 OK. Today we continue studying the consequences of the Cauchy integral formula for holomorphic functions. And this is something I have to tell you before continuing. Remember that we have characterized the sets where the closed curves we are interested in are the good curves. Simply connected domains are, in fact, good. Any closed curve can be deformed. It's almost up to a constant. This is one of the characterization of simply connected sets in the plane in general. Then for the plane, we have also the characterization that the complement in the extended plane is connected. This is only for plane domains. But then we have also domains which are not simply connected. For instance, if you take an analyst, you take a domain like this with several holes, this is not simply connected. However, we can extend the formula for curves which somehow behave well. So we give this definition. We define, well, first, we define a cycle, like in homology. A cycle is the sum of closed, sufficiently regular, rectified, or whatever curves. Sum, in the sense of homology, we have this closed curve here, alpha. That may be this closed curve beta. When I say 2 alpha plus minus beta, I mean this two times and then this, a closed curve starting from the same starting point. This is a generalization of that. So normally, we define sums of curves by juxtaposition. You put one curve, you move along the curve, and then how many times you move, you put the coefficient in front, and then you choose the other. With an orientation, the sign minus means that the orientation is reversed instead of the serve. So we have a cycle, which is a generalization of closed curve. It's just a matter of terminology. And it was Emil Artyn, his famous mathematician, who discovered that there is a very important characterization of those curves, which are important for the Cauchy formula. Because we consider normally circles, and this is an example of a very simple cycle. And locally, this works fine. But then in general, you can have something which is strange. And this domain, which is not simply this example, it's a stupid example. This curve is not a curve which can be shrunk to a point. But these two, alpha and beta, here, they come. So it's not true that any closed curve in a non-simple kind of domain cannot be homotopic to a constant. So what is the characterization? Well, the characterization is the following. So to give this definition, a cycle is this homologous to 0 in this domain called omega, in omega, if and only if, the index, a cycle gamma. Cycle, now for a cycle, we have also the extension of the notion of index. By considering the integral several times along the curves, which form the cycle. So if this is 0 for any a, which does not belong to omega, this is a characterization. And we can characterize, in fact, those domains for which this is true for any a inside and outside. So that omega is simply connected if and only if for any cycle, we have that n gamma a is 0. So n for any cycle is homologous to 0. Sorry, is homologous in omega. So if you want, this is another characterization of simply connected domains. So what I'm saying is that, well, we're not very much interested in this generalization. But since we want to focus our attention to the local behavior, however, I want to just keep the information somehow more general. I want to say that it is possible to extend the Cauchy formula for those curves in general domains, which are homologous to 0. So for instance, any closed curve in a simply connected domain is perfect. But for other domains which are not simply connected, we have to check. So the general form, I'm just sketching you the idea as this following. General formulation Cauchy formula is the following. So if f is holomorphic omega and gamma is homologous to 0 in omega, gamma is a cycle. Then the integral of a gamma of fz dz is 0. So if f is an holomorphic function and gamma is any cycle model to 0, we have this. And remember that this was enough to define a primitive, which was analytic, and then all the other consequences. Just to make a small remark for real valid function of one real variable, continuity is a sufficient condition to guarantee the existence of a primitive, even less. But continuity is enough. So is continuity sufficient condition for complex value functions to have a primitive? The answer is no. You have to require the function to be holomorphic. If we restrict our class of functions to a holomorphic chart, of course, they are more than continuous. Then we have this characterization. So continuity is not sufficient for existence of a primitive function of a complex value of complex variable. And I invite you to think about this, OK? Good. And now, as I said last time, before continuing in the investigation of zeros and singularities of holomorphic functions, let me go back to the Cauchy formula and this very general z, right? And