 So okay, now we are at this step. We have a small neighborhood of the point a, radius two epsilon is chosen in a suitable way in order to have these two conditions fulfilled. Derivative of f differ from zero and f of z minus alpha has no zeros in this punctured disk. So this is lesson number 10, one. Now I take the curve gamma of t to be the circle of radius epsilon. So it is n side of the punctured disk. This is a, the radius is two epsilon. So this is the circumconsidering. And then I define sigma of t to be f of gamma of t. And this is a curve where, where it is a curve somewhere else. If this is alpha and this is sigma, I can say that sigma of t is different from alpha for any t. Is that correct? Because we are assuming that f of z minus alpha has no zero in the punctured disk, in the entire punctured disk, in particular along this curve. So now this is alpha, then I take a small neighborhood of alpha here, say, radius delta, so that b a alpha delta is such that b alpha delta does not intersect the curve sigma. But this notation is, of course, as usual a bit, there is an abuser notation in this. But what I'm saying is that there is an intersection of the image of the curve, so as a set, and the small neighborhood of delta I'm taking here. In other words, I say there exists a delta positive such that this ball, open ball, is not intersecting the curve. In other words, for any beta in this suitable neighborhood of alpha, beta belongs to the same path connected component of the complement of sigma. In our picture here, it is in the unbounded component as well, but it is very close to alpha in such a way that for any beta I'm taking in this neighborhood we are in the same neighborhood. This is important for the calculations of the index, if you remember. All right, so in fact, what I'm doing here is the following. I'm calculating now this, the index with respect to sigma of alpha, which is the same as the index with respect to sigma of beta. This is because we have already proved this fact. If they are in the same component path connected or if you want in the complement of the curve sigma, then the index is the same. But how can we calculate it? Well, this is the integral 1 over 2 pi i, sorry. I always forget this. I apologize now for the past and for the future times. So please, when I say, well, this is because sometimes this 2 pi i is included. Well, this is not an excuse, actually. Sometimes it is included in the formula. So you say, well, this is the index up to 2 pi i. But this is not an excuse. Of course, it is a mistake. The last time it was pointed out correctly that I forgot to put 2 pi i. But then, luckily, I could add it before making the copies for you. So you probably didn't notice here, but in the copies I gave you, it is correct. But just in case, it's not important. We want to be somehow more familiar with the integral part than the constant part. Good. So I'm calculating this as it is expected to calculate. So the integral over sigma of what? So I have this. I have d, I have sigma here. So I have 2 minus beta or alpha. Well, this is alpha here, but this is beta. And what is sigma? Sigma, remember, is f composed to gamma. So when I write this line integral, I have to replace everything and make it somehow well, I write it in terms of a line integral, not a curve integral, but the line integral. So this is, if I replace sigma with this, this is precisely what I have here. So the integral over 0, 1 of f prime of gamma t, gamma prime of t. And then I have here f of gamma t minus alpha dt. But what is this number? Well, if you look at this number, this is the number, in the number of times, f takes the f, or say, f of z minus alpha as 0 in A. So this is m. On the other hand, this is, I write this here, 1 over 2 pi i integral between 0 and 1 of f prime gamma of t, gamma prime of t, f of gamma t minus beta dt. And remember that this is the numbers of inverse images of beta. And since we are assuming that f prime z is different from 0 in the punctured disk, we can say that the number of zeros is m, and each is different from the other, because the function is injective. So what we can say now is that in a neighborhood of a value which is taken m times, which is considered as m value, the function, which is holomorphic, so we can take a small neighborhood in such a way that any point beta in this small neighborhood has m inverse images in a small neighborhood of a, because this counts precisely the numbers of zeros of this function here inside. This is the index we defined last time. So this is somehow what we have to prove. And while this result tells us something more, and I want to emphasize this in particular, and this goes back to. So we have the following. We have, say, graphically, we have punctured disk here of center a, then I have alpha here, and I take a small neighborhood of alpha of radius delta with the notation correct. And we actually restrict our self to a smaller neighborhood of radius epsilon, not to epsilon, but epsilon. And what we can say is the following, then you take that this is contained in the image of the ball, center a t, of radius epsilon, because alpha goes to, sorry, a goes to alpha. It's mapped by f into alpha. Then I take a small neighborhood. Each point in this small neighborhood