Fundamental theorem of algebra, mapping properties of holomorphic functions
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Complex Analysis10 / 15
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00:00
DifferentialgleichungFunktionentheorieHolomorphe FunktionMathematikNumerische MathematikOrdnung <Mathematik>PolynomZahlensystemModulformKategorie <Mathematik>IntegralAusdruck <Logik>AggregatzustandArithmetisches MittelBeweistheorieFundamentalsatz der AlgebraGruppenoperationKomplex <Algebra>KurvenintegralTermTheoremReelle ZahlGewicht <Ausgleichsrechnung>ExistenzsatzPunktKlasse <Mathematik>Sortierte LogikKoeffizientEinfügungsdämpfungArithmetischer AusdruckMinimalgradMultiplikationsoperatorTourenplanungLOLA <Programm>ComputeranimationVorlesung/Konferenz
10:00
BruchrechnungOrdnung <Mathematik>PolynomDreieckBeweistheorieFunktionalLeistung <Physik>Lokales MinimumTermTheoremParametersystemGebundener ZustandSortierte LogikSchätzfunktionRadiusKomplexe EbeneMultiplikationsoperatorDreiecksungleichungStandardabweichungSpieltheorieLatent-Class-AnalyseGammafunktionGleitendes MittelVorlesung/Konferenz
19:19
FunktionentheorieGeometrieKurveNumerische MathematikOptimierungPolynomMengenlehreZeitbereichGanze FunktionDerivation <Algebra>Analytische FortsetzungFunktionalKompakter RaumLokales MinimumStetige FunktionTheoremZentrische StreckungWurzel <Mathematik>Gebundener ZustandPunktInnerer PunktKreisflächeKartesische KoordinatenDifferenzkernSchätzfunktionRadiusAbgeschlossene MengeDifferenteMinimalgradMultiplikationsoperatorRechter WinkelDiagramm
28:37
Natürliche ZahlOrdnung <Mathematik>PolynomVollständige InduktionFaktorisierungErwartungswertLeistung <Physik>TermTheoremReelle ZahlTeilbarkeitPunktKoeffizientJensen-MaßDifferentesinc-FunktionMinimalgradMultiplikationsoperatorTopologischer VektorraumDreizehnGammafunktionComputeranimationVorlesung/Konferenz
36:57
FunktionentheorieGraphHolomorphe FunktionMaß <Mathematik>ModulformKategorie <Mathematik>ZeitbereichBeweistheorieFundamentalsatz der AlgebraFunktionalIndexberechnungLokales MinimumProjektive EbeneRechenschieberTermTheoremTrigonometrische FunktionTeilbarkeitGleichgewichtspunkt <Spieltheorie>PunktInnerer PunktHauptidealMultiplikationsoperatorRechter WinkelEinsUnendlichkeitGroße VereinheitlichungVorlesung/Konferenz
45:17
MathematikMengenlehreMittelwertZeitbereichIntegralAusdruck <Logik>Analytische FortsetzungArithmetisches MittelBeweistheorieDiffusionsgleichungEinfach zusammenhängender RaumFunktionalHarmonische FunktionLaplace-OperatorLokales MinimumKonstanteNichtlineares GleichungssystemReelle ZahlNachbarschaft <Mathematik>PunktInnerer PunktKreisflächeMultigraphCauchy-IntegralAbgeschlossene MengeSupremum <Mathematik>MultiplikationsoperatorRandwertGesetz <Physik>Lemma <Logik>Lokales MinimumTrigonometrische FunktionViereckGleitendes MittelVorlesung/Konferenz
55:09
Ordnung <Mathematik>MengenlehreMittelwertZeitbereichIntegralAusdruck <Logik>Algebraisch abgeschlossener KörperUngleichungAnalytische FortsetzungArithmetisches MittelBeweistheorieFunktionalKompakter RaumKorrelationskoeffizientLokales MinimumMereologieParametersystemNachbarschaft <Mathematik>PunktInnerer PunktKreisflächeSortierte LogikMultifunktionsinc-FunktionRandwertRechter WinkelIkosaederLokales MinimumGleitendes MittelVorlesung/Konferenz
01:05:01
DifferentialgleichungFunktionentheorieGeometrieHeuristikHolomorphe FunktionNumerische MathematikOrdnung <Mathematik>MengenlehreZeitbereichGanze FunktionAlgebraisch abgeschlossener KörperDerivation <Algebra>AggregatzustandEbeneFundamentalsatz der AlgebraFunktionalGruppenoperationKompakter RaumKomplex <Algebra>Lokales MinimumMereologiePrimidealSkalarproduktTermTheoremZählenNichtlineares GleichungssystemKritischer Punkt <Mathematik>Spannweite <Stochastik>Wurzel <Mathematik>Nachbarschaft <Mathematik>PunktKlasse <Mathematik>Offene MengeKreisflächesinc-FunktionExplorative DatenanalysePermutationRangstatistikPrädiktor-Korrektor-VerfahrenE-FunktionVorlesung/Konferenz
01:14:53
DifferentialgleichungHolomorphe FunktionPolynomModulformDerivation <Algebra>GrenzschichtablösungDrucksondierungEbeneFunktionalGeradeMereologieMomentenproblemTheoremWinkelÜberlagerung <Mathematik>PunktOffene MengeKreisflächeMultifunktionRadiusKomplexe EbeneMultiplikationsoperatorMinkowski-MetrikRechter WinkelComputeranimationDiagramm
01:18:54
Holomorphe FunktionOrdnung <Mathematik>PolynomMengenlehreKategorie <Mathematik>ZeitbereichDreieckAggregatzustandAnalytische FortsetzungBeweistheorieEinfach zusammenhängender RaumFunktionalGeradeLemma <Logik>LogarithmusMereologiePotenz <Mathematik>RangstatistikRechenschieberStetige FunktionTheoremKonstanteParametersystemWurzel <Mathematik>SchwebungOffene MengeArithmetischer AusdruckKomplexe EbeneAbgeschlossene MengeDifferenteMultiplikationsoperatorRandwertRechter WinkelLemma <Logik>VerschlingungVorlesung/Konferenz
Transkript: Englisch(automatisch erzeugt)
00:05
Okay, welcome, complex analysis, last class before Christmas. Let me start with a correction, I believe, which is, which are also in the course notes,
00:23
and it was Sebastian Bechdel who pointed me to the problem. So, this is the formula showing you that a holomorphic map has infinitely many derivatives,
00:43
and it works by differentiating the Cauchy integral formula, and in fact, we have to differentiate with respect to d z, and not to d zeta, this was wrong in the notes, and it's completely wrong, so please correct it
01:00
