 Okay, today I want to show you that what we have found for simply connected domains in the plane, different from the plane itself, this is to say that two of them, any pair of such domains are in fact biomorphic is not true for non-simply connected domain in general. In fact, we will find very strict conditions. Okay, so I always start with a simple case, we can imagine, so instead of considering simply connected domain, we consider an annulus. We already use the annulus, okay, for the Voron series for which is the natural domain of definition of a meromorphic function, right? Good, so I want to prove the following. So consider two annuli, so for instance centered at the origin with this is a and this one and this is a two, okay? R1, Rj, say, and are like this, positive real numbers, right? These are the two annuli. From the point of view of topology, that's obvious but it's better to remember this. Well, I drove them like this, right? From the point of view of topology, you cannot distinguish one from the other, okay? You can deform one into the other, right? So topology doesn't classify, okay? And apparently, this works fine for simply connected domain. No, not apparently, but for sure this works fine also for simply connected domain. So you can take very odd domains but simply connected different from C and they are in fact deformed with some function which are not only homomorphism but even more biomorphism, okay, with a unit disk. So what if I try to find a biomorphism between two annuli? This is the natural question, right? Is it always possible? The answer is no. Well, first of all, we consider this simplified version of the statement. So instead of considering R1 and R2 to be two real numbers, positive real numbers, we can always deform one of the both annuli in such a way that R1 and R2 is equal to 1, right? And the deformation is okay because it's also a good function from the point of view of complex regularities of R, deformed but not. So we can assume that, okay, without loss of generality, the two small radii are equal to 1, okay? Just to simplify things. And assume that F from A01R1 into isobiomorphism, all right? So assume that you have this. No obstructions in terms of topology. So for instance, if I try to use the annulus and annulus with two holes instead of one hole, okay, this cannot be possible, right? Because there are some restrictions given by topology. But here topology doesn't tell you anymore than say, okay, you can do this, okay? We'll see that this will imply something important on the two external radii. Okay, all right. We know that F is a holomorphic function and for the holomorphic function, okay, biomorphic of F is holomorphic function. So with the inverse, which is also holomorphic, the open mapping theorem holds. That is to say, if I take Z and the first annular, the first annulus, and consider a Z which approaches this boundary here, what you have is that F of Z cannot be internal, right, in this second annulus. It has to go to the boundary. So it is either converging to one or to R2, right? This is the only possibility. Are you with me? Because otherwise you would, right? All right. So I assume that by contradiction this number here, which is a real number, is in between. So it means that as Z approaches the boundary in this way, the points tends, F of Z tends to an inner point. Now this would be, okay, a value which is a neighborhood, right? Remember that F is biomorphic. So that also the image of F minus 1 should be, right? So this point which is a boundary point becomes an inner point, which cannot be. I'm assuming that F is a biomorphism, right? As I said, it's holomorphic and biomorphic, right? So it can only be that this number is either 1 or R2 because this is the boundary of the second annulus, right? So it's not, okay, this time I have to write 1 here. Without restriction, I can assume that F of Z actually tends to 1 as Z tends to 1 because otherwise I would take F tilde of Z to be R2 over F of Z, right? So I made some simplified, simplifying assumption that is to say that the first two radii are equal to 1 and then because of this, I can also assume that this is the case, right? Otherwise I should consider F tilde instead of F, right? Might consider. So also this function is well-defined. You see that because this is not 0, well-defined, it is, of course, a biomorphism as it is F. So we can use this notation that F is a biomorphism from A 1, 0, 1, R1 into A 2, 0, 1, R2 such that F of Z tends to 1 as Z modus tends to 1, okay? And then we define a function U of Z to be logarithm of modus of F of Z minus M. M is a constant number, logarithm of modus of Z. So this logarithm is the real logarithm because this is the real number. M is a real number, so that is to say U is a function from A 1 into R, okay? This is just a new function I'm considering. And we have, well, actually without saying anything, I might, I have to say that M defines here for any choice of M another function so that we have here an infinite family of functions like this. Okay, observation. U is well-defined and for any M this is true, right? No problem because this number here is never 0 and this number here is never 0, right? Furthermore, since, well, this is the case then you are composing something which is never vanishing, okay? So you can write it also this way if you wish. 1 half logarithm of F of Z times F of Z bar which represents the modus of F of Z square, right? So I put 1 half here because of the square, right? So the rules for logarithm works fine and if I write this way you notice that the function U whatever M is is in fact the C infinity function when