 OK, so today I want to continue talking about the properties of holomorphic function connected to the value of the integral of a holomorphic function over a closed curve in particular. So last time we have shown that for a holomorphic function on a rectangle, the value of the integral over the contour is 0. This is the statement of Gursath theorem and we also have considered other examples of holomorphic functions for which this similar property is valuable with a generic curve. Generic means a bit more strange than a rectangle and we also showed that when the function is not holomorphic, one point, the example was remember this, which is a function with only one singularity in A. It is not defined, not even continuous in A, so it's not holomorphic, but elsewhere it is holomorphic and we have calculated the integral of gamma, the circle of radius r, center of A, r positive. We obtain that this number is not 0, t0 to pi. So it's not always the case that any complex value function as integral equal to 0 over a closed curve. On the other side, we have seen that if the complex value function has a primitive, complex primitive, then this is true. Now, we will investigate a little bit more, what are the consequences of this Gursath theorem, right? Of this and we will then study a bit more in detail in this case, right? So I will start from considering this possibility. Now assume that the function f is holomorphic in a disk, d, I sometimes use also b for the disk because b is the letter used for balls, okay? So the ball in the plane is called disk, right? b being balls centered at r at A and radius r. Now we can take given z and the disk, this is the open disk, right? Normally when I use this parenthesis, I mean open disk. So just a matter of notation. If I want to indicate a closed disk, I use either the overline, okay? Which cannot be confused with the conjugate because it's a set, right? But in any case, when I want to consider also the contour, the boundary, then I use either this or in some books you can find here other parenthesis, you know? Like for the intervals and on the real, okay? I will often use also the terminology as follows. When I say domain, we consider open and connected sets, okay? So domain, d means an open and connected. So the disk is somehow the prototype of any interesting domain. But in any case, when I say domain, I mean something more general, something which is, for instance, not a disk, but open and connected. Now in a disk, I can start by considering something which we already done, okay? The function f of z being capital F, huh? We have the olomorphic function f being the final form. So take z and this will be the potential. But how? This is the integral over sigma of f xcdxc, sigma being this curve, okay? The polygonal path connecting the center to z with segments parallel to axis. This is some, imagine that this is, right? So this function is defined. And of course, because of Gursat theorem, we also have the following same situation. So I move this way, first horizontal and then vertical and connect the point z, a to the point z, okay? So this is sigma for me. But well, since I know that the integral of f of zd z is zero when here I have a rectangle because of Gursat theorem, then I can also consider equivalently this path with two segments parallel to the axis, but in the opposite order, so first parallel, first vertical and then horizontal. Now remember that this, well if you only have zz is the integral of f of zdx plus i f of z dy, right? So when I take the derivative of f with respect to x, we have f of z. When I take the derivative of f with respect to y, I have i f of z, like it was in another previous argument used for the definition of the potential. Which means that the function f is, in fact, complex analytic or holomorphic. It satisfies Cauchy-Riemann equation. So since this is well defined and f has a primitive, holomorphic primitive, then I can conclude that the integral over any closed curve gamma inside the disk is zero. Okay, I have a holomorphic primitive and this is, and of course this works fine also for another class of open sets, not only the disk. If the contour is an ellipse, it works fine because you can always put inside the rectangular repeat diagram. So we have a very general case, but not the most general case. Okay? However, if you consider just locally, and the holomorphic function can be studied just locally, you can say that locally this property is valid. So the integral over a small closed curve around the definition, so in a neighborhood, in an open disk, okay, of a point, well, you can always say, well, the integral of a closed curve is zero for a holomorphic function. And as I already recalled, this is not true if the function is not holomorphic at one point, correct? So, but let me show you that we can somehow extend also for the class of holomorphic function in an open ball or in an open set without holomorphicity defined all over, but with some singularities, final singularities, and additional good geometric properties of these singularities in order to have the same the same fact about the integrals. So you take f holomorphic rectangle. So this is somehow another general version of Gursat theorem, weakening a little bit the conditional regularity of functions. So in a rectangle r, and as I already said