if it's wrong there. Also, I changed the text a little, since I became aware that what I presented was too short, so when we apply the Leibniz formula, which is not present on this slide,
01:21
when we apply the Leibniz formula, we should think of not applying it directly to what's written here, but rather, since this is a line integral to the explicit form, the explicit form would be insert a parameterization, and then you have a term
01:41
where you insert your path into f, and have the derivative, the tangent vector, and then you convince yourself that differentiating this expression with respect to z gives precisely, well, precisely yields a second derivative,
02:01
and looks exactly like this, and then you reformulate. So after you've done this, you're allowed to write this, but you should be aware of what you're doing, and so I corrected this in the notes, and it will be on the net later today.
02:21
Okay, so this is a small correction, and so let me now start the class. Well, I don't want to give a huge review of last class,
02:43
but there's one thing I want to recapitulate, which is Liouville's theorem, which is simple to state since it only says that, all it says is that f is entire
03:04
and bounded when f is constant, and here entire means that it's defined on the entire plane, all of c,
03:24
and that f is holomorphic. Okay, that's the famous theorem by Liouville, and it certainly should make you aware that complex differentiability is not anything
03:44
on the real Jacobian, but it's really a partial differential equation, and so strong properties come from the Cauchy-Riemann equations, and this is one of them. There's nothing in the real world without such a differential equation
04:01
which resembles this theorem. Okay, today we have one of the most famous theorems, which we actually prove as a consequence of the Liouville theorem, namely the fundamental theorem of algebra. So fundamental theorem of algebra,
04:31
and I mean, there are hundreds of proofs of this theorem, but the one in terms of Liouville's theorem
04:41
is particularly easy, so let me present it. So it's following theorem if you have a polynomial which is non-constant, so each non-constant polynomial
05:09
let's give it a formula, p of polynomial of z is, where do I start with the lowest order term,
05:23
plus a n z to the n will be the highest term, okay, where the coefficients a, j or a k are complex numbers,
05:42
okay, each such polynomial has a root, has a zero, has a zero, what's the zero? Well, it's a point where this b and c
06:01
such that f of b equals zero, p, sorry. Okay, I'm pretty sure you know the statement and have seen it a number of times. What is difficult about it, well, as usual,
06:21
what's difficult is there's no explicit formula in general, it's the real sort of the effect which really belongs to algebra is that you don't get an explicit formula, an explicit general formula once the degree is five or more.
06:40
You have up to degree four, there are formulas due to Cardano where you can explicitly solve it and those formulas led to the complex numbers, discovery of the complex numbers, but from five onwards, there can be no general formula, so this is an existence statement and doesn't tell you how to obtain the zeros.
07:04
Statement of pure mathematics, not of applied mathematics in this sense. Okay, so yeah, I should just draw one easy consequence which is this is about the polynomial having value zero,
07:23
what about having an arbitrary value of c? Well, that's the same thing since you subtract c from a naught and then a zero of a subtracted polynomial will be the point p of z equals c of original polynomial, so we see that consequently, p is either constant
07:52
or subjective, okay, so it doesn't even miss one point
08:04
which would be allowed by Picard's theorem, okay. So here's what most proofs, or there's a number of complex analysis proofs and they all use the same fact for a start
08:23
and so this is what I want to do first, but let me just to make notation rigorous, let me say I can assume that this number is non-zero else I have such an expression with a lower n, so we can certainly assume, can assume a n is non-zero
08:49
for an n in n, okay, yeah, if I have only a naught being non-zero, then I'm constant, yeah,
09:03
so I can really assume n is at least one, okay. So now comes the fact which everybody needs, well, everybody in complex analysis needs to prove this theorem, so what we show is that the highest order term
09:20
in some sense dominates the behavior of the polynomial. Okay, so highest order term dominates the polynomial the behavior of p in the following sense
09:44
that for p and large, say, and large z, values of z, large z, okay, in the following sense there exists an r and we may without loss of generality
10:04
assume it's at least one, such that p of z in modulus is, well, is at least half of this modulus, half a n z to the n
10:29
for z's which are at least r modulus. For all z with, okay, so this is easy to believe,
10:49
but let me give the proof, what is the proof? Okay, so indeed, with the following we want to estimate,
11:04
well, perhaps it's easier to write the z to the n onto the left hand side, so we get p of z over z to the n. Okay, so what is this? Well, actually, I want the modulus outside
11:21
so that I first divide, so I get, yeah, let's just do this, okay? So what do I get once I divide? Well, I get a naught over z to the n plus how many terms will I write?