considered as a function from R 2 into R. So the components are C infinity. You cannot say that this is homomorphic because it is real value function, right? And we already said that well if the imaginary part is 0 so the function has to be constant and this function is not supposed to be constant, right? This is a biomorphism, right? Good. So now the question is how can I describe some properties? Well I can use one of the, in one of the first exercises I ask you to prove that the Laplacian, the real Laplacian is associated to the Cauchy operators, right? And so I want to calculate this which is up to a constant the Laplacian of U. This is four times, right? If I'm not mistaken with a constant, in any case this is a constant times the Laplacian. So if I calculate this and remember that the U of Z was defined as 1 half logarithm of f of Z times f of Z bar minus m over 2 logarithm of Z times Z bar. Well since the functions involved here are all analytic, okay? In particular the function U is C infinity but the derivatives are continuous so that it can also apply Schwarz theorem, right? So they can invert the order of derivation. So in particular when I calculate this, all right, either you write this way or you write this way, right? That's what I wanted to say yesterday. It's a second order differential operator, right? So what I mean is differentiate first with respect to Z bar then with respect to Z but you can of course change also the order of differentiation. So I try to follow this rule. So first differentiate with respect to Z bar and then with respect to Z. So this is 1 half the logarithm of f of Z which becomes U over C and here I have 1 over 2 f of Z bar, right? Because I'm using the chain rule, right? So this is and then I have to differentiate this with respect to Z bar which apparently is a product of functions, okay? So I should imagine that I have to use the Leibniz rule but deal with the Z bar of f of Z is 0 because f is holomorphic. So the only part which remains here is f of Z times, okay? And similar here I have minus m over 2, okay? This and times Z, correct? So Z and Z comes because it's not 0, f of Z comes because it's not 0. And then I apply the other derivation and have, this is, sorry, yes, D over DZ of what? Of 1 over 2 Z bar, D over DZ bar, f bar of Z minus m over 2 D over DZ bar, 1 over Z bar, sorry, this is not Z. Which means that this is 0, obviously, right? Because this function here is antihelomorphic which means, okay? And here I can use the fact that, well, I have, this is a product of two functions, right? I have the derivative of this which is 0 for the same reasons, right? So I have DZ times because this is antihelomorphic and this means that the derivative with respect to Z or an antihelomorphic function is 0. So this is 0, okay? And plus then I have 1 over 2 f of Z bar and then I have the derivative of Z derivative and this is already 0, right? Whatever m is, sure, sure, right? So this is not 0 in principle. No, this is not 0. This is the derivative, the antihel derivative, okay? Of this antihelomorphic function. But since I have these two operators here, right? And I can exchange the order of differentiation. I can also say that this is 1 over 2 f, but since this is 0 of D over DZ bar, D over DZ f bar Z. And this is 0 because this, as I said before, this is antihelomorphic and this is the bar, the operator, this is the bar. So this is 0. So it means that U is harmonic. U is harmonic. So conclusion is, okay, this is, that is, D U is 0 and this is true for any m, right? Okay, so I have this function here is harmonic and furthermore, as Z tends to the unit circle, that is to say when the Z as modules which tends to 1, here we have that this is the modules, right? The square over 2, sorry, right? That is to say this tends to 0 and this tends to 0 again because we are assuming that modules of f of Z tends to 1 as Z tends to 1. This implies the following, that is to say U of Z tends to 0 without any assumption on m. Now as Z tends to, in modules to R1, that is to say if Z tends to the boundary of the first annulus but not the interior, but the smaller, but the larger radius, what happens? Well, we have that this number here becomes, well, we have that this becomes log of R1 square, right? 1 over 2 and then I have a ham here because Z, this is modules of Z square, right? You can write it also logarithm of R1 because this is the square root, right? And here I have similarly that this tends to 1 half logarithm of R2 squared. As we noticed, since it is a biomorphism, it cannot tend to 1, right? R1, right? And so I take m, since I have the freedom of choosing m whatever I want, m to be logarithm, let me check if it is correct, of R2 over logarithm of R1. With this choice, up to now, all the conclusions are true for any m but then I take this m. This is a real number, right? And this real number makes the following that Z tends to 0 as Z tends to either 1 or R1. So we have annulus A1, a function u defined in the annulus which is there, harmonic. And the values on the boundary is 0, okay? So how does it look as a harmonic function with boundary values equal to 0? This constant equal to 0 because for harmonic functions, I think that the harmonic function has the real, take for instance this as a consequence of the fact that the real imaginary part of a holomorphic function, in fact, harmonic, right? Then you can show that given a harmonic function, this is related to holomorphic and anti-holomorphic but what we have is that we have a maximum principle for a homomorphic function, right? Okay. So up to now, we have just a function u which is harmonic in the annulus and