is such that it has m inverse images and this small neighborhood of a, m distinct. We have proven this, which means something else. Since this is an arbitrarily small neighborhood of a point that whenever u is open and f open in omega and f is holomorphic in omega, then, necessarily, f of u is open. You see this? How can I prove this? Well, to prove that something is open, it's easy to show that any neighborhood, any point has a neighborhood that's contained in this. But any point in the image is, in fact, such that this works fine. In other words, this property means that any holomorphic function, and I'm assuming that it is non-constant, is an open function. Topologically, a function is open if the image of an open set is open. Pardon me? Now, take, for instance, the function sine, cosine. Cosine on the reals. Cosine is real, nullity. It's whatever you want. So I see infinity is very good on the entire real line. However, it is bounded between minus 1 and 1. And if you take some open set, for instance, from 0 to 4 pi, the image is a closed set. It's not an open set. Other consequences of this simple fact, simple but in some sense, we are considering special function. So this gives you a topological property of holomorphic function. We started from, say, an analytic definition, the existence of complex derivation. We proved that this is equivalent to complex analyticity. We gave a equivalent condition in Cauchy-Riemann equations. We gave a characterization in terms of conformal mappings. Now, we have also a property which is topological property for function. This result is also important for the following fact. Assume that f is holomorphic, as usual, non-constant. Non-constant is somehow, because constant functions are not interesting for the configuration we are taking. Topologically, they are not interesting. And for the statement, I'm going to prove it's not interesting. So assume that there is holomorphic, sorry, I didn't say it, in omega. Then I said there is omega is an open subset of Ca, as usual. There is no z0 in omega such that f of z0 is modulus greater than f of z for any z in omega minus z0. So if you consider the composition, if you want, you have omega here, then here is c. This is the function f. And here you take the modules. You can say that this function associated to any complex number are no negative real number. Take the composition. The complex are not ordered, but the reals are. So we simply are considering the possibility for this composition to have a maximum. This is the maximum inside omega. This is not possible. So this is as it is normally mentioned, the maximum modulus theorem. And I will show you why. Maximum modulus theorem, OK. So if z0 in omega is such that f of z0 for any z in omega f holomorphic, then f is constant. Or this statement can have a larger formulation in the following way. Assume that the function f can be extended up to the boundary, continuously up to the boundary. So in this case, omega is bounded. So the closure of omega becomes a compact subset of R2, or c, right? So for sure, this composition of two functions, this is infinity, analytic, whatever. But this is continuous. So this composition is at least continuous. So you have a continuous function on a compact set, real valid. So definitely it has a maximum and a minimum. You have a continuous function between, assume that f can be extended up to the boundary. This is not in general true, but assume that it is holomorphic inside this bounded open set omega and continues up to the boundary. As a continuous function, this composition of two continuous functions, these are continuous functions, define on the closure of the bounded omega. We assume that omega is bounded, and the closure, of course, is closed. So it is closed and bounded. So it is compact. As a compact set, you have a function which is continuous on a compact set. It has a maximum and a minimum. So it has a maximum. So this means that the maximum cannot be in omega, but it has to be on the boundary. This is not very much surprising for those of you who have seen this same principle for harmonic functions. Harmonic functions cannot take any maximum inside the domain but on the boundary. And since we have pointed out that the real imaginary part of holomorphic harmonic, this is not so surprising for those of you. But well, how can we see this? Apparently, this is something completely different from the real KGTs, in fact. And how can we see this? Well, this is quite simple. And this is a consequence of the openness of the function. So I just draw a very elementary picture without using invoking many other properties of holomorphic function. We know that if alpha is defined in A, it is actually defined in a small neighborhood of A. The definition of holomorphic function is, in fact, a local definition. We have to define what it is now equivalent to analyticity. Analyticity is something which works fine in a neighborhood, in the sense that we have to have a positive radius of convergence of the series, otherwise we have just a point-wise definition. And this is the value. So for sure, if I take a small neighborhood of A, the function F is defined in this neighborhood. And its image contains alpha. This is true for any point inside the domain of