so in particular, we have used this formula. In the case that gamma is a circle centered at the z or at a, whatever, and proved that any function which is holomorphic is, in fact, complex analytic, OK? Remember this? We have also characterized and calculated the coefficient of the power series functions in terms of the nth derivative of the function f at the point a is centered, OK? So remember that it was a lemma we used. Lema just to calculate the integral of the limit as the limit of the integrals, right? And this was done because the functions involved in our calculations were all continuous and the limit was uniform, right? This is a general statement. Now, what I want to do is, is it possible to have an expression of the derivatives of f in terms of the integral formula like this one, like similar to this one, right? What are the hypotheses I have to use to ask in order to have? So if we were allowed to do the following, it's natural to think that, well, if I differentiate on the left hand side this expression, well, I might differentiate. Of course, I can differentiate on the right hand side. But what is the next step? Well, the idea is to pass the symbol of the derivation inside the integral, somehow, with respect to z, so that here would come out c minus z to the power minus 2, and so on and so forth. So the answer in general is yes, you can do this. And let me use the same notation I have here. Assume that, well, this, in fact, is a proposition, which is more general than what I'm doing. And the whole assumption we are taking is that the function which appears here has to be continuous of the curve. So notation is as follows. Take phi continuous gamma, gamma curve. It's a curve, not necessarily a cloud, it's a curve, right? And define fn of z to be the integral over gamma c minus z to the power n. So this family of function, because it depends on n, right? The sequence is, in fact, is then. For any n, fn of z is complex, differentiable, or holomorphic. But I want to show that, well, there is a derivative, right? In this complex sense, or holomorphic. And the derivative of fn with respect to z is n times fn plus 1c, all right? So if I show this proposition, which is more general, and I apply this to the case we are interested in, we have an expression of the derivatives of the holomorphic function in terms of the adapted Cauchy integral formula, which means that if this proposition is correct, then this would imply that nth derivative is 1 over 2 pi i n gamma z n factorial integral over gamma fc c minus z to the power n, sorry, n plus 1, sorry, right? Now, let us prove this proposition, which is, as you can see, it's very general. So let's go to the. First, we prove that the function f1, so we'll prove it by induction. Remember that I define this function. I copy it. Integral over gamma is an integral function depending on n. And pi s continues along the curve gamma. Now, we'll prove it by induction. So the first step is to prove that f1, for f1, the formula I have to prove is correct. So we'll prove first that f1 is continuous. If f1 were not continuous, then there is something which is not correct. So I consider f1 of ze. And f1 of z is minus f1 of z0 is the integral over gamma of c, c minus z plus, sorry, minus the integral over gamma, dc, right? This is the definition. Now, observe that this can be also written as the integral over gamma of c, c minus z, c minus z0, c minus z0, dc. Why? Well, you multiply. You put the common denominator. You multiply this times c minus z0. And then I have minus c minus minus z, plus z. So z appears here. Minus z0 appears here. And c times pi cancel. So this is the expression. And of course, this factor does not depend on c. So it can be taken out from the symbol of integral. So this is z minus z0 times the integral over gamma c minus z, c minus z0, dc. Now, I have a curve. And of course, what I'm taking here is that, well, I didn't say it properly. But I'm saying that z is not on the curve. Otherwise, this is not reasonable, right? Like in a formula we are considering in the center or inside, the circle or the curve or outside depends on what we're interested in, but not on the curve, correct? Because otherwise, here we have something which is 0 at a certain t, right? So it means that z is here, for instance, and z0 is here. And this is gamma, right? So take delta, delta being the distance of z0. Take delta to be the distance from gamma to z0. So this is maybe this. It's the length of this segment, right? And we can take also z close to z0. This is local, OK? So we will make then the limit as z tends to z0, correct? So we're interested to a neighborhood of z0. So that we can take not a z like this, but a z very close. Such that z minus z0 is smaller than delta over 2. That is also z minus z is greater than delta over 2, OK? When z, of course, belongs to gamma, right? So I have the