for any m, it turns out that as Z approaches the unit circle, so it is the smaller, the inner boundary, okay? If there's any meaning, okay? We obtain that since we have assumed that f of Z tends to 1 as well in modulus, then u of Z tends to 0. No, no, what I said is, okay, probably I didn't say, okay? These are the two possibilities but if the second case occurs, as I said, I can always take f tilde to be this function here which is also a biomorphism and in such a way that this is true. So I'm assuming this, since it is a biomorphism, well, I didn't say it explicitly but maybe it's, well, thank you for the question. Now I'm taking without loss of generality a biomorphism from one annulus to the other with the property, well, the two annuli of the smaller radius equal to 1. And as Z tends to 1 in modulus, okay, so it tends to the unit circle in the first annulus, f of Z tends to the unit circle in the second annulus, okay? So this is the assumption which is without losing generality in our consideration. And I then constructed here a function u which turns out to be harmonic with this parameter here. So I have a family function and since this is the assumption I made, I also said that u of Z tends to 0 as Z tends to 1 in modulus. When Z tends to R1, f of Z in modulus cannot tend to 1 because it is a biomorphism, right? So it has to tend to the other boundary. That is the same modulus of f of Z tends to R2. That is, u of Z tends to this. Since M here is not chosen up to now and for a suitable choice, that is, if M is this real number, it turns out that u of Z tends to 0 as Z tends to 1 and as Z tends to R1. So the boundary value of the harmonic function is 0 but the maximum principle applies also for harmonic function, okay? So if a maximum is taken of the modulus, it has to be taken on the boundary. So it is that the function is constantly equal to 0 with this choice of M, okay? So let me probably write it down and another slide, so fine. Maximum modulus principle applies to u since it is harmonic. This is because in general, it is true for any harmonic function but in our case, it is even easier because, well, the maximum modulus gives, well, this function here is monotone, right? Since R1 and R2 are greater than 1, right? What I'm saying here is that the maximum of this number here depends on the maximum of these two functions, right? And these two functions are not harmonic but they are holomorphic, so they have to be on the boundary. Good. So maximum modulus principle applies also to u and u of z is 0 as z tends to 1 and z tends to R1. So we have the annulus. u is defined here. It's harmonic and the values on the boundaries from the boundaries of the annulus is the payoff of circles on this boundary, the function u takes value 0 or tends to be value 0, right? That is to say, the function u is constant. With the choice of m that I made in this because this is not true in general but with this suitable choice, I obtain the second fact. The first fact is independent on m, which is harmonic. But if u is constantly equal to 0, I have that logarithm of f of z minus m, logarithm is 0, okay? Or I have that logarithm of modulus of f of z is m, logarithm of modulus of z. This is true for any z, which can be also written as logarithm of modulus of z to the power m, correct? But the logarithm is the standard logarithm, okay? No tricks up to now. The standard logarithm is as I said monotone, right? So it is invertible and then we can say that from this it follows that modulus of f of z is modulus of z to the power m because this is the real logarithm. I didn't put r, but it is obvious. This number here is a real number. So if you want, it's the restriction on the reals, okay? So no problem of the termination of the principal argument and so on. All right. Now, we take in the annulus, well, the annulus is not simply connected, obviously, okay? Step zero. I didn't say it, but I can always take, say, a disk which is simply connected, say, disk D while contained in the annulus, correct? Well, I can take many disks, okay? As soon as I take a disk, the disk is simply connected, a subset of z, and it never, well, it has to be simply connected, right? So it cannot go around the zero. It can take also something, say, not as a disk, something odd, but biomorphic to a disk. No problem. So what you cannot want to have is another annulus around here, right? So that on this domain D is simply connected while contained, okay? Even something weird like this, but simply connected. In any simply connected domain, I can define the logarithm. Remember, this was a lemma which I proved several weeks ago, okay? You have a simply connected domain. You have a function defined on it, which never vanishes. Then you can define a function g such that x of g of x is the function, and then it is also uniquely determined as soon as you fix the one value, right? And the termination is unique, which is important. Then I define, if it is simply connected, let me say, define a logarithm, complex logarithm, of course, and D, okay? This can be done. Okay? Such that since f of z is not 0 for any z in this domain D, even in an entire annulus, we can write f of z to be x of g of z, okay? This is in fact the lemma because D is simply connected. Good. Now I have that modules of f of z is modules of z to the power m, right? This is also the modules of e g z, okay? I write it also this way. The exponential is the complex exponential, of course. And from this, I also have that modules of z is the modules of e g z over m. Remember that m was a real number and it was determined as the ratio of the two logarithm, logarithm of R2 and logarithm