 So, today we continue studying the consequences of Cauchy integral formula and Cauchy representation formula we had. But before continuing, but before saying a few words, let me just mention what is known as Morera theorem, which states the following. Assume that f continues function such that in d such that for any triangle t, the integral over the triangle of f z d z is 0, d of course is a domain in C. Then there exists the function f which is holomorphic and such that f prime of z is f of z. In other words f of z is also holomorphic. This is nothing special in some sense but what I wanted to show you is that the idea well we have an equivalent more general statement of this Morera theorem already. We prove that when this property is valid for any closed curve, then necessarily we do have a primitive and then the function is. But I wanted to show you the very elegant idea in the proof of the theorem. So what you don't probably in this statement you don't see the very striking idea of the proof. So like in Gursa theorem, there was a geometric interpretation of the condition on the triangle of the rectangle for Gursa and now on the triangle. So the idea is the following. So you define in this domain d, you take point a and consider the segment connecting a to z. And you define f of z to be capital f of z with the integral over the segment and I write it this way just to indicate the segment of f w d w. Then I take another point z naught and I consider f of z minus f of z naught and this is the integral, the difference of the two integrals but since we have a triangle here, this is the triangle we are considering. This difference is in fact the integral over the third side so it is the integral between on the segment d z naught f w d w, this is sketch of the proof. As I said, the general statement has been already showed but this is just for the proof. This is another proof, another way to prove the similar result. So now we can also consider this difference over z minus z naught and this becomes one over z minus z naught of the integral over the segment of f and then I take this increment ratio minus f of z naught, f of z naught does not depend on w, so it's a constant. So I can put it on the right hand side of this expression, I can also put it inside the integral because we are integrating over the interval. So in some sense I make this kind of, when I consider the modules, sorry, of the left hand side, I can also take this estimate, one over z minus z naught which is the length by the way of the side. So this is not, there is no parenthesis here, right, there is no parenthesis, but when I consider this length I can also take the integral of the modules of the difference over d w because this is a constant, right, and the constant over an integral over, sorry, the integral of a constant over an interval is the constant times the length of the integral which is over here, but this number tends to zero if z tends to z naught, correct, because we are assuming the function is in fact continuous. So this can be made smaller than epsilon and this is the proof. So I wanted to show you this proof, so it's somehow an elegant proof and you can repeat it of course in the general case. So this in fact is smaller or equal to the maximum of this when w is in this compact set, okay, so it depends on w and then you can make it as small as you want because this is precisely the distance, so the distance taken for continuous function over an interval, okay. But just because of the sake of completeness in the subject, and now let us continue by, okay, this is number three. Let us continue by considering the consequences of the Cauchy integral representation. I want to remind you that if we have a function f, holomorphic, and I take a in d and a curve, gamma, closed curve, so the general statement we have is that the value of f in a is one over 2 pi i n gamma a index over times the integral over gamma of f c, c minus a, c. So in particular, as I said, if this is d and this is a, we can take as a curve a circle around a, centers a, okay, inside d and the radius r, so that gamma of t is a plus r e i t and t varies between 0 and 2 pi, so that is this index is in fact one, right, good. Now, so this can be done locally at any point, so this if you want is a local version which is enough for us at the moment, all right, so now consider, this is the notation I need, as I said, take this circle of radius r, center of t, and consider, take the general point z, right, and z is any point here, so the aim of this calculation will be to show that any function which has this integral representation is in fact analytic, complex analytic, so that we somehow close our description of holomorphic function as functions which are complex differentiable and in fact complex analytic, we started from examples from complex analytic functions and prove that they are effect holomorphic and now we prove the reverse. Any holomorphic function satisfies the Cauchy integral representation and then we prove that this implies that it is complex analytic. We do the following, so we take, okay, x c varies here, right, so x c minus a as a modulus r, constant, okay, I take z such that z minus c is in fact smaller, sorry, z minus a, right, smaller than r, so inside the disk, the center of the disk is a and I take z inside, remember that x c, because we are integrating over gamma, I didn't say it, okay, this is my preferred gamma circle, so each point x c varies on the circle, whereas z can be taken arbitrarily outside, but I take it inside so that z minus a is smaller than r and then I write this, equality, z minus c, x c minus a plus a minus z, obviously true and which can also written as x c minus a times 1 plus a minus z over x c minus a or equivalently x c minus a times 1 minus z minus a, x c minus a, right, what can I say with our assumption here, we can also say that this number, this ratio here is smaller than 1, which is important because we recognize