one over R is remaining, correct, C over R, C over R, okay, so this is one

okay, hopefully correct now, so what does it mean, well, this is valid for all, for all R larger than zero, well, that means F prime of Z is zero, F prime of Z must be

zero, well, but this is for arbitrary Z and C, well, this means F is, F prime is identical to zero, and if F is identical to zero, then F is, and defined on the

entire plane, then F is constant, and that's it, so it's really, if you like, it's a simple consequence of the Cauchy integral formula differentiated once, yeah, this

comes from differentiating the Cauchy integral formula once, and then you plug in the length estimate, and see what you get, and there you are, it's an amazing theorem, however, here, that there is no, there is no bounded entire function unless, except for the constants, question? Mr. Kleine Gleichen, you say, where do I, well, if, oh, okay,

yeah, right, sure, since it's sort of, it's these things here, yeah, it's this here, all right, thanks, yeah, okay, so, there are many interesting consequences

from, which you can draw from this theorem, and one, one I will show next time is the fundamental theorem of algebra follows easily, so next time we will have,

so next time we'll have this famous theorem, oh, let me also make an announcement,

which I forgot to make in the break, physicists asked me if they can, whether they can register for the English course, complex analysis, they seem to have trouble to register for

the English version, but can only, could only register for, according to Mrs. Muhoiser, this is not true, so try hard, if you want, if you want to register for, if you're a physicist and want to register for complex analysis, it should be

possible, and if it's not possible, ask C. Maybe Mrs. Klink, some, from physicists, yeah, who should know, but it should be possible.

Okay, so, let me draw, let me draw consequences, well, one thing is, it's a problem, which is, show that f entire is either constant, or,

if you look at the entire range, it's dense in C, the closure is C, that's,

that's, well, that's just a small trick, which leads you from Leo Wilt's theorem to this statement, so, also, well, this means, such an f, holomorphic f,

could miss many values, in fact, there is a theorem, which we will not prove, so-called Piccard's little theorem, there's also great theorem, but, this is the little theorem, telling, telling you that f entire misses,

at most, misses, at most one point, at most one point. One point, or is constant, or is constant, so,

yeah, same alternative, if it's not constant, the entire function, which is not constant, so, which is, or, implied by which is not bounded, can, must take all values, except for one. Yeah, what's the value missed by every exponential? Zero, right?

That's an example, yeah? It may be possible to miss one value. Yeah, I mean, polynomials take all values, as we will see, but, but one point can be missed. Okay. And another consequence is, consequence, which I would like to mention,

that if you consider f on the entire complex plane, say, to the unit disk, B1, well, then such an f cannot exist by Liouville's theorem, this is a bounded set, yeah? Entire, not entire, but holomorphic and bijective,

holomorphic, bijective, does not exist. Yeah, this is an immediate consequence of Liouville's theorem. Why is this interesting? Well, this, geometrically, this means there's no angle-preserving map of two spaces,

which, topologically, from the point of holes, cannot be distinguished. Yeah, the complex plane and B, the unit disk by some, well, these are holomorphic sets. Let's put it like this. Homomorphic, but if you restrict yourself to angle-preserving maps,

holomorphic maps, then you can distinguish these two sets. It's also interesting, since the Riemann mapping theorem, Riemann mapping theorem, mapping theorem says that if you have a set U,

which is a proper subset of the complex plane, and which is simply connected, simply connected, then, contrarily, there is a holomorphic map. Then there exists F, holomorphic, from U to B.

So, as a holomorphic in the sense of bijective and holomorphic. And holomorphic. So, that means that any nice, simply connected set, for instance, this one here can be mapped in an angle-preserving way to the unit disk.

Yeah, also the slit domain, yeah, whatever, whatever you like. All these nice, simply connected sets, which are not C itself, however, C itself is the only exception where such a map doesn't exist.

So, you can distinguish different domains and classes. You can make an equivalence relation from that by saying that subsets U are equivalent if such maps bijective holomorphic exist. And then all these sets come into one class, but C comes into another class.

Surprisingly, although it's holomorphic, the conformal type is different. Okay, so this is an outlook and next time we will prove the fundamental theorem of algebra.