of R1. It was chosen properly in order to have vanishing the boundary on the second boundary. Good. Now I define this function g of z, capital g of z to be e g of z over m over z. So this is holomorphic in D. Why? Well, because this function g is holomorphic in D. It is defined in the simply connected domain. Z is a holomorphic function. We have a ratio of two holomorphic functions and the second one, so the denominator never vanishes in D for sure, right? So we have no singularities. Holomorphic function, good. Furthermore, I have that modules of g of z because it is chosen in this way is 1. And this is true for any z. This means that the function is g is a holomorphic function in a simply connected domain and the modules of g of z is constant equal to 1. Well, this is one of the exercises I left you. If the modules is constant, then the function is constant itself. You can see it in several ways. Not because this is the exponential function. In general, assume that you have a holomorphic function in a domain and assume that you know that the modules of this function turns out to be constant. Then necessarily the function is constant. How can you say this? Well, in several ways, but a simple way is that we have the maximum principle for g, right? So this is true not for the points on the boundary, but for any point. It means that any point inside is a maximum for modules of the function, right? So it is constant. Another point is that, well, you have well, this is related to the open mapping theorem, right? It cannot be unless the function is constant that the modules reduces to a point, right? It has to reduce to an open set or a tool, right? Correct? So in any case, what I obtained from here is that g is lambda. g is constant. Because of maximum modulus theorem, right? Then I have, these are stupid inequalities as I said, but they have a g of z is lambda, right? Remember that g of z was defined to be e g z over m over z, which means that e g z over m is lambda z. z is not 0, so I can multiply or divide everything. Now I take the derivative on both sides, right? And I obtain that this is e g of z over m times g prime of z over m, yeah? On the right hand side, I have lambda. But lambda is what? This, e g of z over m over z. That is to say that g prime of z is m over z. I don't know anything about g, but g prime of z is m over z. So I have the derivative of functions, right? Very well. Now what is this number here? If I consider f prime of z over f of z, remember that this g, sorry, e g of z is f of z, right? This comes in here. We have defined g z to be, in some sense, the logarithm of f, right? And then we obtain from this equality is that g prime of z is m over z. m is the fixed m we have chosen, huh? So now f prime of z is, well, the derivative of z over f of z, right? Which is g prime of z, right? Times a of g z. So it is m over z, all right? And I take this, the integral of this number here over a close, 1 over 2 pi i, of course. I always forget this. 1 over 2 pi i times the integral of f prime over f. And the integral is along a curve, say a circle. This number here is between, right? Between 1 and R 1, right? What is this? This is the, well, this is the argument principle, okay, but what number is this? Is the index of, well, if you call this curve gamma, this circle gamma of 1 over z, with respect to curve gamma, capital gamma, which is f composed of gamma, right? The index of which point? Of 0. And 0 is inside, okay? So this number is 1, right? Is it okay for you? Because this is where I can translate, okay? Call this curve gamma, okay? This is a circle. This is the integral up to a constant, which I always forget, of the curve gamma, which is f composed gamma, of 1 over z, which represents the numbers of times. So the index, with respect to gamma, capital gamma of 0, which is 1. No, it is 1. This is independent of, this is a problem of saying times the curve gamma, one time. Well, on the other hand, on the other hand, that's correct. That's correct. This is the integral. Well, you just forget this, okay? And look at this on the left-hand side. We are just computing the index of a point with respect to a curve. And this number is 1. On the other side, this is m over z, 1 over 2 pi i is in front, of the same gamma, right? And this number is m, right? So m has to be 1. But m was the logarithm of R2 over logarithm of R1. That is to say, R2. So if the biomorphisms exist after this long excursion and harmonic analysis, and, well, it's not because I wanted to make this proof so long. There is no other way. So the consequence is that the two, if you assume that R1 is equal to R2 is equal to 1, also the capital R's, so the two radii, the big radii are the same. So in general, what you preserve is the ratio R1 to the smaller over the bigger, right? So it is very rigid, the request to be two-annuali with the same, with the biomorphism, okay? It's not true that you take two-annuali and say, well, one is somehow related to the other quite easily, okay? So the conclusion is that R2 is equal to R1. R1 is equal to R2 and it is equal to not capital ones, okay? R equal to 1, right? And I think that with this we have seen that the tools we have are very, very powerful in some sense, okay? Now for the second part of today's lessons, I won't go back to the notion of univalent functions and I will use these slides if they work. Is it gone? Okay, okay. So let... Ooh, okay, I do it like this. So let S be the class of homomorphic functions injective in the unit disks but with this extra assumption, so that A0 in the power expansion is 0 and A1, the first coefficient in the power expansion is equal to 