for the Gursat theorem, the statement of Gursat theorem, we are considering a function which is holomorphic in the neighborhood of the rectangle, then consider the integral over the contour of the boundary of the rectangle. But holomorphic this time is not in the entire rectangle, except for a finite number say x1, xk of points. So these points are called singularities. So they are not, what is important to consider is finite number of singularities. And this is not enough because if we consider again the case of the function 1 over z minus a, this function has a singularity only at a, and we know the integral over a rectangle over a curve is not zero already. But we assume that we have this good property that the limit as z tends to each singularity of z minus xj f of z is zero. And this property is not satisfied by 1 over the function f of z 1 over z minus a, correct? This has to be true for anything. Then the integral over the contour of the boundary of the rectangle of f of z is zero. Okay, as for the Gursat theorem, we'll use just geometric consideration. Remember that the proof is somehow not related to analysis. We divided the rectangle, subdivided into small rectangles and then make consideration about the limit of this process. So first of all, we can make a consideration just on one singularity, which is of course the interesting case. And then repeat the same argument for a finite number of singularities. And we also can assume that the singularity can be isolated say in subdivision, I'm sorry, of this time, such a way that the singularity say j is inside a square. Instead of subdividing, as we did last time, the rectangle in four rectangles, similar to the previous one, we adopted this, the composition in such a way this is, this singularity is the center of a square of radius say l, of radius of say, of side l, of course. This can be done. And as we observed last time, the fact that we are calculating the integral over the boundary of r is equivalent to considering the integral over the small sub rectangles and sub squares, whatever here. Because the contribution you have on this side is what contributes to the integral over r. But what you have here, putting this orientation, is balanced by considering the contribution here. So each time you have opposites. Okay. So at the end of the story, what is left is just the rectangle, sorry, the integral over the square centered at xj. If we show that this is zero, we are done. So if there is no extra contribution when considering this special case, we reduce our consideration to the previous case, which means without singularities. Let me just use the same notation I have here. Okay. So the integral over dr, q is the square. This is q and this is r. What do we have? We have the assumption, the assumption that, well, the modules of f, right, the modules of f is more than epsilon. I am sorry, as I said, well, this is. Now we have this, right, limit as z tends to cj, fz times z minus cj is, right, so this is zero. So I have this or modules of f of z is more or equal to epsilon. On the other hand side, remember that we are in a square, q, this is cj and any point to z we are considering for the integral is on one of the sides, right? So that the distance between z and cj is at least l over 2, l being the length of the side, correct? Listen, I see it this way, right? So any other segment connecting the center of the, is longer than l over 2. The integral over r of fz, z, as I said, as a modules, which is the integral as a modules of this. And this is smaller or equal to the integral of modules of f of z to z. Remember this was one of the inequalities we proved last time and we used several times. But now f of z from here has a module which is smaller or equal to epsilon over the distance of z minus cj, right? In other words, putting together our inequalities, we have the integral over r of f of z to z, smaller or equal to the integral for the q of the modules of f of z to z. And this is smaller or equal to epsilon, the integral over the q z minus cj z. But remember that z minus cj is greater or equal to l over 2. So one over modules of z minus cj is smaller or equal to 2 over l. So again, I put this integral over the q epsilon 2 over l dz, correct? So this is a constant number, which does not depend on z. You can take it out from the integral and what is left is the sum of the length of the sides of the square q. But the square q has, of course, its side is l, so it is 4l, right? So this is 4l times 2 epsilon over l, which comes as l and gives you 8 epsilon, so it is smaller than epsilon prime, so it is infinitesimal. And this concludes the proof, because you can repeat the same argument by considering squares surrounding the finite singularities we are considering, okay? Good. And with this in mind, we can also say the following. Well, start from a function f, holomorphic, the disk. But I omit, as for the rectangle, a finite number of points where the function is not holomorphic anymore, right? Call it d prime, which is punctured disk. Well, the situation is like this, a and then I have some points where the holomorphicity fails to be satisfied, so it is not. However, I can, so let me show it this way. Like in the previous case, I can always restrict my consideration to sub-squares. I assume that this is the case, right? And