freedom of choosing a small neighborhood. So I give you another sketchy. So I take a neighborhood of z0 in such a way that any point here has a distance from the curve, which is at least this half a distance, right? So now I have this. Modules of the difference is smaller or equal to z minus z0 times the integral of c, c minus z0 times c minus z, right? Mixing. We are using the inequality. So the integral of a product is the product of the, sorry, the integral of the, sorry, the modulus of the product is, we're saying something, which is wrong, but the modulus of the product is equal to the product of the module i. But then we apply also inequality. We use several times for the integral of, so the modulus of the integral is smaller or equal to the integral of the module. Now, and this can be made smaller than z minus z0 times. Well, we have a gamma. Gamma is a curve. So it is the image of a closed, it is an image under, say, sufficiently regular. So at least continuous, a closed interval. So it is a complex subset. So phi of c is a maximum on gamma. So I can take m, m being the maximum of module of phi of c. I remember that phi is continuous on gamma. Gamma is a closed curve. And now I have that this is 2 over d square because of the inequality I have here, right? z minus z0 is smaller than delta over 2. And c minus z0 is a modulus greater than delta over 2. So that we have a reciprocal, right? Now, this number here is smaller than a constant times z minus z0. So that we have that f1 is continuous. And this, you see, this is a constant, which is independent in some sense of c. So we take the limit when z tends to z0. And we see that, well, actually, f prime of z of f1 of z, which is this, or the limit, when calculated z0, more correctly this way, is the limit as z tends to z0 of f1 of z minus f1 of z0 over z minus z0. This is the limit as z tends to z0 of what? After dividing by z minus z0 in the expression here, which we use in this estimate, we are left with the integral of a gamma phi of c, c minus z, c minus z0, dc. And when z tends to z0, since we are considering uniform convergence, this is the integral of a gamma of phi of c, c minus z0 square dc, which is precisely f2 of z0. So the first step in our induction proof is done. Let me just remember that this is the fifth page. So now, assume that we have this proposition showed for the case fn. I want to prove it for the case fn plus 1. So assume proposition is valid for n. We want to prove it for n plus 1. So I consider this, as I did before, take fn of z minus fn of z0 and write it explicitly. This by definition is, so I write it integral of gamma phi of c, c minus z to the power n dc, I know the integral of a gamma phi of c, c minus z0 to the power n dc. And I consider another expression I want to use. So I will apply what this. I write this, this first term, and this equivalent way. Phi of c, and I c minus z to the power n minus 1. And then I put c minus z0. Then minus phi of c, c minus z to the power n, times c minus z0, times z minus z0 and dc. Let's see if it's correct. So I put everything together, so I have z minus z0 here, and I have z minus z0, right? It's better to put it this way. So like in the previous case, yes, I have to multiply times c minus z here, right? So I have, yes, should be. So I have c minus z0, c minus z, sorry. And then I have minus, so we have to cancel this, right? So as we did before, let me check. So let me check. I have c, c minus z, and then I have minus. And this is c minus z, times power n, c minus z0, minus c. And this is what? No, I'm just considering something wrong, because I have the plus here, right? Let me check. This is c minus z0, and then I multiply times c minus z, minus z counts as z here, and I have c minus z0 remains, which counts as this. So this is correct, with a plus instead of a minus. Let me just sketch here. What I have, phi times c minus z, then plus phi z minus z0. And then I have c minus z to the power n, c minus z0. C minus z0 counts, so this is phi of c times c minus z to the power n, right? So that this expression is like this. And then I have a minus here, minus, remember, we have also this second part here. So the second part is here. So I put a plus here, minus the integral of phi of c, c minus z0 to the power n, big c. So in other words, I have that fn of c minus fn of z0 is in fact the integral of the gamma phi of c over c minus z to the power n minus 1 times c minus z0 minus phi of c over c minus z0 to the power n. And then I have plus phi of c over c minus z to the power n, z minus z0 over c minus z0, integrated over c. Are you still following the calculations or not? Just some, well, this and this is the integral of phi of c, c minus z to the power n, big c. And I invite you to verify it. So if you put the same denominator, you have to multiply this time c minus z. Then you sum phi of c, z minus z0. So what is