1. You can always reduce this case because well, otherwise you can subtract A0 and divide what you have a suitable constant, right? In order to have this what you know by the hypothesis of injectivity of the function guarantees that f' at 0 is different from 0 so that you can divide, okay? So after... So this means that you are translating the image keeping 0 fixed and rotating in such a way that the derivative and dilating if you need, it is multiplying by a complex number divided by f' of 0 is equal to 1. So this is the class S. This is normally called class S because in the literature in German this was called schlichtfunctions class. Schlichtfunktionen schlicht means smooth and of course this notion relate another class of functions and the function are like this, okay? The class of function is indicated by sigma. This is a function defined on the complement of the unit disk as you can see 1 over z, 1 over z squared and so forth appears in the Laurent expansion. It is defined in a neighborhood of infinity if you want and the functions are supposed to be injective in the complement of the unit disk and you can see that if you look the half hemisphere in the Riemann sphere which is outside projection so you take the upper instead of the lower hemisphere you are taking what what is outside the unit disk, right? Remember that we projected from the north pole each point of the sphere to the plane except for the north pole which was sent to infinity. The upper hemisphere went to the plane without the circle and the disk and the lower hemisphere was mapped into the disk. So you are considering this function and if you take the inversion which is what you have to do you want to see this the singularity of this function in 0 you see that in 0 the conjugated function as a pole and its residue as written here is equal to 1 so at infinity this meromorphic function has residue 1 so the choice of the coefficient in the principal part it is reversed if you are working in the complement of the disk what is normal is to have z over z minus z to the power minus 1 minus 3 so on so forth. So what is the singular part so the principal part is given by z and then you can also reduce the case such that g of z is different from 0 for any z in the complement of the disk because well this is not a big problem if by chance you have the image it's an open set and a plane you can translate everything in such a way that the origin is not contained in the image but this adjustment doesn't change the natural of the function because the function still remains injective and meromorphic just add a translation what is also interesting is that when you start from a function in the class s it is slicked and you consider what is natural to do when considering what is inside outside you take 1 over z which maps the outside inside and take the function f and then take again 1 over so inversion over the inversion of the function you obtain the power x function which makes some calculations not very difficult and this class this g belongs to sigma prime so it is holomorphic with a single idea infinity of residue 1 and it doesn't vanish and what is also nice to know is that when you consider this transformation which is natural in terms of projections but it is natural in terms of power series expansion so you take the function g and sigma 1 and compute at z squared and then you take the square root of this because of the properties of the power expansions as I said the pioneers in the study of complex value function were very experts in the power series so g of z as this expansion which turns out to be good in terms of preservation of the of remaining in the same class because z multiplies 1 and something else and this something else is good it's like the g of z good now don't be worried about the notation what is surprising is that if you consider a univalent function on the complement of the unit disk then the coefficients, the moduli of the coefficients of the power expansion the Lorentz expansion satisfy this very important theorem so the sum of all moduli of coefficient times n where n is the index is smaller or equal to 1 this is known as aria theorem and it has also some geometric motivation I cannot get into now but assume that we have this simple case that the first coefficient is equal to 1 in modules so in this case it means that the others are all the others have to die because otherwise there is a contribution which fails to get on to this this estimate so the function reduces to this okay b0 and b1, b1 is modules 1 and z is the principal part right so in the general power expansion I showed you here what is it here b2, bn sorry yes so b0 and b1 survive if modules of b1 is equal to 1 b0 cannot be controlled by aria theorem because there is a 0 in front of it okay but if b1 as modules 1 b2 and bn all the bn's from 2 to infinity die so they vanish so the function is like this and this function is a conformal mapping of the complement of the unit disk onto the complement of the line segment of length 4 in some sense it's like if you consider topologically this means that well if you consider an infinite annulus well this function marks injectively what is the infinite annulus with r1 infinite into the plane minus a slit of length 4 and the calculations are here if you want to check so you take the boundary and the boundary is mapping to the boundary you can see that this distance here depends on a real cosine and 2 is in front so this gives you a direction and cosine is from 1 to minus 1 so that it covers a segment of length 4 at this number 4 will come out in several other statements now pardon me you want to tell