in this expression what the sum of a series, not over series, the series of a power series generated by this ratio, right, so with this in mind continue here and we have that, so what we are integrating as this, which becomes this, since z minus a over c minus a is smaller than 1, strictly can be also written as f of x c over c minus a summation and greater or equal to 0 of this, z minus a over c minus a to the power n, right, so this in fact is this, right, good, so now remember that this was the integrand, we have the following, 1 over 2 pi i with the integral of this, in fact now with the new, if you substitute becomes 1 over 2 pi i integral over gamma summation, sorry, maybe, well I put it this way, f of x c minus a then summation n greater or equal to 0 of z minus a power n over c minus a to the power n big c, correct, just substitute the substitution of this with this and this is consistent, but now I notice that z minus a does not depend on c, so I can take out of the integral and I can also add 1 to this exponent in order to have c minus a and this summation, in other words, I can write f of z 7, f of z is, I write it this way, 2 pi i integral over gamma summation n greater or equal to 0, f of c c minus a to the power n plus 1, remember that we have an extra coefficient, sorry, extra term, right, and then I can also integrate here and take z minus a to the power n out, which is very close to what we are looking for, right, if we prove that this is a, sorry, if you can take out the symbol of summation and we obtain that this is a number, call it not here, but well, the integral of this, okay, so forget the symbol of summation and this call it for n, for h n, call it a n, okay, without this, we are done, but remember that with this we meant a limit, the limit of the partial sum so that here we are taking the limit, or we want to prove that the limit of the integral is the integral of the limit, is it true in general? Well, not in general, unfortunately, but in this case in particular, it works fine, it should. Let me remind you one simple result we should know from some, from general calculus, which applies to this particular case, so we are left to this point and then we continue, okay, we cannot conclude, but we have all, so the feature of what we are looking for is very close, this is z minus a, so to the power n through select power c, and this is very close to coefficient, good, so let me just recall you that when you have, when we are in the case, we are dealing with essentially gamma rectifiable curve so that you can calculate the length of the curve, reasonable curve, and c, and they will take f, we call it un, just to be okay, un sequence of continuous functions on gamma. In fact, we are considering function restricted to the curve gamma, gamma is a circle in our, in our prove, right, so a circle is rectifiable, so this is more general statement, and we take then u to be the uniform limit, u limit as n tends to infinity of un, so this is a uniform limit, so the uniform limit, okay, independently from the parameter we are considering on the curve, is a continuous function, because it's a uniform limit of continuous function, and we can take the integral of a u over gamma of u, and this is in fact, as the limit of this numbers, okay, you are, are you with me, so remember that, well you say, well what is this relate, what is the relationship of what we are doing, we were doing before, well remember that here on the right hand side we have the the limit of the integrals, and inside here you have the u limit, so a uniform limit, right, so the limit of, of the value of the single, of each integrals converges to the integral of the uniform limit of the functions, which were, is our, our task, right, so this depends on the fact, well this, I think that this is, should be obvious to you, you should have already said this somewhere, some of the causes, however if you take say, well any z, okay, this can be made smaller than epsilon, right, for any z in gamma along the curve, so for any epsilon positive there exists a not or n such that for any n greater than n, this is true for any z, right, this is in fact the, the hypothesis we are taking, so u is the uniform limit of the u n's, it is continuous, and you have also in particular we take the integral of a gamma of u n minus the integral of u, and you take this and this is the integral of gamma u n minus u, this is smaller or equal, remember one of the inequalities we proved explicitly and we used several times, this is smaller equal integral of gamma of u n minus u, okay, and this can be made smaller than epsilon because well this is what, epsilon prime is this epsilon times the length of the curve gamma which is finite, right, that's why I'm assuming that the curve is rectifiable, so it is as small as we want, so we are done, so going back to this case, special case, then we can say that this is in fact summation 1 over 2 pi i of the integral of the gamma of x c, x c minus a n plus 1 d x c, and this whole stuff is will be called a n times z minus a to the power n, all right, which is more than what we want because we have an expression of the coefficients in terms of an integral and if we are smart enough to understand the meaning of the coefficients we can also say well this gives you also information on the higher derivatives of the function f itself, okay, so for the time being what we have showed here is that indeed starting from a function which admits a Krushy integral representation, particularly if it is allomorphic it has this representation, we can always find coefficients which we explicitly wrote here of the power expansion in a neighborhood of a point where it is defined so that this guarantees it is locally analytic and analytic, complex analytic function, all right, so that there is a complete characterization of allomorphic function, those functions which are complex analytic, and as we will see the fact that this local is not