definition. This kind of game is true for any A in omega. Assume that F is defined in omega. Take A, consider alphas F of A, take a small neighborhood. Omega is open, so small neighborhood in omega. And then take its image. Its image contains alphas. It is like this. It's a neighborhood of alpha. So I can say this because we have proved that holomorphic function is, in fact, open. This is an open neighborhood of alpha. So assume that Z naught of the statement is, as I said. Assume that Z naught in omega is such that F of Z naught is greater than F of Z. For any Z. Now we get a contradiction because then we restrict our consideration to a small neighborhood of Z naught. And repeat the same picture over here. This is F of Z naught. And assume that, well, I enlarge a little bit. This is F of B Z naught epsilon. I enlarge a little bit. What I'm saying is that, well, since this is an open neighborhood, there is a small neighborhood of F of Z naught, or rather you say delta. So B F of Z naught delta is contained in F of B of Z naught epsilon because of the open mapping theorem we have proved so far. But then, as you can see, assume that you can visualize the modules. This is the origin. This is the modulus of F, the length of this segment. Well, of course, this point here has a larger modulus. Now this is not because I'm cheating you with these pictures. Because every time I have the value of a point inside the domain of the definition of homomorphic function, I can take a small neighborhood and ensure that this is in the image as well. Therefore, F of Z naught cannot have the maximum modulus of all the other values of the function F. So this is contradiction. For any point inside, this is not allowed. So I consider this an important step and an important geometric characterization using the anthropological properties of homomorphic function. So these functions cannot have the modulus which takes a maximum inside the domain of the definition, but they're different from the real case. Now after starting the zeros, let me just mention that we want to study the singularities. We have already encountered some singularities. We have noticed that when you divide a function times something, which is a polynomial, we have to be sure that the quotient is well-defined. So that since we are dealing with complex numbers, for instance, if you have a polynomial dividing something else, and polynomial has as many zeros, or many roots as its degree. So for sure, this quotient, this ratio, is not defined whenever the polynomial vanishes. And this in the complex number means that it always vanishes in as many points as the degree counting the multiplicity of them. So I want to go into the details of the studies of singularities a little bit more. And using the tools we have developed so far for the study of the zeros, for each zero of our homomorphic function, we have defined a multiplicity. So in principle, we want to extend this to the singularities. And we will see that this is not always possible. So first, definition, a function has an isolated singularity to a in omega f. f is homomorphic in a small neighborhood of omega, in a small neighborhood of a in omega. But it is not defined in a. For instance, as an example, 1 over z is defined for any z different from 0. But in 0 is not defined. And it is, in fact, homomorphically. Well, I can take as epsilon the radius, whatever I like. But well, this is a singularity. Another singularity, apparently, is this. Remember the exercise. This is not defined in 0 for the same reason. This is a homomorphic function. But this is, like here, a ratio of a function of a homomorphic function and a polynomial. And this polynomial vanishes at 0. But as most of you, because I haven't checked all the exercises, but most of you have noticed, then this function has, in fact, power series expansion also in 0. And so this function, in fact, is homomorphic also in 0. Well, it is not defined so as to be clarified somewhere, somehow. Well, it is not defined at the beginning, but then, in some cases, it can be actually extended, also, the singularity. And so, in fact, the first definition is the following. So we start classifying the singularities. And we say that, so this is the definition, again, is removable singularity. So we are taking a singularity, and we are considering a special class of singularities, so-called removable singularities for f, f that exists, holomorphic g, defined in a neighborhood of A such that f of z is equal to g of z in the punctured disk. So this is the situation. We have a singularity, and the singularity, in general, is defined for a function which is holomorphic in a small neighborhood of this point, but not at this point. And then we say, well, this singularity is, in fact, removable when we can have a holomorphic, and so complex analytic function, which agrees with the function in this small neighborhood, but not in A, and is actually also holomorphic in A. Well, this is not very much surprising because the example I gave you before is, in fact, an example of a removable singularity. You can consider the expansion where you have calculated in your exercise to be an analytic function, defined also in A. It agrees with this function here, in a small neighborhood of 0, and the identity principle