11:41
The last one perhaps, last but one to power one plus, and since a n has the z to the n coefficient, is the z to the n coefficient, there's nothing there, and now I apply the reverse triangle inequality,
12:03
reverse triangle inequality to separate these terms and can say that this is less than a n, sorry, this is what I am at, a n,
12:20
and now comes sort of all the remaining terms, so I can rewrite this as a naught over z to the n, plus, plus, and the last but one is this one, a n minus one over z to the, z to the one, okay?
12:46
So far, so good, and now I will, since I want a nice estimate, I want to separate these terms as well, and that can be done by the standard triangle inequality,
13:01
so, and using the fact that I'm working with z's with modulus one anyway, so if I take, if I apply the standard triangle inequality, if I apply it, then I can write,
13:22
I want to, actually, I want to take one of these z's out since that will tell me that I will use for, to estimate with r, so one, I pull out, and the remaining terms here, then I have a naught over z to the n minus one,
13:45
plus, emphasis, plus, and so forth, last term is the plain term a n minus one, and since r is larger than one,
14:02
this is, a modulus is at least one, so this gets only smaller when I erase this, yeah? And I'm using r at least one. Okay, and I, similarly, I erase all the denominator z's
14:22
for all these terms in between, okay? And so, I want this, I want this to be, okay, now it's almost visible, right? If I can estimate this by a half a n, then I'm done, so I want that this is, the whole thing
14:43
is less than a half a n, and why is this true? Well, okay, so this will be true if and only if,
15:03
okay, what do we have to do, okay? Oh, so we subtract a n from this inequality, then we get minus is larger than minus a half a n, perhaps it's better to rewrite this,
15:20
minus one over z, times, times everything, times a naught plus, plus a n minus one, must be larger than minus a half a n,
15:40
by subtracting minus a n modulus, and so this is equivalent to, now I want z on this side, multiply by minus one, so this is z is less than, it's greater than, I multiply by minus one,
16:01
greater than two times this thing over a n, whoops. Okay, and so this is just a complex number here, and so I set this equal to r,
16:22
and if it comes out less than one, I set r to be the maximum of one, and this number, yeah? So set r equal the maximum of one,
16:40
and this here, okay, and then I'm done. Okay, so this is not really surprising, and now what makes the proof nice is how we use this fact, and as I told you, there are various ways,
17:01
we want to use Lyoville's theorem, so we do it in the following way, so suppose, okay, as usual, this proof is indirect, so we suppose p has no zeros, has no zeros, okay?
17:24
So the idea now is, well, if this is the case, then we can consider one over p, and one over p is the function we want to apply Lyoville's theorem to, okay? So then one over p is again an entire function,
17:49
and is holomorphic, is holomorphic, okay? And in order to apply Lyoville's theorem,
18:02
let me prove that one over p is bounded. Okay, why is this so? Well, for short, the argument is the following. On the ball, for those z which are less than r, on the disc of radius capital R,
18:20
I say, well, this is a nice continuous function, so it takes a maximum, so it's bounded by something. On the exterior of this disc, so for z's modulus larger than r, I have my estimate there. What does the estimate say? Well, it tells me that the polynomial is larger than a half a n z to the n.
18:44
Well, if I consider one over p, again, I get an estimate. It's less than one over the thing on the right-hand side, okay, which can be estimated in terms of r. So let's do that, okay? So there exists,
19:05
how do I say this, there exists such that f one over p, a modulus, is less than or less than or equal than, okay, c,
19:24
for on the closed disc about zero. Why is this? Well, by continuity, by continuity, and the theorem of the maximum, right?
19:41
On this compact set, a continuous function takes a maximum. Okay, and I don't know, perhaps this is clear now. Yeah, by continuity and by compactness. Okay, and the other estimate is one over p is less than, now I have less than two over,
20:03
two over a n z to the n, and I can estimate this. Well, since z is at least r, one over z is at most one over r. So this is two a n r to the n,
20:21
which is again a bounded, which is just a number. And so this is a bounded, this gives you a bound on the function one over p, and makes Liouville's theorem applicable. Okay, so that means that one over p is bounded.
20:48
One over p is bounded, so by Liouville's theorem and its entire, yeah, it's an entire function defined on c.
21:02
So one over p is constant. Well, that means p is constant, and we are done. So the main thing is really to get this estimate
21:22
on the exterior of the ball, which is not so hard, really. Perhaps it's good to visualize what's going on. Let me see. Before I keep on going, please.
22:21
And we will return to entire functions later. So, okay. So here's a complicated program. Basically we are, right now it does two things. It shows you for small values of z
22:42
and for large values what's going on with polynomials, and the polynomial is written here. So this is a fourth-order polynomial. And what we can do now is, okay, here you see the pre-image, the domain,
23:00
and here you see the range, okay? So now we can take little circles, yeah? First I want to look at the image of circles. But why is everything visible? Yes. Okay, now I increase r, okay? Whatever, okay, what happens, we will come to that later,
23:23
the red points are zeros of the derivative. And when you have zeros of the derivative, the function will turn twice or many times around them. Okay, now we cannot quite see what's going on, so we have to change scale. Okay, so now the setting is a little different.
23:46
So on the left-hand side, you see the domain circle in real size, so I will increase radius from, I don't know, five to 40. On the right-hand side, the image is scaled automatically.
24:01
So in fact it grows, but it's scaled back so that you can see what's going on. So if you scale on and on, what you see is exactly my main estimate here, that the polynomial looks almost like highest coefficient, like z to the n.