left is phi of c, c minus z0 over c minus zn, times c minus z0, which you cancel out. So it is the same. Now this is also, I split. I consider this difference under this integral and then this. So this is integral of the gamma phi of c, c minus z0, c minus z to the power n minus 1 minus phi of c. And I write it this way. c minus z0 to the power n minus 1, big c. Plus the integral of phi of c, c minus z to the power n, times z minus z0 over c minus z0, big c. So I'm sorry, I have to continue on another new slide. But then you should recognize in here something which reminds what we have here, something very similar to this. So in particular, if I divide everything times z minus z0, this is the integral over gamma c, c minus z0, z to the power n minus 1 minus phi of c, c0, c minus z0 to the power n, big c, over c minus z0, plus the integral phi of c, c minus z to the power n, c minus z0, which is something which is missing here, minus 1, right? Is that what I'm doing? Or something which is not convincing you? All right, so I divided times z minus z0 on the left hand side, I have z minus z0 here. And the factor z minus z0 is canceled here, right? Now I should recognize here something which reminds what? The incremental ratio of this is f n minus 1, z minus f n minus 1, z0 over, sorry, this is z minus z0. And then I have here something which is the integral. So this, because I have this, right? This expression, right? And here I have this other expression, correct? But I have also c minus z0, right? c minus z0 is here. You see this? So I have to consider one and more stuff. And so this would give us something which is, when I take the limit n times f n, z, which is not what I want, right? And this would converge to plus f n plus 1 of z0, right? So I have to, sorry, this is n minus 1, right? This is something which converges to this, right? According to our hypothesis, the hypothesis is that the, right? This is the statement for the case n. And we assume it valid for the case n minus 1, right? On the right hand side here, we have n plus 1, right? But here, if you think a little bit, we can rephrase it in terms of these functions, f n minus 1 or f n. I have to, sorry, I have to check it a bit more carefully about the index. Maybe you are lost in the notation. But no, this is correct. This is n minus 1, right? Correct. This part here. But we have also the integral, right? 1 over c minus z0, dc. So this notation is a bit horrible. So let me rewrite it this way and use this notation. So I finally have f nz minus f nz0 over z minus z0. And I have integral of a gamma. Then f of xz, xz minus z to the power n minus 1 minus f of xz, xz minus z0, n minus 1. Then I have 1 over c minus z0, which is the common factor. And then I divide times z minus z0, right? Plus the integral over gamma of xz, c minus z to the power n times c minus z0, dc. So when I take the limit, this is, in fact, what I was writing before, this is f n minus 1, sorry, f minus 1 xz and f minus 1 z0, f minus 1 z, f minus 1 z0, right? And so this tends to, when I take the limit, f n minus 1 f n z0, which is, by definition, f prime of n minus 1 z0, this is by induction, which is the integral over gamma xz over minus z0 to the power n n minus 1, right? So when I then take also this in consideration, I have plus 1. And on the other term, which appears here, I have, this is the integral over xz of xz minus z0 to the power n. So n minus 1 plus 1 gives you n times f n plus 1 at z0. So when I take the limit as z tends to z0, I have that f prime of n of z0 is, in fact, what I have to prove, that is to say, that this, so f prime of z is the limit as it tends to z0 of f n of z minus f n of z0 over z minus z0. And this is the limit as it tends to z0 of z0, sorry, of this, right? n minus 1 integral over gamma, phi of xz minus z0, this is not a limit, I have already taken the limit on the right hand side, so n minus 1 plus 1 is n times integral over gamma, phi of xz minus z0 to the power n, sorry, n plus 1 and plus 1, which is n times f n plus 1 by definition. So we have actually the formula we were looking for, the nth derivative of a function which is represented in this internal way is given by a similar expression integral over the same curve gamma of n factorial of phi of xz minus z to the power n plus 1 dx, and this is what we were looking for, okay, sorry for this longer calculations, but I cannot find another way to show it to you, right? So in some sense what we have shown is that any holomorphic function and its higher derivatives have a Cauchy integral formulation in terms of what? Same functions along the curve over a power of xz minus z, up to a constant constant. Now, this is just for the sake of completeness, but I want to focus our attention to the zeros, remember the last time we have shown that a holomorphic function either is identically zero or the set of zero cannot have a