guarantees that this function is, in fact, the same. And how can we characterize this fact? Why? And these two very simple examples, one has to be considered different from easy minus one. Or so we have, you know, the same denominator, and this, in this case, I can assign this, apparently you cannot. So I have this proposition, as usual, I'm assuming that the singularities are isolated, so that I can make this consideration in small neighborhoods. A is a removable singularity for f, f, and only f, limit, as z tends to a of z minus a times f of z is 0. How can we see this? Have we encountered something very close to this statement? Well, we have used this condition once, and so generalized Gursa theorem. We assumed that, well, the function was not defined, but with this condition, which guarantees, for instance, what? That the function is bounded, I tell you. In a small neighborhood of 8, it's bounded. So in fact, I have this alternative proposition. Assume that f has bounded, so that modules of f is bounded in, and this, well, put epsilon here, right? A small neighborhood of, or in the punctured disk, small neighborhood of a, or the singularity itself. Then, A is a removable singularity for f, right? So this explains why 0 here is not a removable singularity. Here, 0 is a removable singularity. So this result is also known as, sometimes it's called Riemann theorem, but it's not the Riemann theorem. There are many Riemann theorems. So I will not use this theorem as the arch. But we'll take this function here to be defined like this. And h of a, z is 0, when z is equal to a. So I define this. I define this function. Well, this is a function. It's a complex value of 5, and now I have to just do, right? So f is defined in the neighborhood. The z minus a is defined wherever I want. And I define this function. What happens if I have this? Well, I'm assuming that the function f is, in fact, bounded in a neighborhood of a, right? This is 0, correct? Because this is bounded. So if you want to take the modules of the product, this is infinitesimal, and this is bounded, so 10 to 0. So the function h is continuous, at least. Good. So outside of a, it is also analytic. In fact, this is, well, a product of a polynomial and a holomorphic function. It is holomorphic. What about the derivative at a? Is it possible to define a derivation? A complex derivation of h at a? Well, we have to calculate it. So I calculate what the limit of h of z minus h of a, which is defined in 0, z minus a, as z tends to a, correct? Well, this is 0 because h of a is 0. So it is the limit as it tends to a, z minus a times f of z. You see this because we have z minus square, right? So this explains why this condition equal to 0 guarantees that h is holomorphic and nothing the punctured disk, but and the ball central at a of radius epsilon because the derivative is defined, see? Now this, if you want, is another way to see that f can be extended also. So it can be extended. So f and h are not the same function. They are defined with this relation. But if I write the expansion of h of z, we have the following f, sorry, sum k greater or equal to 0, the expansion in a can be taken. So I take bk, z minus ak because it is holomorphic. Then I have that h of z is z minus a square of f of z, where? Well, this is for any z and ba epsilon, but this is in da epsilon, so in the punctured disk. In other words, I have that, so I consider g of z to be defined as same coefficient of h, bk. And then I take z minus a k minus 2, is it correct? Plus plus 2, this is h plus plus 2, plus 2 of course. But this is another function, which is holomorphic in a. And well, I notice that g of z is f of z, whenever z is in the punctured disk. And this extends f, right? Because we have that h of z is summation of bk, z minus ak. And h of z is z minus a square times f of z. So I have an extra 2 here, which counts as extra 2, and so they coincide, right? So in fact, this is the definition of a removable singularity. We have a holomorphic function g defined on the entire. And g is holomorphic in ba epsilon. And this is explicitly written. In fact, this is the power expansion at a, right? Furthermore, g and f coincide outside of a in a small neighborhood in the punctured disk. So a is a removable singularity according to the definition. So this is, if you want, an analytic, rephrased version of the Gursat theorem again. Remember that we have that integral over a closed curve is of a function is 0 over a rectangle. We proved this. And then we said, well, a rectangle without some points, finite number of points. And then we extended from this to the case of disk of the domain by using the rectangle properties of the Gursat theorem and the extended Gursat theorem. When we have that integral over a closed curve gamma is 0, when then we define the primitive, and then we have a holomorphicity. And this is another way to say this. Now what I'm saying is the following fact, which is not a statement to be proved, but a fact. I'm saying it. Given a holomorphic function f in dA epsilon 1 and only 1 of the following possibilities must occur. 