24:20
If it was exactly z to the four, what I show you, then on the right-hand side, you would see a circle, an exact circle of radius r to the four, travel through four times, right? Z to the four means a little circle. Take the image under z to the four, it goes around four times, and the modulus is r to the four.
24:45
But it's not exactly z to the four, it's z to the four with some, well, some garbage, right? And so that is responsible for the fact that you don't see exact circles here,
25:00
but you see something converging to circles when the radius gets larger and larger, yeah? So we can actually observe this, yeah? It gets more and more closer to circles. Don't worry about the little corners, this is just to speed it up.
25:21
It can be made smoother, but then it would take longer to show, so we decided not to worry about that, yeah? So it gets closer and closer to circles, and that means really this estimate here, yeah, you can, this estimate here gets better and better the larger z is. And why, so why at all do you expect zeroes now?
25:43
Well, think of the disk. This curve, what we see, bounds the disk. So let me go back to smaller radii once again. So perhaps even start with small radii, and now I look at disks. If I can see anything, let's increase r a little.
26:03
Okay, so now we see the image of the disk, which is bounded by the circles, which you saw before. Okay, and now you see once I hit these zeroes of a derivative, whose images are also marked in red, so on the left, the points you see in the domain,
26:21
the red points correspond to points where the polynomial has, the derivative has a zero, and on the right, you see the respective image points. Actually two agree here, so three points move on to two. Don't worry about this fact now. And if you increase, then you see an overlap, okay.
26:41
Now you see that the zero, the value zero on the right is attained at least, it's hard to tell, two times, four should be more than two times. Did you follow what was going on? Let's turn back. Okay, so unfortunately there's still a little artifact here,
27:01
which makes it harder to follow. Right now we have two pre-images of zero, two zeroes. Yeah, if the value z equals zero is covered twice on the right, it means I must have two zeroes of my polynomial, two roots. Okay, if I increase radius, I will see more of those,
27:23
so now we get to at least the third root, and now comes the fourth root. Now the value zero is covered four times. Let's go to bigger radii, and this. Okay, so think of a curve you've seen before,
27:42
which I said converges when we scale to four-fold covered circle. Now you see the disk, which is bounded by it. Well this disk, it doesn't have a hole, right? There is no, take a needle, and okay, no, it's not allowed. This is a complex function. Yeah, it covers all the interior.
28:01
Well, it must cover the value zero also, right? This is the geometry which is going on, yeah? Outside at infinity, it turns around n times, if it's a polynomial of degree n, yeah? And it bounds the disk. Well, this disk will cover the origin also n times.
28:24
Okay, and this is what I wanted to show you at this stage. Okay, so as I claimed, we have not only in this example, we had four zeros.
28:41
Well, up to now, we've constructed one zero. Let me show you that they're in fact n zeros. This is now easy, okay?
29:06
So I can say gain all zeros by a factorization theorem.
29:24
So it's the following corollary telling you that a degree n polynomial has in fact n zeros, n complex zeros. So for each, for each, well, polynomials as polynomial,
29:49
polynomial p of z equals, same thing as before, a naught plus a n z to the n.
30:00
Well, let me assume that a n is non-zero. There are n numbers, actually, there are numbers. Let's write it like this. C is a multiplicative constant,
30:23
and zeros be one to be n. So as many as the degree is such that, and now I can factorize p of z equals constant times, and I write z minus b one times z minus b n.
30:48
Okay, so here, the bs must, need not be distinct, right? They can coincide, so you can have multiple zeros here.
31:00
Yeah, in my, it may be that some of these agree. So the bk are not necessarily, not necessarily distinct. Necessarily distinct, okay.
31:26
And, but they are, yeah, perhaps I should write, but they are unique up to ordering. Yeah, but let's write it like this. The bk are not necessarily distinct,
31:41
but unique up to ordering. So I'm saying that a degree n polynomial has n zeros as expected.
32:02
So the main point here is that the complex numbers, we can actually factor out, factor out polynomials, something we cannot do in the reals.
33:00
Okay, so let me prove that. So this will be a proof by induction. We split off one of these factors by the next.
33:21
So that means by the theorem, by the theorem, we have one zero that exists, zero, say, zero b one of p. And now all I need to know is that I can rewrite
33:42
the polynomial as a development with respect to b one and get p of z equals c times, no, sorry, a naught, I want a development with respect to a one, a naught plus a one.
34:02
I should have different coefficients. Alpha I used here, z minus b one, and now higher orders of z minus b one. Okay, why can this be done?
34:21
Well, think from the back to the beginning. If it has degree n, you can consider this term with the same, with alpha n equals a n. Well, the difference to a n z to the n is given by lower order terms where z comes out
34:42
with power n minus one at most. So then you rewrite the term with degree n minus one, taking this into account and so forth. That's the way to do this. It's basically the Taylor series, Taylor development of the polynomial
35:02
at a different point also. Okay, so once this is done, yeah, then, okay, then, well, since b one is a zero, alpha naught must be zero. Otherwise, b one wouldn't be zero.
35:23
And so, and taking that to account, if this is not there, we can actually divide by z, or factor out the one factor z minus b one. So p of z equals z minus b one times q of z,
35:45
where q is now a polynomial of degree one lower. Yeah, where q has degree one n minus one. N minus one, so again, my fundamental theorem,
36:02
q has a zero, factor it out, and make an induction proof, and this gives the statement, yeah, fine. So induction, induction gives the claim.