I have a singularity here somewhere. Well, I can take this with a square and then define the function this way. I single out the singularity and surround it with a square and then I repeat my consideration, starting from a from here, right? Because the contribution of the square is nothing for the for the integral, of course. And I can define starting from here, well, a function which is a potential again. So what I'm saying is that with the argument we used before, even if the function is some singularity, but find the singularities, together with the previous lemma I showed you geometrically, then we can extend the same result. So if you want the integral over a closed curve, which does not pass along the singularity, guarantees that this, only over this path, closed path, the integral over a holomorphic function is zero. And this is very important, because from this we can have, finally, the Cushier representation formula. But on the other side, we have, since we are missing to put in the class of function we can consider from now on, the function which gave us an interesting counter-example of the general case, we want to investigate it a bit more in detail. Namely, this seven, take a closed path, closed curve, gamma, I take again the function z minus a over, one over z minus a, right? And assume that the closed curve is such that gamma of t is not a for any t, third draw, okay? Gamma is defined. Now this function is, as I said, holomorphic and the complex plane, but not in a. So in the complex, in the punctured complex plane. And assume that this closed curve we are considering is not passing through the point a. Some irregularity, well, gamma is piecewise differentiable. It's a complex value function with the real variable, and we assume that at least almost everywhere there is the tangent vector, correct? Now I want to show you that, well, this number here, integral over gamma, one over z minus a, d z is a multiple of 2 pi i. In fact, in the example we considered as a closed curve, the circle center dot a over radius r, which of course is not passing through a because r is positive, right? Remember? It is a differentiable curve, and the result of this integral was 2 pi i was done by hands, huh? Okay, I will use this notation here. So I have 7. So gamma, assume that gamma is defining i, i as the interval minus, well, alpha beta and c, okay? Use alpha and beta because I don't want to use a and b. A is already used for the singularity. And I define h of t to be the function integral from a to t of, well, precisely this. I have to write it explicitly, means we are considering the integral over gamma over the function 1 over z minus a up to the time t, correct? And I write it in, by substitution, by substitution input s here and t is here, correct? Now I can also, well, where gamma is continuous and gamma prime say is continuous, we can also say that h prime of t is gamma prime of t over gamma of t minus a. This is a fundamental theorem of calculus. This is not a real variable. And then I define another function g of t is e minus h of t times gamma t minus a. If you want, this is as gamma t minus a over e h of t. t varies in between alpha and beta. What is the derivative of g with respect to t? Well, we use standard rules. We have, if you want, e minus h of t times minus h prime of t, this derivative of this times gamma t minus a. So, library's rule applies, plus e minus h of t times gamma prime of t, right? All right. So, remember that, okay, I said, oh, I write it down again. So, this is minus e minus h of t h prime of t gamma t minus a plus e minus h of t gamma prime of t, right? And remember that h prime of t is gamma prime of t over gamma of t minus a. So, when I substitute, when I'm allowed to do this, okay? So, I substitute h prime of t here. I cancel this and this. And what I have is this is what? Zero for any t. So, g is constant. And in particular, g of alpha, remember g of alpha was defined by using the function e minus a. But this is zero because the integral between alpha and alpha. So, this is one and so we have this. You see this? Thus, from the fact that g of t is e minus h of t minus a, I conclude this, right? When I put as t alpha and remember that h of t was defined in this way, the integral from alpha to t of gamma prime s over gamma s minus a ds. So, the integral is clearly zero when you take as t alpha, all right? So, we can also say the following. g of t is constant. g of t is gamma alpha minus a for any t. That's what we have discovered after considering the derivative of g and proving that the derivative is zero. But e minus h of t times gamma t minus a, which is g of t is, in fact, gamma alpha minus a. So, that I can also say that gamma t minus a over gamma alpha minus a is, and this is, this number here is different from zero, right? Because we are assuming that the curve stays away from a for any t. Gamma t is not zero, it's not a, sorry. So, gamma alpha minus a is different from zero and this is, from this quality, you see I put here this over this is over this, right? It's one over e minus alpha t, h of t, sorry, right? Is it correct? Yes. And this is e h of t. But remember that, is it correct? Remember that the curve was closed, was assumed to be closed. So, gamma of alpha is gamma of beta. In other words, gamma of beta minus a over gamma