1, a is a removable singularity, which will be of no interest in some sense, because it's like considering a fake singularity. The function can be extended if you want, and holomorphically extended to this singularity. So it's not very interesting. Second, there exists a minimal n greater than 1, such that the limit as it tends to a of z minus a to the power n of f times f of z is bounded in a neighborhood of a. And finally, there is no minimal n with this property. So for any n, z minus a to the power n times f of z is unbounded. So we have characterized, we want this is equivalent to saying that the limit as it tends to a of z minus a times f of z is 0, so in particular, this is another characterization. But when I say minimal, when I mean the following, I mean that if I take n minus 1 instead of n, this product, this limit when I take the limit of z minus a n minus 1 f of z is unbounded. So minimal means the smallest n such that this occurs. So examples again. So in fact, let me say again, this is not a statement. I'm just saying, OK, if it's not the case, this is the other possibility. Now I want to show you that all the cases may occur. And then I try to characterize them. So the first is already somehow clearly justified and characterized in terms of the limit as it tends to a of z minus a times f of z equal to 0. And then we'll study the two other singularities. So I'll show you one example of the first case and one example of the second case. Well, one is quite simple. So 2, when I take z minus a to the power n, take n. And if you take n minus 1, multiply this by this. And of course, this is like 1 over z minus a to the power 1 or k or teva. So this is unbounded. But then for n and then for any n greater than this n, the limit is bound is actually 0 for n is strictly greater than this n. The surprising fact is that we have also functions for which it is impossible to make the product of z minus a to whatever n you want times f of z bounded. And the classical example is the following. And remember the exponential as power expansion 1 plus z plus z. So 1 half of 1. So this function is in fact not defined in 0. But if you take, so a is 0, right? So the singularity is in 0. Here the singularity is in a. But well, I can take 1 over z minus a if you want. But 0 is just for the sake of simplicity is taken. So what I have to consider, I have to consider this, right? Which corresponds in the statement I gave you before to z minus a, a is 0. And this is, well, it is z to the power n plus z to the power n minus 1. You see plus, plus. And this part here is a polynomial up to the degree n. The coefficient, sorry, to the to the nth coefficient, to the nth, sorry, not coefficient, but the nth term in the power c expansion. This is plus 1 over n factorial. Then I have something which is a polynomial. And here the powers increases, right? So I have 1 over n plus 1 and then z plus, sorry, factorial plus 1 over n plus 2 square plus and so on and so forth. So this part here is, in fact, bannet as it tends to 0, but this is not. And you can take n as large as you want. You can always find infinitely many terms in the power expansion of e to the power i 1 over z, which have this property. So this is an example of the case three, say, and this is an example of case two, okay? So just for the sake of terminology, we say that case two occurs. If case two occurs, the singularity is called pole, okay? And the n associated is the order. Case two means not this example specifically, but the example listed in here in this fact, okay? So if there is a minimal n such that this is bounded, then a is a pole and this n here is called the order of the pole. Whereas if case three occurs, the singularity is called essential singularity. And I have just shown you that all three cases may, now let me, okay, this is, now let me characterize and this proposition, the two singularities, poles and essential singularity, pole and essential singularity. Because the more singularities, as I said, are not very much interesting for us. But I have also characterized them. So if you know something like the limit like, so if you know that the function is bounded in a neighborhood of the singular, then necessarily the singularity is removable singularity. If you want it limit, as it tends to the singularity of C minus A, the function is in fact zero, okay, equivalently. Now, so this is the proposition for the other two cases. I'm interested in. Assume that f is holomorphic in dA epsilon. That is, equivalently, f has singularity and k is two is equivalent to say. So k is two means that the limit as it tends to a of z minus a n. f of z is bounded. There exists an n such that this is true. This occurs if and only if there exists c1, cn, complex numbers, cn different from zero, such that I continue here, okay. Such that I take this f of c minus sums from 1 to n of ck z minus a k. This is another function, okay. It has a removable singularity to a, and we know what it means, okay. And then, I say case three, that is to say that there is no n such that the limit as it tends to a of z minus a to the power n of f of z is bounded. So this characterizes the singularities regarded as poles, and this is the essential singularities. This is interesting, for any positive r, the image of, sorry, dA r, take its closure topologically, this is c. So it is dense. So equivalently, this can be said like this. The image of any open neighborhood punctured