36:20
Okay, so this basically didn't work to be done in order to get from one zero to n zeros, that's easy. Let me also remark that if you're interested in the real case, if you have a real polynomial with real coefficients, looks the same, of course,
36:43
but now think of the a n's as real numbers. Then you can, again, you apply the same theorem we had. Well, with noting that if b is a zero,
37:01
then also b bar is a zero, so the picture of zeros is symmetric with respect to the real axis, then what you obtain in the end is that you can factorize it,
37:20
with, in the form, perhaps I should actually write here x to make it look real, to in the form x minus b, or, and now if you take two, if you take, say, bk and bk bar together,
37:41
and multiply two of these factors out, here, z minus bk times z minus bk bar, then you get a term in the form x squared minus, or plus, whatever, a real term, say, I need a x plus z, yeah?
38:05
And you have terms of this and that form, yeah, you need indices now, and I don't want to do everything here, but it's clear what I get, yeah? So, in particular, for all degree, I must have at least one of these factors in the factorization of my polynomial,
38:23
since the other ones have even degree. Okay, so this, well, the fundamental theorem of algebra has a long history, and I wonder if I should tell you
38:40
about the original proof of Gauss from 1799, which I've prepared a few slides of, but perhaps I do this at the end of the class, depending on how the time works out. Yeah, so there's a long history of proofs, and actually, the first proof of Gauss was not complete.
39:02
Okay, let's come to that nice point later, and let me start with the next section, which basically tells you something about mapping, properties, mapping properties of holomorphic maps,
39:25
or holomorphic functions. Okay, so one thing I want to start with
39:44
is the discussion of maxima, of extrema. Well, a complex function cannot have extrema, since it's R2 or C-valued, but what you can consider is, you can consider the modulus of F and discuss extrema.
40:14
Yeah, remember that in standard analysis, real analysis, the discussion of extrema
40:20
is one of the main subjects. Well, here, it's not a main subject, since we can only consider mod F, but there's still a nice theorem on it, which is the following. So-called maximum principle,
40:43
or actually, I read, I saw it called maximum modulus principle, which is a nice name, principle, maximum princip. So it's a principle of a maximum for mod F.
41:06
So let me first write it out, and then discuss it. Okay, we take, let U be a domain, which is important here, so I emphasize it once again,
41:20
and F from U to C, holomorphic, as usual. And now, what's the statement? Well, the statement is that the modulus of F cannot take a maximum. So what does it mean? If mod F attains a maximum,
41:47
a maximum at some point within U, say at, and let me call it an interior point, although this is, right now it's meaningless,
42:01
interior point, since a domain is, a domain is open, so each point is an interior point, and an interior point B, or how do I want to call it, yeah, B and U, okay, then, which means that, that means F of mod F
42:25
is less than mod F of B for all Z and U. Yeah, I'm not talking, as usual, I'm not talking about strict maxima, or extrema, but just weak maxima.
42:42
So it's the largest value which is around, then F is constant. Okay, so that means it's impossible for F to attain an interior maximum,
43:02
and let me just show you how Wikipedia explains it, which is okay, which is the, okay, so what you see there is simply the unit disk
43:24
and the graph of a function cosine Z on the unit disk. Still not visible, okay, it's too large, too large, okay. And I, actually it's also hard to see, yeah, but what you see is that actually there's no,
43:43
I mean, it's a saddle-shaped surface, right? On a saddle, you always have maxima at the boundary, right, in fact, for this function here, it looks also as if the minimum is attained only at the boundary, namely the minimum would be attained, well, probably only there.
44:03
The maximum, I guess, the maximum is attained there. But this is not true in general, so in fact, this is not the best picture. If you take a larger disk, this is the unit disk, but if you go to pi over two, then, I mean, since the modulus of a function
44:21
is always non-negative, and cosine has a zero at pi over two, then you would see a zero as a minimum. So this is, I guess there are better functions actually to show, but due to lack of time, we didn't get it done, really,
44:41
in the visualization project. But the main thing is, think of modF as a function which cannot have a maximum, saddle points and minima are fine.
45:01
And, actually, I guess the best is if we have the break right now, since the proof takes a little time. Okay, so let's make the break right now. Okay, I would like to continue.
45:23
So actually, there's another statement missing, which I forgot. And perhaps I scribble it in here. To, okay, if U is bounded, if U,
45:41
no, this is not enough, sorry. So let's write it here. If U is bounded, then, and F can still be defined
46:04
on the boundary of U, and F extends continuously to DU, particular, perhaps, it's actually holomorphic
46:21
on all of DU, or a slightly larger domain, containing U bar, then we have, then we can estimate the values of modF
46:41
by the soup of modF on the boundary, which is exactly what you saw in the image. Yeah, the maximal values are attained on the boundary, not in the interior. Or, in the case, the function is constant,
47:01
they are attained everywhere, but they're certainly attained on the boundary. Okay, so I would also, before giving the proof, make two remarks. I mean, for one, for one thing,
47:22
certainly this is not true in the real case. Yeah, I mean, there are many real functions. So certainly, certainly not true for real functions, or in real case, real setting.
47:40
Yeah, there are many functions which take their maximum in the interior, graphs like this. So the question is, why can it be true? What's the magic here? Well, again, it's the equation. And so I can say, for instance, however, it's true for harmonic functions,
48:03
equation Laplacian equals zero. And it's true, actually, without a modulus. Yeah, a harmonic function is the one with Laplacian u equals zero. So this is a scalar-valued function,
48:20
so it's really true that u of z is less than the supremum of u, or that if u attains the maximum at interior point, blah, blah, then u of z is less than u itself is constant. Okay, and actually, there are more equations which you can come up with the maximum principle.