me that I have to do this right yeah sure yeah thank you without rolling the slides it's better yes this is the stupid calculation but it is important so this is the theorem which probably consider the birth of geometric function theory for this class of functions because beaver back a century ago almost a century ago said well if you have the power expansion and I can prove that the second coefficient is necessarily smaller in modules smaller equal to 2 and there is an an extremal function for these estimates it is known as the KB function in fact this very simple function this is well defined it is injective it is in the class s because at 0 takes the value 0 and at 0 the derivative is equal to 1 well if you write a power expansion it is exactly this so the an is n so you cannot improve this for the case of a KB and what beaver back conjecture is that well in general the nth coefficient is in modules smaller equal to m and it took a long time well first let us see how this can be proved now it is easy to see this is true now it is easy for a long time it was considered very difficult result of so you apply one of the transformation formula I have and G of z is in fact like this use the square transformation and G is in sigma as I said so that well for instance from the area theorem a2 over 2 which is the second coefficient is smaller equal to 1 so actually more some all the coefficient times n each an times modules of an is smaller equal to 1 so in particular this is true and this is obvious you see this is one line proof but it requires a lot of calculations in the past well if a quality occurs then we have something related to the previous example then well a quality occurs when we have this and this is one of the cases we have already encountered so it is essentially the remember the function which maps the complement of the unit disk and to the complement of a segment of length 4 and you make this well substitution you go back make the substitution you obtain it this is up to a rotation the k-bed transformation and this conjugation say you first rotate consider k and then re-rotate the opposite in the opposite of the opposite angle doesn't doesn't make any influence in the modules of the coefficients kebe is the name of kebe is related also to this fact which is quite surprising to me you have the map in the class with some regular analytic conditions it is defined in the unit disk holomorphic and injective well you can also say well it is normalized good well for sure the image contains this disk you cannot avoid to have this disk in the image entire disk which is surprising and well this can be seen in the following way and assume that omega is an omitted value by f take this function here this is by definition another function which is still in z and this is the power expansion and then you obtain from the previous calculation that this is smaller or equal to 2 the first coefficient right but since a2 is smaller or equal to 2 then you have that modules of w cannot be smaller than 1 over 4 so that's why it's called one quarter theorem and well apparently in this calculation you are using just the fact that this is the power expansion related to this function and f has a power expansion z plus a to z square plus a so on and so forth so once might wonder if univalency can be avoided and it cannot if you take this class of functions and this sequence of functions here they are all in s sorry they are all normalized but fn omits this value these are not injective because enz is not injective right the kebe functions which have this also this representation say and here you should recognize something which relates the disk to the right of plane yesterday I called this transformation z which the function which maps z into 1 plus c over 1 minus z is called the kebe transformation so we have one quarter so w square minus one quarter so the one quarter is c and in fact the kebe function is the map which maps the unit disk t onto the entire complex plane minus an infinite half line starting from minus 1 4 so it is extremely close in this sense because it contains the disk of radius one since this function is in s you cannot hope to have a larger disk contained in the range of any function I'm sorry while there are other results concerning the estimates of the derivative of a function in s and this is known as kebe distortion theorem which relates the value of the modules of f prime in terms of r where r is the modules of z when the z is in the disk this is the growth theorem and this is well an estimate of something which is related to this and to this well as usual kebe the kebe function which realizes the the quality in these estimates so we have a conjecture is the following so assume that you have a function and in the class s then a n in modules is more equal to n for n n and the secret quality occurs if f is a rotation of the kebe of the kebe function the the techniques involved in the first attempts to prove these conjectures and some of the experts consider these conjectures false so they try to find a counter exam for a long time so it was finally proved only in mid-80s by the branch but using another theory so the Levener-Chains theory which is also interesting but you're not getting let me just give you some of a view of the class of function which have been starting because as usual for the general case if the general case is difficult what you can do is that well maybe for some simple cases or for some restricted classes I can have I can have some partial results so for instance it was proved almost immediately that the coefficients are all real then the conjecture was true then