point of accumulation, right? And this implies the identity principle. So if two functions which are holomorphic in a domain agree on a set which has a limit point, accumulation point, for instance an open set, then necessarily they coincide elsewhere. And this is also known as uniqueness theorem in some sense for a holomorphic function of, it is a unique way to prolong a holomorphic function. If it is defined locally, then you know that another function can be extended to some small neighborhood of the points of the boundary, then thank you itself is extended, it coincided with the previous one in the, in the, in the section, okay, good. Now, yes, today I want to, well, to, yes, to tell you something more about this zeros, but in particular let me show you that given a holomorphic function such that f of a is zero, f of z is different from zero, so f of c is different for some c. What I want to say is that f is not identically zero, okay. Then there exists an n depending on a, an integer such that f of z is c minus a times n a times g of z where g is holomorphic and does not vanish at any, okay. So this enforces the, the idea that, well, in fact, a holomorphic function can be considered as generalization of polynomials. If you have a root of a polynomial, then you have also factor and then you consider also the so-called algebraic multiplicity of the root of a polynomial, right. Well, this is not true for real valid functions. There are functions which vanish at the one point, but there is no integer or any linear factor times something, okay, because the function vanishes at this, this point. It's not separate, it is not necessarily an n such that z minus z not to the power n times g of x is equal to f of x, but for holomorphic function this is the case. And in fact, well, of course, this implies, so this definition follows from this property and say n a is the multiplicity of the 0 a for f, of course, right. So the proof is of course a consequence, it's an obvious consequence of the characterization given last time. So remember that we have proved that well, either the function holomorphic which is holomorphic is identically 0, the set or, right, the sets of zeros has an accumulation point and the third characteristic is that there is a point such that all the derivatives of the holomorphic function are a vanish, okay. Now, assume that n is the largest integer such that the n derivative at a is 0 for any n smaller equal to n a minus 1. Well, I'm assuming that the function is not identically 0. So I'm sure that at the point a you cannot have all the derivatives vanishing. So I consider the n a I want to define as the largest integer such that all the derivatives calculated at a evaluated at a are 0 up to n a, okay. And then I define g of z to be, this is a function I have to define, right, to be z minus a to the power minus n a times f z. This is for z different from a and g of a being the nth derivative, the n a derivative of f at a over n factorial which is the nth coefficient in the power expansion of f, right. What do we have? We have that, well, g of z is certainly holomorphic in omega minus a because, well, f of z is holomorphic. This function is not holomorphic only z equal to a, but elsewhere it's holomorphic. So we have a prior to holomorphic function and we know that the Leibniz rule also we can say immediately g of z is holomorphic. But remember because n a is the largest integer such that this is a case, then we also have that the limit as it tends to a of z minus a times g of z, this is 0, correct. So the function can be actually extended. Remember that it has only one singular, do you remember the generalization of Gursa theorem? This hypothesis guarantees that the function in fact is bounded, okay. And if you look at the power expansion of f of z and of g of z, the do function in fact coincide up to this. So we have the g of z is, sorry, times z minus a to the power n a is in fact f of z. And this number is not 0 because we are assuming that all the coefficients are, it's already the derivative of f at a vanish but not at n a. So this is not 0, so we have the theorem. So n a holomorphic function which is not identically 0 and a is a 0 can be written as factor z minus, as a 0 to z naught of a, z minus a, z minus naught, certain power times a function which is holomorphic and not vanishing at the same 0. So if this were a polynomial case, this would be obvious, this is the multiplicity, right, of the root a, okay. The other term is the part which does not vanish, okay, good. So we have this definition. Now assume that the function is in fact more than 1, 0. So assume holomorphic in omega has z1, zm, 0s. So which means that f of zj is 0 and j varies between 1 and n. So not necessarily 1, 0 but several 0s, right. So if we repeat the same argument used before, we can say well, you see, I have that this factor can be taken