of alpha minus a is one, right? This is the left hand side when I substitute t with beta. So, on the right hand side, I have that this is e h beta. Then I have to remember that the h of beta was the integral from alpha to beta of gamma prime s, which is precisely this from the definition of the integral over a closed path. So, what I have here is that this integral provides a number whose exponential is a complex number, whose exponential is equal to one. We already noticed that the exponential of zero is one, but the exponential is periodic period 2 pi. So, this number here has to be a multiple of 2 pi i. Hence, we conclude that the integral of a gamma is 2 k pi i k and z, which is very important. So, we associate to a point and a curve an integer and this integer has a name definition index closed curve gamma is defined in the following way in short n gamma a. So, it depends on gamma, it depends on is just the integral over gamma or one over z minus a is z. This number is also, it is called index of the point a and it is also known as winning number. Sorry, sure, definitely this is not an integer. Thank you. One over 2 pi i of course, the integral, right. So, thank you. So, the integer we are dealing with is the integer which represents the times of the number of times, sorry, the number of times you are multiplying 2 pi a in the calculation of this entry. And this represents if you think of the function one over z, obviously, the number of times the curves wins around the point a in the case of one over z is the point 0 because you have the logarithm to be considered as primitive. Now, good properties and then come back to Cauchy integral formula. Good properties of this winning number or index properties. Well, one is obvious. Consider the integral over the curves with the reverse orientation is to say consider this. As we know, this number is obtained from a calculation of an integral and when we reverse the parametrization of the curve gamma, we obtain the opposite of the integral. So, this is elementary property. Now, this is property number one, say, property number two. Assume that gamma, this is a and the curve gamma consider as letter, change letters, I put z naught as a center, sorry, take the disk centered as z naught or radius r. I assume that gamma is containing d like here. Of course, I use some abuse of notation, which is very common in any textbook. When I say gamma, I mean sometimes the function, sometimes the set of values. Well, in this case, I mean, well, if you want to be very precise, gamma y. So, the set of values, the subset of c, which represent the values of gamma function, in any case, I think that it is quite normal to interchange the roles of the function and of the curve as a set. Anyway, assume that a is outside of this disk. It is a is not containing d. What can I say about the index of gamma a is zero, correct? The answer is quite simple, if you think a little bit. Well, a outside of the disk means that the singularity is not considered where we are sticking our function, one over z minus a is nothing. So, it is holomorphic in the disk. It is holomorphic. So, the integral of a closed curve is zero, period. So, the next property is probably the one which guarantees that this definition is important. So, n gamma a is constant on each component c minus gamma and n gamma a is zero in the unbounded of c minus gamma. Well, I omitted to say that, of course, gamma is to be a curve. So, nothing pathological like piano curve or something like this. Somehow, I am assuming that the curve is at least a Jordan curve. Well, at least. It can be a bit worse, but it cannot feel an entire region of the plane. So, it is the image somehow of a regular function of the interval and the values at the end are the same. So, it can intersect itself several times, but it cannot be too pathological. So, the reason unbounded path kinetic components and what I want to prove you is that, well, this index is the same for any point in the same path kinetic component. Let us consider two points a and b and the same connected component so that we can join the one to the other with the path. So, in particular, we can join like this with polygonal path and we restrict our case, we prove the invariance of the index b joined by a segment. So, then we can repeat on each segment of the polygonal. Now, a segment, of course, which does not intersect in the same path connected region of the complement of gamma, so it cannot intersect. So, we can join two points with a curve, with a path, which is not intersecting gamma in general and in particular we are restricting our consideration to just to a segment and then we prove it from this, you can conclude the same for a polygonal path and that is it. Now, I take z minus a over z minus b. Well, this is a and this is b and this is the segment and if z is not here, the segment call it t. If z is not in t, this number here is not a real number, it is not multiple. It is not only a real number but it cannot be a non-negative real number so that what do we have? We have that when z is not on the interval connected t and a, connecting a and b, sorry, we can consider the principal branch of the complex logarithm of this. Well, this is a, this is b, this is z here, this is z minus a, right? This is z minus b, it is not a real number, it is not