48:42
For instance, the heat equation I've mentioned before. So I could say, all four solutions of the heat equation. Yeah, if you have the heat flow, then it's not true that suddenly
49:02
at time five and locus six, you get a maximum of heat, but it spreads out. So we'll not have a maximum only on the boundary. So it's, once again, you should think of the Cauchy-Riemann equation as being responsible for the maximum principle to be valid.
49:24
Actually, there is a very close-cut connection to the Cauchy integral formula. Here is, once again, the Cauchy integral formula. Remember that once you plug in the center of the disc into the Cauchy integral formula,
49:41
then you get the mean value formula. F of b is equal to one over two pi integral from zero to two pi f of b plus r to the EIT, which is exactly the mean value of f on a circle. Okay, taking this formula into account,
50:04
it's no surprise, I mean, this formula into account, f of b is less than its average on a circle. It's no surprise that the maximum principle holds. So, and so this is actually the key idea of the proof.
50:22
So let me just start without the technicalities and say that f of b by the mean value formula is one over two pi, well, perhaps I don't need to write it up as it's written there,
50:45
but I can just estimate it by the modulus of f of b plus r to the EIT. One little, say, I will need it with r to the little i t later, dt. And so if a quantity, I mean,
51:06
yeah, if a quantity is equal to its average, it will not, and if it coincides with its maximum at an interior point, well then the only way is that it's constant. Yeah, that's no surprise,
51:21
but let's make a mathematical proof out of it. So the proof is a little technical. So for one, let's do the following. Consider, so in one, we say there's an interior maximum of a function. We want to show that f is constant.
51:42
The idea here is look at all these points where f takes its maximum. So let's call this set m, for instance. Consider m, the set of points in u, such that f of z is maximal,
52:05
is equal to the value taken at b, and b is the largest value. Okay, what do we want to show? Well, we want to show that f is constant, so m is all of u. What's the standard?
52:20
So let me also say that this is a non-empty set since b is obviously in it, right? So what's the standard way of showing that m is u for u domain? Well, show it's open and closed. Then it must, since moreover it's non-empty, it must be everything.
52:40
So why is it open and why is it closed? The easier thing is why is it closed? Well, this is a closed condition, being equal to something. So the set is closed right away. So f or mod f continuous means that m is closed.
53:05
Formally, continuity means pre-image of a closed set is closed. Well, this is a point is a closed set. So this is a pre-image of the point, the real number mod f of b, the inverse image.
53:22
So it's a closed set. Yeah, that's the formal, formal reasoning. Why is it open? Well, to be shown, yeah. M open is to be shown. And if it's both, well, since u is a domain,
53:40
u is a domain and m is non-empty, this shows that m is u, which is the claim, i.e. claim. So what we need to show is that the set is open. Yeah, so each point in the set has a little neighborhood.
54:02
So each point where f takes its maximum, like b, has a little neighborhood where f also takes its maximum. And this will follow from the mean value formula.
54:28
So, so we have r larger than zero such that b r of b is contained in u.
54:44
Actually, I want compactly contained in u. So in the notes, the of b is always missing. I'm sorry. Okay, so in u, within u, I have my point, actually, I have an arbitrary point, z.
55:04
Actually, I want, let's show this first for b and then do it for every point, and then say b is arbitrary. Why not? Okay, so if b is in the set, then I want to use the mean value formula for a circle.
55:24
Okay, so I pick and actually, in order to show that it's constant, in some neighborhood, I take an arbitrary circle. So pick a little r
55:40
in zero r such that, not such that, just do it. And then the claim, then the claim is that on d b r of b,
56:03
f is also mod f is also maximal. So on the entire, so this is r, and so on a little circle, little b r, the function will coincide with its maximum value, f of b.
56:22
This is the claim, and we want to use the mean value formula. Okay, so let's assume this is not the case. So we are on a little circle, and there's one point on the circle, this is actually bad drawing, sorry.
56:40
There's one point on the circle where f is not equal to its, f mod f is not equal to its maximum value. Well, that means mod f is a little smaller, right? So, I suppose not. Suppose not, then there exists a t naught
57:04
in zero to two pi, such that mod f at b plus r to the e i t naught is strictly less than f of b, okay?
57:25
Well, if it's actually strictly less, well, by continuity of this function as a function of t, then there's a little neighborhood on which this is also less than mod f of b, right?
57:45
So if this corresponds to t naught, I have a little neighborhood on my circle where this is also smaller, okay? Then by continuity, by continuity,
58:02
same inequality holds in a little neighborhood. Same inequality, same inequality here holds in a neighborhood, neighborhood of t naught.
58:24
And that means looking at my mean value formula, then mean value formula, mean value formula gives,
58:45
well, on one hand, we know that f of b is less than the average of one over two pi integral of f of b plus r to the e i t dt, okay?
59:06
On the other hand, we know that this, on an open interval of t, well, this is always less than b, f of b, and at certain points, t in the neighborhood of t naught,
59:21
it's strictly less. So in fact, the integral must be strictly less than, strictly less, sorry. This is strictly less than the integral of f of b. But that is, so I could write, yeah,
59:44
using this fact here. If I integrate a function which is strictly less by continuity of, by monotonicity of the integral, if it's strictly less than this value, then it's strictly, then I get the strict inequality for the integral.