by hood of a is dense in c. And this part of the theorem of the proposition I'm going to show you is, in fact, quite well-known result, which has also an extension, but it's known in this version like the Casorati-Vaistra theorem. So that I will recall it here. Casorati, which sounds very Italian, family name, Liarstras theorem. To be more precise, the best statement, this part of the statement is what is commonly known as Casorati-Vaistra theorem, that is to say that the image of a neighborhood, punctured neighborhood of an essential singularity is dense in c. This is just a statement. I don't want to say that it is obvious to see this. It's quite surprising, okay? The image is up to something, okay? So let us go to the proof. The proof is quite simple, actually. So I want to give you the idea of the proof. As I said, there are three possibilities. The removable singularity, Paul, essential singularity. So removable singularity means that we are dealing with something which is a fake singularity. So we are dealing with the whole of our financial no singularity. So what I'm saying is, assume that what is stated and point in Casorati-Vaistra theorem is wrong or is false. So by contradiction, assume that the image of a small neighborhood of an essential singularity, the singularity is not dense. What happens? So I'm assumed by contradiction. Assume, or not by contradiction, I'll assume. In some sense, by contradiction, I'll assume. Well, if I'm proving that if c is not fulfilled, if c is not fulfilled, then if b occurs, I'm not saying something which is in contradiction. This is by contradiction, when you just prove Casorati-Vaistra theorem, take an essential singularity and assume that the neighborhood of this singularity is not an image which is dense in c. So assume there exists a positive r and a positive delta such that, and exists w and c, such that, well f of z minus w as modules greater than delta, sorry, as z belongs to, this inequality means that I'm assuming that there exists a point such that the image of the punctured disk centered at a of radius r, and there exists also this, is far away from this point w. So it's not dense, there is some distance, there is some open set. So I'm assuming this. This is f of d a r, and if I also add the closure, I'm still far away. So I'm assuming that the image of the punctured open disk centered at a of radius r is not dense. So I'm sure that this function here is well defined and holomorphic in a r. Why? Because well f is holomorphic, right? This function g of z, one over f of z minus w is well defined for any z in here because this difference as a modules greater than delta which is positive, this f, this means that f of z minus w never vanish. Furthermore, it is composition if you want. It's the ratio of holomorphic function, it is holomorphic. But we also know from this is the same inequality that g of z as a modules more than one over delta. So it is bounded. So g is holomorphic in punctured disk and bounded. So this implies that g has a removable singularity at a. Correct? We are saying that g has a singularity at a, but g is bounded and g has a removable singularity. Okay, assume that, okay, g of a is different from zero. That are two possibilities. Either g of a is zero or g of a is different from zero. If g of a is different from zero, then then g of z is different from zero for any z in, I'm sorry, then I can consider f of z minus w, one over g of z in this neighborhood of a. Therefore, I have that f of z is w plus one over g of z. Then it's bounded in a neighborhood of a, or if you want, or f can extend it, extend it a, r, right? From this, it is to say, f then f has removable singularity at a. Now, consider the case g of a is equal to zero. How can we write it equivalently to the g? z is z minus a to the power whatever m times g one of z, right? Because we have characterized the zeros of, g is homomorphic, okay? And z minus a to the power n is the factor I can take out from g of z, and I also say that g one of a is not zero. It's not zero at a, but after reducing my neighborhood is not zero in a neighborhood of a, right? So, in a suitable neighborhood of a, g one is not zero, g one over z is not zero. So that I can define h of z to be one over g one of z. Therefore, I have f of z minus w is one over g of z, which is one over z minus a to the power m times one over g one of z. Or if you want h of z after, sorry, over z minus a to the power m, okay? This is not just rephrasing and changing the name because g one is not zero in a suitable neighborhood of a, this function h is homomorphic in the same neighborhood of a, okay? G one of z is not zero, this is a homomorphic function. So that I have that f of z minus w is h of z times z minus a m. But if f, sorry, if h is homomorphic, write h of z in its power expansion, center of a. Center of a, summation, let's say bk z minus a to the power k. We are just using the definition of homomorphicity. In other words, I have that f of z, which is w plus h of z over z minus a to the power m is w plus some b of k, same coefficient, but then z minus a k minus m has to be considered because of this ratio here, all right? So I have w constant plus then I have p zero, z minus a to the power m plus b one, z minus a to the power m