restrictive because then we show that if you know something locally for homomorphic function then it will be somehow globally extended, good, all right, what okay, what we have to prove in some sense is well to conclude like this we have to be sure, well just to be very precise, we have to be sure that this number here can be calculated and that well the the integral can be taken, well this is in fact the gives you the sequence of function, not the integral, okay, this is the sequence of functions, so the sum, the partial sums up to the order n is the sequence u n and the previous lemma, we have to prove that the limit is uniform just to be on the safe side, but this is done quite easily because well remember that we have f of c over c minus a to the power n plus one, then I have also z minus a to the power n to be considered and well you take the modules, this is in fact a modules of f of c and then I have c minus a to the power n plus one and then I have z minus a and we are taking the integral over a curve, a compact curve, a compact subset, so in particular this number here can be controlled, it's bounded, right, that is m times modules of c minus a to the power n over r n plus one, remember that c minus a represents the the radius of the the the circles, the circle, I have to consider because c varies on the curve gamma and well and since we are now in here, this is m over r times z minus a modules over r power n and this number here, this number is less than one, so this is this guarantee, this is this is if you want the m test application to guarantee this well, all right, so everything works fine and we are now at the certain stage that we can conclude something, in particular we can say the following, well remember that coefficients the in the in the in a complex analytic function, but in general for function which are analytic and c infinity and c infinity regularity, well you can calculate the derivatives and term by term derivation is what we applied for our case and it works fine because we have approval of this fact, if you differentiate term by term consider another another series with the term by term the the terms of the the series as the difference variation of the previous term, so you obtain the derivative which was proved to be true, it was proved that this is equivalent to consider the derivative in the sense of the complex numbers theory, all right and the value of the nth derivative at a point a was related remember the value with the coefficient with nth coefficient up to a constant, remember this is n factorial of a m, we will start from say f z to be summation z minus a to the power m, so this is an n factorial, this is in general true because the other terms have something like z minus n to the power n plus something and so when you calculate it at a they they they vanish, so this is the first term but then we also can we can also say then this is in fact 1 over 2 pi i the integral of a gamma of f of c c minus a correct because this was defined to be a m and the previous expression which tells us something interesting all right as I said before okay gamma is the circle because otherwise I have to put n gamma a right gamma is the circle and then as I said before f of c is smaller than m on gamma because it is a closed curve it's a complex set f is continuous so it has a maximum its modules has a maximum right and c minus a is r since gamma is a plus r e i t and t is interval is 0 to pi so we are taking the circle so that the modules of f and a is n factorial smaller equal to m over r m you see this while you might say well but there is something else here and the n plus 1 is not here but we also have dc remember this in fact when you calculate well let us do it as an exercise okay this is to be checked this is cushy estimate right this will be cushy estimate when I prove it for the moment there's a statement okay a claim so what I'm saying is that well we have to consider this integral over gamma when gamma is a plus e i t r correct so that this becomes and as I said t is in between 0 to pi so this is 1 over 2 pi i integral 0 to 2 pi and f to substitute f of gamma t then gamma t is this right so a minus a 0 so r e i t to the power plus 1 and then I have here gamma prime of t dt and this is 1 over 2 pi i integral from 0 to 2 pi f of gamma t and then gamma prime of t is what e sorry r e i t because I'm differentiated with with respect to t so here I have r i t and here I have r n i t and plus 1 sorry r n plus 1 right and then this can be okay if I take the modulus so here r i n appears on the on the denominator the integral of d t is controlled by the mad modulus of this and this is a constant okay and so i disappears and so we are done right so this is an important estimate and as as we said this r here represents the radius of the circle we are we are considered from the very beginning centered at a but assume that you have very general complex analytic function so it has a power expansion written and it has a radius of convergence r which is the largest radius where the function is defined so this kind of game can be repeated for any r inside the disk of convergence of the function itself independently in some sense as we as we know if we take r strictly smaller than capital r where r represents capital r represents the radius of convergence of the series of the functions analytic function then we are sure that the convergence is actually uniform remember this was proven in Adam or theorem so that this estimate can be repeated as I said for any r and so the best estimate is when r approaches capital r because the number increases right denominator so it is smaller so the best estimates in fact is with r being the radius of convergence so this is correct then applies to any radius inside the the disk of convergence and in the general form is as follows so Cauchy estimates sorry I'm sorry this is 12 right Cauchy estimates complex analytic a capital r so f of z this power expansion