01:00:00
dt but this is f of b and a modulus and this is since it's the average so it's a contradiction right so it cannot be true that on a circle about on a circle contained in you about my given point B in the set that on the circle the
01:00:26
function is actually a smaller modulus than f of b it must be equal right so and since this was for arbitrary R I get I get for all little r's in this interval
01:00:42
I get that f of this here f on the circle is equal to f of b so okay so how do I write this out so on DB R of B we have f of z equals to f of b for all r
01:01:13
I mean for all r in 0 to R so that means that we have on the entire disc we
01:01:22
have this and also by the same argument we can replace B we can replace B here by any point in the set M where F is maximal yes use the same argument and show that it's open and so that tells us that the set M is open yeah same for
01:01:46
any set in M and so in place of B place B and so that tells me that M is
01:02:03
open each point in M has a little neighborhood contained in in M and that comes from the mean value formula okay so this is sort of a little technical use of the mean value formula since since you have a problem is that we want to extend a point wise inequality for function really to an integral
01:02:25
inequality and then we need continuity and this is this is the tricky bit here okay and so for two two is now easy so you is you bar is compact
01:02:45
meaning that f mod f takes a maximum takes so what if I mean I required that f extends continuously to du so actually also mod f extend is a continuous
01:03:04
function on the closed domain on the closure of a domain and so f takes the maximum back by compactness f takes a maximum on you on the compact set u
01:03:22
bar and now let's look at the two cases I need a point say at B and you
01:03:43
okay so now I have an alternative either B is in the interior or B is on sits on the boundary while both cases I'm done why is this so I have a B is it's in the interior well then part one tells me part one tells me that in
01:04:06
fact f is constant f constant and then a constant function certainly takes its soup at the boundary so we are done or B is in du well then we are
01:04:24
done anyway yeah then the maximum is taking of a boundary well this is the very thing we wanted to prove okay so this gives you proof of the maximum principle let me remark that that actually for there's not a similar
01:04:48
principle for the minimum of mod f however it's a little just a little more complicated problem you can show and it's not hard to show just take
01:05:00
one over the function that if B is an interior interior minimum minimum of mod f then either f is constant as before or we are at a zero of f f
01:05:25
constant or we're precisely at a zero so the only zeros of the modulus can occur at zeros of a function and certainly there must be it both at
01:05:43
zeros any zero is a minimum of mod f clearly okay so now for remaining part of the class my goal is to show you a little more dramatically well how how complex functions look like there's some
01:06:05
theorems also but the main thing is to get an idea of what's going on
01:06:24
let me see what I can achieve okay so let me let me write this down as a
01:07:13
heuristic principle it's actually true but I don't want to prove this which is easy well easy to state for f entire so let's I can look at the f
01:07:32
functions holomorphic functions on domains but let me just concentrate on entire functions although this is not necessary but then it's easier to see what I want to say well then my claim is then okay then I claim that f
01:07:51
has the same number of pre-images pre-images however this number may be
01:08:21
point in for any point in range except when I hit when I'm at zeros of a derivative so say for any point in the range say W in C such that for all
01:08:44
said in the pre-image and I have f prime is non-zero okay so only for critical points critical points where the derivative is zero I don't want to
01:09:04
count images but outside I won't think of a polynomial we've seen outside the red dots I have the same number of pre-images was for in the example I showed you this is the statement I want to make here for an arbitrary function and also I want to say that if okay perhaps I can say the following
01:09:38
okay actually this follows yep follows from implicit mapping theorem implicit
01:09:49
mapping theorem why is this so let me give you the idea for this so here's
01:10:00
my domain well here's my a part of my domain in order to draw something and look at some F well perhaps this is multiply I don't know this is multiply some some set of a complex plane is multiply covered some set is singly
01:10:25
covered and some other sets are not covered so this is not for f entire but let's just look at f from u to C right so this is and this set here is for you say come you with complex with compact closure okay why is this so
01:10:47
well take any point here this is the point W say my assumption is that the pre-image contains no values which are derivative zeros so I may say
01:11:00
perhaps I want to take take it here in order to have two pre-images and for z1 z2 well I know I assume that I'm not in points where the derivative is zero how does the image look like well by the implicit mapping theorem little circles map to little circles here so if I have two pre-images at W I have
01:11:24
two pre-images in some open neighborhood of W same thing with n1 or 0 pre-images right okay so this is this is where this principle comes from and it at points okay at points with F prime F points Z with F prime of Z equals 0
01:11:51
the heuristic idea is write out a Taylor series then when we have F of
01:12:01
okay I need a Z naught or let's call it B when we have F of Z equals okay then if I have a zero derivative zero of a derivative when the Taylor series doesn't start with a constant term but it starts with higher order right so
01:12:20
I have a and Z to the n plus a n plus 1 Z to the n plus 1 and so forth okay for n and n and that means okay if I'm sorry is it minus B I should write
01:12:40
that minus B I'm sorry since B is that point I'm I do my her a and Z minus B to the n okay now it should be correct if I'm close to B then I can heuristically I forget and forget about this then my function looks like
01:13:03
this how does this function look like well it has n roots right so if I look for instance at Z squared yeah then I see going turning around the origin entering once around the origin sorry sending once around the origin I get I
01:13:28
turn here twice around the origin actually on any little circle I can do this and so that that is the fact that I have and this is also effect from the fundamental theorem of algebra if you like it if you forget about
01:13:41
all this stuff but here you get n roots of the equation so there may be points here with a single image but those are the points where where F prime so I may have points Z such that I have only one image here but then
01:14:00
within the two but then F prime of Z is zero and the Taylor series would start with a second order term so think of the values being punched faffed it at one point and to start to tell you the geometry let me do the
01:14:20
following let me show it show you some images of the computer and then I will see what I can do formally in terms of theorems okay let me first