out and I put n1, the multiplicity of z1 is expanded and I have time at g of z, but g of z of course does not vanish at z1 because this is the hypothesis but probably vanish at z2 or z3 and so on and so forth. So I have this factors times a function, I call it g of z again, which is not vanishing at z1, z2 and zm. So it is not vanishing at all, correct. Now let us make some exercise in derivatives and apply Leibniz's rule several times. We have a product here. If I calculate a derivative, well, the product has a derivative which is related but in a specific way by Leibniz's rule to the sum of the products of the derivative, right. So I take the derivative, notation is like this and I start differentiating this, okay. So I call it n1 z minus z1, correct times and I want to call this say g n1 0, okay and then plus I have z minus z1 to the power n1 times the derivative, correct. This is Leibniz's rule but then this part here is as before the derivative of this times gz2 plus something. So the procedure goes like this and if you do calculation in a row in a sense, okay. Repeat, repeat, repeat, repeat. At the end of the story you have this, okay, n1 z minus z1 n minus 1, n1 minus 1, sorry, times g1 g n1 z plus, I write it this way, okay and 2 z minus z2 and 2 minus 1 g n2 z plus plus and m z minus zm and m minus 1 g m z. What is gm z? It's in the previous notation was g, so the last part, right plus I have z minus z1 and 1 z minus z2 and 2 g prime of z, right. So if I consider f prime of z over f of z which in general is not something reasonable but I'm considering well where this is meaningful that is to say where z does not vanish. So except for the point z1 z2 zm I can always consider this. Well you see, take this, okay, first summand. Remember this is everything in the function except the first factor after derivation, okay. So if you divide times f what is left? Everything cancels. So g f z minus z2 with power n minus 2 z minus z3 and so on and what is left is z minus z1 to the power n1, correct. So I have n1 times z minus z1 to the power minus 1. So I write it this way. Similarly I have n2 and nm z minus zm plus and the last part is g prime of z over g of z. So you probably see this much better for one point for one zero, right. And then you repeat it several times. You want it to see it in detail or is it sufficiently clear at this stage? Is it? So assume that you are well just taking z minus z1 times g of z. After the first derivative you have this and this, right. And this is g. This is the g, okay. This is the g n1. You then consider the ratio f prime over f. You have g n1 cancel here but you have z minus z1 to the power n minus 1. So one factor comes out. And here g prime over g is the part which reminds, okay. So after repeating this procedure for all the zeros you are left with this. And here it is interesting because you have precisely what is needed to calculate the index of the zeros, right. Times the multiplicities. Each multiplicity for each zero. And then if you think what you are asked to do now is to calculate the integral of this ratio over a closed curve gamma. And up to a constant one, so up to 2 pi i, here you have what? The index of z1 with respect to the curve gamma times n1. The curve, the index of zm times nm with respect to gamma not the index of zm with respect to gamma. Sure, you are correct. What is missing here is z minus z1 to the power n1, right. Yes, sure. So what I didn't say it but what I meant when I said g n1 was everything which is not in the factor containing z1. So everything which is not in the factor containing z2, right. You don't know what is on the right. What is in the expression without, okay. You are correct. But if I said that g nj is f of z over z minus zj to the power nj, I'm correct. So I'm canceling the factor z minus zj in the expression, right. I didn't want to mean the factors on the right. Even though the first is probably the misleading example because in fact what is left is the right factors. When I start off on the second, I consider the first and from the third to the end. Okay. But at the end of the story I arrived to this, okay. This is the g, this is the g. This is the g I have here. Should be. Because when I finish considering all this, I've just differentiated this. The gm is g. Yes, gnm probably. gnm means everything but, oh, you're right. You're right. You're right. So yeah, that's why you are, yeah, sure. Sure. The two notations are not, okay. This, but this is correct. Then I consider everything which is, okay. So, correct. So if I'm taking this, this is not correct. Thank you. All right. But when I divide, what is left for this last sum is just the derivative of g over g. And this does not give any contribution for the calculation of the index of this ratio because the function g is not vanishing in omega. So that if I take 1 over 2 pi i, the integral over