proportional, we want as a vector unless z is here or here or here, so on the line connecting a and b. So, this number is not so not even zero but it is a negative. So, the log of z minus a over z minus b is well defined. What I mean by well defined? There is no ambiguity for the choice of the arg of z minus a over z minus b, right? Because this number, this ratio here is not surpassing the negative real axis. So, it is if you want the image of this function here is contained and c with a slit from zero to minus infinity. So, the logarithm is well defined, there is no problem of making a tour around the zero, right? Okay. Now, remember that we want to prove that the integral of, so the index, the integral over gamma over, yes, over gamma is the same as the integral with respect to b. Now, this is the integral of gamma one over z minus a d z from the definition and this is the integral, this has to be proven, right? What we want. On the right hand side, we have this. On the other hand, we also consider this function here is well defined and holomorphic when we stay away from the segment connecting a and b. Now, it is holomorphic in its derivative is one over z minus a minus one over z minus b. Okay. So, we have all the ingredients to conclude because on the right hand side, we have just the difference of the integrand we are considering here. On the left hand side, we have, when taking the integral, the derivative, sorry, the integral over a function which has a complex analytic also a holomorphic primitive. The primitive is written here. So, the integral over gamma of the function log of z minus a z minus b d z is clearly zero because we have a holomorphic potential of this function. It is an elliptic. On the other hand, this is the integral over gamma one over z minus a minus integral over gamma one over z minus b. And of course, I forgot to put one over 2 pi i in front but we want to prove just that they are the same and this is zero. So, this is now just to conclude if a is an unbounded path connected component of the complement of gamma. It means that well, we can take a as any point very far from gamma. So, that gamma can be surrounded by a disk where the function I consider one over z minus a is in fact holomorphic because the index is invariant for points which are in the same path connected components. So, I can take a to be very far. So, take a prime very far from gamma. Take the integral, gamma this is zero because of the previous considerations. So, with all this new properties and information together, let me point out that we have all the details for the important tool of Cauchy integral representation formula. Now, consider function f holomorphic d the disks and consider capital F of z to be f of z minus f of a over z minus a. F is holomorphic in the disk but not in a. So, if you want we are in a situation where we have a function holomorphic with the one singularity but luckily enough the limit as z tends to a of f of z times z minus a is the limit as z tends to a of f of z minus f of a which is zero because f is in fact continuous in a f is holomorphic and then continuous in a not capital F the other one the starting one. So, in other words we have the hypothesis to apply the version of the theorem where we have the good sir weakened version right. So, the integral of a gamma of f of z capital f of z the z is zero. So, gamma closed curve which doesn't pass through a and well you can also rewrite this as the integral of a gamma of f z minus f a z minus a d z right. And then we apply additive property of the integral remember that the integral of a curve over a segment before and over a curve was obtained by repeating the consideration of the integral of real integrals the real imaginary part of the function. So, all additive properties of linearity were satisfied and then I have here sorry minus f of a z minus a over d z. So, since this is zero and zero is the integral of a gamma f z z minus a d z minus integral of f a z minus a d z which was in fact the integral of gamma of capital f of z d z okay. But f of a is a constant with respect to z so that I can also write that f of a times integral of one over z minus a over gamma d z is the integral gamma f of z z minus a d z. And this is exactly what we call n gamma a times 2 pi i. So, that f of a is one over 2 pi i yes n gamma a and this is a Cauchy formula. Well this is one version of the Cauchy integral formula which applies of course for the case of the punctured disk because in this specific case we have the weakened version of Pulsat theorem. It applies also for more generic subsets take for example the ellipse or something else. But with the tools we have we cannot as I said we cannot give the very general statement. So, in fact I will not give it in well I just give it or not prove it all right if you don't mind. In any case this is the local version say you have a disk and this is enough. In particular this formula becomes also very interesting when we consider as a gamma the circle and the circle which just surrounds the points a once and with the positive orientation because this number becomes one. So, they could in some sense this gives you an integral mean value property for holomorque function. So, the value at the center of the disk the value the center for holomorque function is obtained as an