an z minus a the power n and right then we have that f and a is smaller or equal to n factorial m over uh to the power n m being the maximum of f of z and okay these are Cauchy estimates and it has interesting consequences in particular well we could also say that an is remember that this is f and a over n factorial right now assume that you know this f is holomorphic and this now can be finally considered to be going complex analytic and c so this equivalent claim means that r is plus infinity correct so in c means that the radius of convergence the largest possible plus infinity and we have examples of function with the radius of convergence infinite for instance the exponential any polynomial correct good and this is something but assume that f has its modulus which is bounded then what do we have we have that an the coefficients and the power expansion turn out to be well this is up to a constant is this is smaller or equal to what m over yeah right and this 10 to 0 since r is infinite so you take a take another r right smaller center at a and then make it go to infinity this estimate works fine okay for any r what do we have that all the coefficients are zero so except the the first coefficient right when n is zero so the function itself is constant this is the real theorem right so it it it can summarize it in the in the following way any function which is complex analytic and its radius of convergence is infinite and bound and its modulus is bounded is in fact the constant function so it means that the function exponential complex is not bounded it also means that cosine and sine not bounded complex cosine otherwise we they have to be constant right good we apply Liouville theorem several times but I want to use the terminology which are which is which is standard in many books so when you say the function f is almost in c we actually mean equivalent to the f is an entire function all right so it's just a name this term entire function okay so for entire function there is this very very important geometric characterization if they are bounded they're constant so let us apply this fact for a proof of the fundamental theorem of algebra so you I guess that you all know the statement of the fundamental theorem of algebra and you probably have already seen one of the proofs of this theorem fundamental theorem in algebra but you probably don't know that it was Gauss who actually proved this theorem with several proofs and in his master thesis so it was quite a smart guy he used well other approaches but essentially he used something related to the index of curves okay winding numbers associated to polynomial so the statement is as follows so take a polynomial p of degree greater or equal to 1 in c well p has a root because this is fundamental of algebra and that's why now we consider the complex numbers to be in fact the algebra closure of the reals because all the reals the theorem fails to be true exactly we have shown examples of degree two not to the degree one of a degree two where this theorem fails to be true right but as one of my teachers told me one when I was your age if you want to miss the beauty of the complex number just restrict your attention to the algebraic properties well this is just one of the properties you want to see all the beauties of the complex numbers please look at the analysis and the geometry beyond algebra well algebra is important of course so the proof of one of the proofs is a consequence of the Lyouville theorem assume by contradiction assume that p of z is not zero for n of z so by contradiction okay sorry I'm saying by contradiction well and then I consider the holomorphic function one over p of z it turns out to be holomorphic where for any c for any sorry for any z so in c right with this so it is entire right it is entire so one of the hypothesis of Lyouville theorem is satisfied can we prove that f of z as a module which is bounded yes we do because f of z as modules we use one over p of z modules of p of z and p of z well it's never zero first well what is this already used but also when z tends to infinity modules of p of z tends to infinity so we can do it like this we can restrict our consideration to a disc and inside this disc p of z as any continuous function beside the closed disc inside of the disc this polynomial as the modules of this polynomial has a maximum right and a minimum as well right this is by the theorem right but outside this so in in the f net region outside this we we know that p of z as a module which diverges to infinity this is because p of z is not the constant polynomial and since the modules you know this you can you can you can you can use the the simple estimate on the modules and since z of z as a module which tends to infinity p of z has a module with tends to infinity but then f of z is bounded sorry the modules of f of z is bounded right so it is an entire function which is also bounded so f is constant QED and so we have a contradiction contradiction you will now we have also examples of functions which are entire and never vanish for instance the explanation it's not a polynomial right now i might be interested in the possibility of extending holomorphically this is some somehow a higher degree question i might be interested in extending a polynomial to the Riemann sphere can i do this it has a value to infinity okay i have to repeat i have to repeat okay so one of your suggestion is correct we can take the limit okay but remember the holomorphicity means that something is known about the neighborhood of a point not that simply you add the value at point so you can continuously extend the function but not holomorphically for instance though this has to be checked right you understand what i mean well it's obvious that since we have this polynomial cannot have a finite value at infinity if you want to extend it at infinity right because otherwise it wouldn't be even continuous so well you can extend it then there is no