there were some estimates by Litt, by others in some other case some power expansion which have only coefficient in the I don't remember probably in the odd in the odd for odd indices it was proved so for some classes and in the meanwhile these classes gave other important results so in particular the class of functions with good ranges have been studied so in particular I remember you I recall you that the star-like domain is a domain with the properties take a point and with respect to a point sorry means that whenever you have any point in this set the entire segment connecting the point to the entire real segment is any right so it is called star-like because you can assume to have an odd boundary but referred to one center right convexity is something which is more familiar probably to you means star-likeness for any points in some sense of you take two pair of points in a set then necessarily the segment connected to points is in the set but we have examples of function of sorry of sets which are star-like but not convex right you have an example well very easy take something with take pardon me take a star okay it is not convex if you take two points in the in the near the vertices of the star okay into near the vertices of separate separate near the separate vertices then it cannot be connected by a segment but the star is star-like with respect to the center of the star right so this is a geometric definition okay which apparently doesn't have anything to do well a function is called convex in this in this setting if it is holomorphic, universal it is in S and if the image of the disk is a convex set and similarly it is called star-like if f of d is star-like well with respect to the origin because the origin is certainly in the range right so there are functions which are star-like but which are not convex and the simple example is the KB function the KB function maps the unit disk into the plane with an infinite half line removed with a slit which turns out to be star-like but not convex of course right so the class of this the class of functions which are convex are denoted is denoted by C and the class of star-like because S is already used is S and a star this is a standard notation okay many books even in old books as I said this is obvious so S star is in S and convex is an S star and the KB function as I said is an example of star-like but not convex function right so the idea in this very short of a view is to show how we can prove the we can prove it has been proved by the expert of that time that the BVAC conjecture works fine so it became a theorem for these classes of functions in particular okay so related to the classes C and S star so convex and star-like function is the class P this is also a standard notation functions which have the property that map the unit disk into the half plane yesterday I gave you an example of biomorphism of the unit disk into the right half plane so for sure it is a function probably up to a constant which is in the class P this class P is called the class the class of carotidory functions it's just a name okay and we normalize this function such a way the phi of zero is one if you remember in the curly transformation we had z mapped into one plus z over one minus z and zero is mapped into one so it is one example and in fact carotidory gave this important result well since phi of zero is one it means that in the power expansion this first term is one right then we consider the power expansion n greater or equal to one and these are the coefficients so the estimate of the coefficients like this is uniform for the class P is equal to two and the equality is sharp for each n and this is how they just I wanted to show you how the techniques we have so far introduced can be applied so you take the function phi and you multiply times this auxiliary function right and integrate over the circle so this integral turns out to give you well this something which dies because it is holomorphic and something which doesn't so this is an integral which depends on n right because n here is changes right so by the residue theorem we have the i n I should have written i n is two minus a n which well you can also say well this is the real part of e and this is what you can calculate right the real part of this one because here you actually considered something related to complex trigonometric functions from this they have that well this number here is positive it follows that this number is the real number two minus a n the real part sorry is greater or equal to zero that is the real part of a n is smaller or equal to two and this is valid for n e n this is an application of not stupid application of of the residue theorem if you want right so this becomes a tool more than in fact calculate this integral which becomes a tool and it is well well and actually as I as I told you this is an example of a function transformation in the class p and if you write the power expansion remember that we added somehow different versions but if you write this down now all the coefficient from a two to a infinity or equal to two I'm sorry what's going on there's something missing well I'm sorry for this well I think that we can stop here and continue with the next slides next time I don't know what happened but in any case the idea is that you can better characterize maybe it is like this I'm sorry so you can say that a function is in class or if something else is in class p for instance you can characterize the classes using these three sets s star c and p okay we'll see you we'll see tomorrow how it it goes on tomorrow morning right we have classes um I stopped here