gamma, gamma closed curve of f prime of z over f of z d z. This is n1 n gamma. So 1 plus nm n gamma zn. And this ratio here gives you certainly zero contribution in the integral because the function g has no zero, right. It's holomorphic and has no zero, okay. Good. Now, let me see how we can apply this fact. One simple example. So consider, that's a good idea to start with this. But consider, well, probably this is one case we have to study, okay. So we are calculating this integral along the circle center of the origin, radius 2. And I notice that what? You see, this is that the numerator is the derivative of the denominator. So if I differentiate this square plus z plus 1, this is precisely 2z plus 1. So, okay, this is something which will be useful again. So in general, assume that I consider the integral over curve gamma, capital gamma of 1 over w d w. This gives you what? The index over gamma of the point zero. How do we calculate this? Well, gamma is a curve. But assume this curve is precisely the image of a curve time, composed with holomorphic function f. So this number here is what? The integral of this and gamma, it's a zero as a alpha beta into c. And I substitute, right? So I have f of gamma t prime over f of gamma t and gamma prime time dt, correct? And so what we are doing is that if w of t is f of gamma t, we are considering the indexes, the index of the zeros of f, right? With multiplicities. So to calculate this, well, this can be done. You substitute, you put everything inside and you make a lot of calculation. But if you have in mind that, well, this is the derivative of the function. So if you want, you can write this also, integral over gamma of f prime of z over f of z d z. And this is known to be the sum of n j and gamma z j and z j's are in the zero set of f. So the multiplicity times the index. This number, this sum is the finite sum because the number of zeros is finite. You say, well, if the curve is a closed curve, then consider the zeros which are inside. And of course, the curve gamma cannot pass through one of these zeros. So the bounded region, if gamma is a Jordan curve or something like this, it is bounded and unbounded. In the bounded region, the contribution is not important for this calculation because the index is zero, right? Inside the bounded region, there can be only a finite number of zeros. If they were infinite because of Wolf-Sano-Weisser's theorem, it would be an accumulation point to the boundary, to the closure. So either inside, it's not allowed, or to the boundary, not allowed as well. So on the right hand side, we have a finite contribution in this summation. In particular here, if we apply the calculation, well, the zeros here are at most two. Well, we have just to check if they are inside or outside this curve, inside, so in the bounded component or in the unbounded component. And then this integral becomes simply 2 pi i times this number here. Of course, I forgot 2 pi i. 2 pi i is missing. So the zeros here are here is, well, take the standard of formula minus 1 plus or minus square root of 1 minus 4, so minus 3 over 2, which is the two conjugate numbers whose modulus is smaller than 2, right? So we have two simple zeros. The curve is, the curve gamma is a circle, so the number of times when, the number of times the curve wraps or is winding is 1 for each. So we have two, right? Simple zeros of 1 times 1 plus 1 plus 1 times 1. So this integrates 4 pi i without making any calculation. This is very simple example in an application of a more general theory about integration, knowing something about coefficients of the zeros and so the multiplicity. You don't see this, right? And I also add this. Of course, the fact that we are considering a zero is just by accident, because zero is simply to define. But if you want to be more general, you can do the following. So consider f of z and take f of z, sorry, minus alpha, alpha being a complex number. And define this function to be g of z. This is still complex, sorry, holomorphic. So f is holomorphic, you know, omega, so is g. Now, what can I say about the zeros of g? Well, this corresponds to the number of times f takes the values of alpha, right? So g vanishes at any point, z, such that f of z is equal to alpha. This is the only possibility. So if I consider 1 over 2 pi i, the integral of g prime of z over g of z, this is what? The number of zeros with multiplicities sum together. But this is also 1 over 2 pi i over gamma of, well, f prime of z, actually f of z minus alpha d z, right? Because the derivative of f and the derivative of g agree, whereas f and g do not. So in this way, I can say, well, this is summation of nj and gamma, say, zj, where f of zj is alpha. So this counts the number of inverse images of alpha with multiplicity. In fact, there are zeros of this function g. Now, this, okay, this is 16, sorry. This together with the