hope to find to to have an holomorphic extension however however what we do when we consider the remember that that we introduced the Riemann sphere the extended complex plane by applying the serographic projection right so if you want you have a hemisphere which represents the disc and the other hemisphere which represents what is outside of this and to reverse the positions you have to consider one over z what is outside put inside and then consider one over p of one over z and see how it works and consider what happens at zero there and extend it there and if you find our expansion anywhere and the good position and in fact you can add infinity in some cases and in some others you cannot why because remember that our function if they are extended to the Riemann sphere since they are continuous and the Riemann sphere is is common they must have exactly modules which is bounded so in particular when we show that locally you have information which comes global then if it is locally constant then unfortunately it's globally constant so you cannot extend it right or you can extend whether it is constant so in fact very few functions are holomorphic on the entire sphere Riemann sphere the constant constant function which is one of the cases the constant polynomials you understand why well if you don't I'm not saying that this is an explanation but if you don't I will give you some extra okay results in order to understand this good and in fact the next step is to start considering and and describing the properties the so-called local properties of holomorphic functions so we'll investigate in the next lessons next lessons zero singularities whatever but first we have to I want to I want to give you a very important characterization how holomorphic functions the so-called identity principle for holomorphic function this is a proposition and I refer to this result as the identity principle which as you'll see has no counterpart in the real case the real even in the real case the real so see in theory the case is no counterpart assume that D is domain and I write it explicitly open and connected subset of C important to be connected then F holomorphic and D so the following statements are equivalent first F is identically zero in D the second is that there exists an A and D such that the F derivative of F F is holomorphic so we know that now it is infinity blah blah blah but at A is zero for any end third characterization as the following the set of zeros Z of F so the set of point Z in D such that F of Z is zero has an accumulation point or limit point so what I'm saying is that there exists a point A and the domain D such that each derivative of F at A is zero including the zero derivative which is the value of the function well of course one implies two and implies three what is difficult to prove is that three implies two and two implies one so that the three statements are equivalent once we have this we can say well if you locally two functions agree on an open set well they agree elsewhere where I define and this gives you the possibility to extend holomorphic function if it's in sense of a germ so you have something in an open set something else in a another open set which intersects in something with an interior then the function can be extended so is it clear what I meant here in the second condition so I'm saying that if there exists A the existing way is important right so as noticed of course well this is one implies two and implies three right now we prove that three implies two and two implies one so by contradiction okay assume that there exists a knot such that a knot of A is not zero for any a and not so that the power expansion of f of z is like this right summation of n greater so take well a knot so assume that a knot is the minimal such that this this occurs and so that we can write this a n minus a to the power n which is in fact summation a over n factorial right z minus a to the power n so take the minimal such a knot now we define another function g of z which is related to f in the following way that g of z be such that be such that f of z is z minus a and not g of z or assume that g of z and summation of a n z minus a to n minus and not equivalently right I show you we have this this equivalently which is now what do we have well remember that a n yes the nth derivative of a at the nth derivative of f at a over n factorial and we also have the four that g of a is a and not which is different from zero because we are assuming that f and not so the nth derivative of f at a is not zero and this number here is the first and the power expansion of g so is the is the term without z minus a right when n is n not okay we assume that f n by contradiction this is different from zero so this is different but g is a continuous function right it is well it is allomorphic it is complex analytic so it is also allomorphic and g of a is different from zero so it is different from zero in a neighborhood I continue it so well this is to be remarked here that g is holomorphic this is the power expansion and this implies a continuity g of z is not zero in a neighborhood say in an open ball say in b a r with r positive and everything works fine but now assumed at a can be assumed the limit point and of the f remember that we are assuming that point three in the statement is true that is to say that the set of zeros has a limit point so this is a limit point of zeros so there exists then b and b a r b and z f right because of the definition of limit point okay a limit point z f is in fact equivalent to say that given any neighborhood of this limit point accumulation point in d e okay there exists an element in z f a set for which a is the limit such that a is in this neighborhood you take as small as you want but there is an element now b is in z f and so this is also true b minus a is smaller than r whatever r is taken but b in z f means equivalent to that f of b is zero so this condition guarantees