minus one, divided, of course. Then I have bm, then I have here k greater than m, bk z minus a to the power k minus m. This is a homomorphic part and this is not, as you see, this is not homomorphic, nay. However, I have defined, after gluing together the coefficients, a homomorphic function, when I remove this, so I say that f of z minus w is w plus bm minus b naught, c minus a to the power m minus b minus a, is summation of k, bk. So after removing this part, well, this can be put also here, right? So here, so after removing a finite number of factors of this form, sorry, of summands of this form, coefficient over z minus a to the power, but only finite number, I obtain that the function f is in fact, homomorphic, that is point two. So if the function fails to have image of any small neighborhood, then since c, then necessarily, there exists c naught, well, b naught, bm. With this, one is not, so this is not zero, right? Otherwise, if they are all zero, we have a removable singularity. And in this case, f has a pole, as you can easily see. So this part here is called principal part, part of f in a, in the pole, okay? And this completes the proof of Casorati-Weisdor theorem. As I said, this is not the most generic version of Casorati-Weisdor theorem, but this is some other statements which are related to Casorati-Weisdor theorem. Casorati-Weisdor theorem is like this, but I can say more, I say, assume that, okay, this is a theorem. f has an isolated singularity and a, and there exist two points in c, such that the image of any puncture disc centered at a is contained in c minus two points. These two points. So we are saying something more, more precise compared to Casorati-Weisdor theorem, okay? We are assuming that the image of a puncture disc is not simply dense, but it omits at most two points, okay? Then necessarily either f has a removable singularity or f has a pole in a. And this is the famous Big Picard theorem, which is very, very deep and interesting in some sense. So the fact that the image is dense, the image of a small neighborhood of an essential degree is dense, is of course not as precise as Big Picard theorem. It tells you that just submitting two points suffices to have either a singularity which is a pole or a removable singularity. In particular, now there is this interesting version of the Picard theorem, which is called the little Picard theorem and states a following. Assume that f is an entire function. Remember that entire means holomorphic in entire plane. And assume that the image of the plane, so that is c minus two points, then f is constant, which is hard to see now. It is strange. Do we have example of functions which are somehow extremal for this condition? In a sense that it is not constant, but entire and omitting just one point. Well, we do have it. Entire and omitting just one point, the exponential. No, no, no, I'm talking about c, not two points in the extended plane. Otherwise, the statement should require three points. Equivalent, sorry, infinite and two extra points. Well, the complex exponential is an entire function and it omits only the point zero. Cannot omit any other point because of the little Picard theorem. Otherwise, it would be constant, it's not, okay? And which is not interesting. Well, not interesting. It's not possible, not interesting. It's not possible. I'm not saying that this is anything you can say, anything you can prove by yourself right now with the instrument that tools we have so far developed. But I can say that, well, this is somehow related to topology, more than to analysis. If we have time at the end, maybe I give you just sketchy ideas related to this. This is a simply connected domain. This is not, you might say, well, this is not surprising. Also, the case of exponential, we have c and c minus one point, which is not simply connected. But in one case, we have, in case of exponential, we have a billion fundamental domain, fundamental, sorry, groups associated to the domains, the trivial and z. And here we have z, free product z, so it's not a billion. This is somehow the reason why all this stuff cannot be compared, okay? Which is strange, you know? So very topological, far away notions come in and interfere with analysis, complex analysis. But to hope this is also somehow challenging and fascinating you. I want you to, well, this is not in a standard course, something you can afford, okay? With simple tools, as I said. But it is interesting to mention it, just to give you hope a deeper insight of the subject, okay? So this, if you want, is related to U-ville theorem. But U-ville theorem, so the conclusion is the same, right? If you have something then, okay, it's bounded, okay? Bounded is of course means that it omits more than two points. So this is a, U-ville is a consequence. But the proof of U-ville depends entirely on, because she estimates, so something related to analysis. Well, where I hear, the fact depends on topology, more than analysis. Just to say that the complex analysis, even in one complex variable, is a very interchanging material for, and a good training for many disciplines, okay, in mathematics. Okay, I stop here. And so next time we continue considering singularities and studying the singularities and related stuff. Thank you.