previous consideration will give us something important. Now, take f, this is a proposition, which would be useful. So take f holomorphic in v, a, r, so in the ball, center of tail, or radius r. So we are considering the very, in some sense, generative, a very local case, okay? We take the disk, center of tail, so for instance, this works fine for any power expansion, center of tail, and whose radius convergences r, right? And define for a the center f of a to be alpha, okay? So if f of z minus alpha has zero of multiplicity m equivalently, if the problem f minus 1 of alpha has m inverse images, right, in L, in A, okay? Condense in A, okay? As m root, as m inverse images in R in A, okay? Can you see this? So this is, well, for sure, A is one of these zeros. Now, I'm going to say they are different, but I imagine that this zero has a multiplicity m, okay? So they might be extra zeros, but well, I want to say that in A, this function has, well, in, sorry, what I'm saying more correctly is this, all right? That's what I want to say. So I have A here, the radius r, and I have alpha here, and f of A is alpha. Well, there exists an epsilon and positive delta such that if beta belongs to the ball centered at A of radius delta, then f of z minus beta has precisely m simple zeros in B, A, epsilon. Do you want to say what I mean? So the function is defined in this ball centered at A, f of alpha, sorry, f of A is alpha. So I restrict my consideration to a small neighborhood of alpha and take the value beta inside this small neighborhood of the value alpha. Then I consider the problem of solving f of z minus beta, so looking for solution of the problem. How many z are there z's and in the domain of definition such that f of z is beta? And the answer is yes, and I can be more precise, I can take delta so small, so the radius so small in such a way that there exists also neighborhood of A where the zeros are precisely m as many as the multiplicity of the solution of the zero A for the problem f of z minus alpha. And they are simple, simple means the multiplicity of each of them is one, which means that locally you can have injectivity of solution. So how do we see this? Well, we just sketch the ideas. This is number 17. First of all, just to start with, let me remind you the zeros of f of z minus alpha are isolated. They cannot accumulate, right? So in particular, we can choose positive epsilon in such a way that f of z minus alpha has no solution in this. I put a dot in the dA to epsilon, so epsilon of course is to be smaller than one half of the radius of convergence. And what I mean with this dot is the following, this is a set of z and c such that z minus A is smaller than two epsilon but different from A. So I'm using this fact, A is certainly a zero of this function g of z, call it g of z, f of z minus alpha, right? But zeros are, for a holomorganizer, are isolated. They cannot accumulate. So there is a neighborhood of any zero which does not contain any other zero, right? Good. And this is the first condition, I'm assuming, so because. And the second condition is the same choice of epsilon. We can also take that f prime of z is different from zero in the same neighborhood. Why can we do this? Well, first of all, the derivative of f is a holomorphic function as well. So its zeros are isolated. So either f1 at A is zero, which means that the multiplicity of A is greater than one. If it is zero, then we apply the fact that f prime is holomorphic, so the zeros of f prime have to be isolated. So we can intersect, if you want, the smallest neighborhood of A in such a way that this is different from zero and this is different from zero. Or if it is different from zero by continuity, it is not zero in a neighborhood of A. So in any case, I can make this choice of a neighborhood in such a way in the punched disk centered at A and radius 2 epsilon. This is just a matter of rotation. So a small neighborhood without considering point A, we have both conditions satisfied. So what we'll do is follow in our considered beta in a neighborhood of delta in such a way that the curve we're considering, the circle centered at A of radius epsilon does not intersect, of course, alpha. And beta is chosen in such a small neighborhood so that it is in the same component of the complement of gamma, of f of gamma. And we calculate and then we will calculate on Wednesday. We will calculate then the indexes of the two, of the, with respect to this curve of the two points alpha and beta. And we show that this is the same. It has to be the same because they are in the same complement of gamma. Making the calculation will show what we want. But remember that this additional hypothesis guarantees that the function is in fact injective. So the numbers of zeros will be m, like the multiplicity of A as a zero, but so each zero has to be a simple zero. And I think that I stop here.