that b is not a so there is an element which is not a in z f close to a as much as i want this is but f of a is zero means also that b minus a to the power n of g of b is zero and this is not zero since we don't have zero divisors the only possibility is that g of b is zero which contradicts the fact that g of z is not zero whenever z is taken in the neighborhood b a r so this is a contradiction are you with me good so by contradiction we prove this now to complete the proof let me assume that all the derivatives at a point a in d is all the derivative each derivative at a is zero now i want to prove that the function is in fact identically zero which in fact gives you the idea of the name identity principle so if in some sense if you know this characterizes completely complex analytic functions if you know the coefficients of a power expansion in a neighborhood then you know the function and there is no other function with the same coefficient take the difference if the difference is zero then all the coefficients are the same which is vice versa if the all the coefficients are the same then they have the same function so coefficients and power series so the coefficients of power series are the same okay locally now okay i want to prove this so that we are assuming point two i want to prove that f is identical is here right so i take a to be the set of point z in d for which this is zero for any n this is a set is it empty no because i'm assuming the two is true so i is not empty because a in the statement of the of the proposition is in a so there exists a okay so if we want equivalently we say this set is not empty this statement two now what we prove is the fault this set will be open and connected and then it is the entire domain d so it follows from topological consequence of the assumption of the domain how do we prove that a is closed consider z and the closure of a i want to prove that z is in a right so z is the limit points in a z n tends to z now what can i say about this well remember that z k is in a so all the derivatives so for any point all the derivatives are zero that's okay but when i want to consider this fix n i want to consider this okay fix n and this is evaluated at the limit but since we are dealing with holomorphic functions we know that the the function the the derivatives the derivatives are itself are also are also continuous okay so this is in fact the limit as k tend to plus infinity z k with n fixed so don't think that this is in the limit so this is zero in other words z is in a because this turns out to be true for any n so it is closed and to complete you know a is open so now take right take a in a it take an open some an open ball center that a in d so we take a point in a and we take a neighborhood of the point in d if we prove that this neighborhood is in fact in a we are done so write f of z and z is in this to be summation of as usual remember that we are assuming that f is holomorphic so in particular it has a power expansion center at a but then remember that also that a n is related to the nth derivative of f at a and the following way right and we are assuming that a is in a so this is zero for any n so f of z is zero in b a r for any z for any z sorry for any z i write it this way if it is zero locally zero so what's what's the what's the end of the story the end of the story if you you take the expansion over that over another point then the all the derivatives are zero so z is in fact in a and so a is open and closed in d this a is now this implies very important facts in particular let me just remind you this version 21 so corollary take f and g holomorphic d such that the set a point z and d what the two functions agree has a limit point so for instance if a is an open set this is certainly true so as a limit point and what i'm saying for instance if a is open this is this occurs okay so any point is in fact a limit point so sorry any point can be considered any interior point is is good but there are very many well then the consequence is that well f of z coincided with g of z and d just an obvious application and this gives you a very deep inside of the differences of a real analytic function a complex analytic function there is nothing similar to this you can exhibit function which agree on an interval which is an open set or an infinite interval but it can be extended but they are not okay so you know the same as a e minus one over t squared and the constant zero and so on it's not zero so this is a very important property of complex and this explains why it suffices to prove something locally in a sense right so to start the local study is very important for the definition is that local you cannot define something point wise you have to if you want to define something related to complex differentiation you need an interval sorry a ball around the point you are considering now in fact in next week we'll investigate locally the the features of points of zero points and it turns out and it turns out that I'm anticipating what you will see okay that the zeros are somehow like the zeros of polynomials so if you have a zero function which is complex analytic then it has a multiplicity which is something not not true for for for for real complex value for real valued functions you can have well each each each each degree of multiplicity you want so each each multiplicity it's a real number okay for homomorphizing this is not true and we'll define multiplicity of a zero and then study the singularities and attributes to to singularity and attribute to a singularity something which will be also multiplicity and then we will discover that will be also some extraordinary strange singularities for which there is no possibility to to give to give a multiplicity which will be called essential singularities okay so I think that today I'll stop here questions so far I intentionally didn't didn't afford the the the calculation which will okay will complete next week