01:14:47
show you just an image sorry okay so here's a visualization of Z squared actually what you see there is fee the points in space of a form said squared
01:15:10
two coordinates and real part of the function and there you see how it turns twice around with a with a point with just one pre image in the middle of a
01:15:24
right you see the same thing for Z cubed then the function turns three times around the origin and it's punched at zero in the actual function you don't see this part so you should think of squashing everything to be XY
01:15:44
plane and you did get a triple on the right-hand side you get a triple cover of the of the complex plane except for this one point where which corresponds to a value of zero and same thing here you get a double cover so in general you get an n fold cover around points where the
01:16:03
derivative is zero and this is the intuition you should have also for the polynomials here so let's go once through this so perhaps we start in
01:16:22
the mode with the disk it's unfortunately little forget about the cone there this is an artifact so if we increase the radius then and we hit oops sorry this was way too fast I'm sorry so let's okay so this much here
01:16:47
is where I don't hit any zeros of a derivative then I come to a point where the zero the derivative has a zero now I see a little spike right I there's
01:17:03
why do I see that well think of think once again of that maps to z squared take for instance via the open upper half plane in the complex plane I mean yeah and map it under z squared to the complex plane what do you see well if
01:17:26
it's the open upper half plane then actually and the angles are doubled then you get everything except for this here so you get the exterior of the you get all those except the positive real line and zero so if you look under
01:17:47
microscope what you see what you see at zero at the second order zero here then you will see exactly such a spike however it's nonlinear so it will open up or do something yeah so now if we increase radius yeah this is the
01:18:07
spike here you which is actually easier to see in the in the circle mode okay if you increase the radius now beyond the zero of the riveted then you get a double covering up to the moment you hit the other zero and you
01:18:26
get the same behavior right and then you get you get a multiple covering and you should think on the of the image of right hand side as several sheets which are punched at the red points and that should give you some
01:18:43
feeling on how these how the holomorphic maps actually look like and let me see what I can tell you about theorems okay now works so what I try
01:19:02
to tell you heuristically now is part of the following formal statements so let me it's actually the theorem open mapping theorem which is yeah has
01:19:20
different names English and German mapping theorem and in German it's usually referred to as the that's ready beats Troyer so what does it tell me it tells us that a holomorphic function a holomorphic
01:19:46
function is either is either constant so if we don't work on the domain I
01:20:01
should say locally constant yeah I'm not assuming connectedness right now or it maps open sets to open sets in particular since our amorphic functions
01:20:24
continuous and connectedness is preserved under continuous maps we have a holomorphic holomorphic image so this should be an in particular in particular the holomorphic image holomorphic image of the domain is a
01:20:47
domain of the domain is a domain okay so this is something you've seen
01:21:02
envy in the computer pictures but let me show you a just to remind you let me show you a map which is not open so in the middle of the slide you see a real polynomial whatever it is I'm talking about this image here it has
01:21:22
a rank to line here whoops a rank to line here so that means that there is a set there is a set in the domain like I don't know like so a little ball which is mapped to this thing here and so the image would be a non open set yeah
01:21:46
it's neither open nor closed right so it's really a prominent property that holomorphic maps are also don't get confused with continuity the pre-image open sets under continuous maps is open here we talk about the image of a tour
01:22:04
of an open set okay so this I mean it's it's hard to see the kind of information you get from this theorem on the start but it's it really tells
01:22:22
you a lot about holomorphic maps and it's exactly what we saw on the computer with the polynomials that the boundary is always on the outside right here the boundary of the domain is not on the outside it doesn't bound this boundary of a square doesn't bound all of the image this triangle here is not
01:22:44
bounded for a morphic map it would be okay I will I can see there will be hard to prove this so let me just tell you there's also a lemma used used for
01:23:01
that shall I tell you about this yeah perhaps it's worth showing so let you be simply connected connected so this is telling you something about taking n
01:23:22
fruits and f from u to c be holomorphic holomorphic with but I'm assume that f of z is nonzero okay if I don't have a zero well then the
01:23:43
statement is I can take n fruits so actually what I showed you with the Taylor series works in general so then then website only state you state what
01:24:00
two says well actually I have a picture for one let's do both cases then exists G to you to see such that holomorphic holomorphic with such that
01:24:27
f of that can be written as the exponential to G of that so this means this means G is is log f right G is logarithm if it can be written as the
01:24:44
exponential of something and now I state the same thing for the nth root outside zero so for all n and n we can write exists exists a holomorphic function holomorphic holomorphic H from u to c such where H to be n equals f so
01:25:14
what this tells you is f H is the nth root H is nth root of f but of course
01:25:27
we are not allowed to do this in general yeah there is no well-defined nth root it's only a function taken to the n equals f and let me show you so let
01:25:40
me show and okay in order to prove the theorem this is this is the key step really but I will not be able to prove it today and probably I will skip the proof and I ask you to have a look at for proof but let me show you what it means in the following picture okay minimize sorry okay so sorry you need
01:26:25
the Beamer mute off light okay so here you see here you see a domain you which is simply connected and here you see what this theorem actually
01:26:43
would tell you for this for this logarithm function of just take the function Z and express Z as e to the something so how does this function look like well it's the logarithm so I can it cannot display the logarithm but I can
01:27:00
show you what the imaginary part is imaginary part is the argument and then you get just a thing like this so this tells you that on simply connected domains and simply connected domains you can take a logarithm which is exactly the statement here so it's it's highly non-trivial although it looks like an algebraic statement okay so I see that I cannot explain
01:27:27
everything to you and you can let me you can let me know if you want to see the proofs next time but in any case I wish you Merry Christmas and good start of a new year see you in January thank you