average an integral average the puncture with this weight z minus a the denominator along the circle and one over 2 pi i again is a weight. So, if you want if you know the value of a holomorphic function in a circle then you can obtain a value in the center. And then do we have any analog property in the real case the reverse? Do you have one example as a general property like this the opposite what do you mean? In frame. Okay in the real case you have to restrict to the values on the ending points right? If you have information on the ending points of the integral well you cannot reconstruct the function inside. Here it is the way and well this is a peculiar property in fact from this property we obtain one fact which is not true in the real case that is to say that any holomorphic function so any function which satisfies this. So, Cauchy-Riemann equations or complex derivation and what we have proved to be equivalent is enough to show that the function is in fact complex analytic all right? So, we have several examples of function which have derivative real derivation valid but not they are not real analytic which by the way guarantees that the function is c infinity. So, I don't think that any c1 function is also c2 cn in the real case actually it's not. So, this is another another word and this mostly depends on this property but as I said well it is important to mention the general proposition which is the following the following number 17. So, I have to recall you what do we mean by two homotopic curves do you know this do you or have you already sent the definition of homotopy the notion of homotopy in topology and algebraic topology okay I will recall it to you. So, take two curves gamma and gamma prime curves with same and points. So, in the plane this is gamma and this is gamma prime right this is the graph associated to gamma and gamma prime these are two okay now we make some okay we of course assume that the function are continuous and we carry scale at the interval of the definition of the two functions such a way that just for the sake of simplicity that the interval is 0 1 okay so the gamma 0 is the starting point and this is if you want to avoid some ambiguity put gamma 1 okay instead of gamma prime because prime means derivative okay you are correct gamma 1 two curves sorry gamma 1 and gamma pardon me I'm sorry yes sure the ending point is gamma 1 so that the function gamma and gamma 1 are defined in the standard interval 0 1 into c and well in general to into a topological space okay with some pros and with this case we are considering curves in the plane in the complex plane and the function h which turns out to be continuous function from i times i into the same plane in this case but topological space in general is said to be a nomod this is continuous such that so imagine this to be the same parameter for the function and the other a deformation of parameters imagine that this is a family of function which continuously the form the first curve into the other this is the sense of homotopy so such that h t of 0 is gamma t for any t for any t h t of 1 is gamma 1 of t for any t in a all right and what we are also requiring is that when s is not 0 or 1 is in between okay so when it is in fact another curve h 0 s is h 1 of s is sorry h sorry h 0 s is gamma 0 which is also gamma 1 of 0 so the same starting point for all the curves in this family of the formation of curves and the same ending point if this is the case we say the two functions are homotopic and we also indicate this by considering this relation which is by the way an equivalence relation you can prove that any curve is in fact homotopic to itself that if gamma is homotopic to gamma 1 gamma 1 is homotopic to gamma to gamma and so on okay and the the transitive property is so about in particular you can also consider the case well this is the generic okay well not generic because here is just see but well you can consider topological space and this provides you information okay so that you can deform curves but in particular we are interested in closed curves which are also called an algebraic topology loops so two loops are equivalent if there is an homotopy function with the same properties but as I was writing here in fact thinking about the loops and not think about the general case well the starting point and the end ending point is the same now the problem in algebraic topology is to what the problem the the interesting interesting tool connected to homotopy is the fact that you can use algebraic property associated to the class of loops to define to to make a topological comparison of sets for just to give you an example it's quite obvious that if you consider the set this set here well and you consider a loop and a loop inside this set here you have the freedom of deforming it to this point here whatever loop you consider even very complicated you can deform it but assume that you are taking a domain like this so with a hole in the middle it's not true that starting from any loop you can deform remember that well I didn't point out here well the homotopy has to be a function which is continuous in the square i times i but its values has to be contained in the set we are dealing with so that the deformation is allowed only inside here okay in this annulus essentially it's like an annulus from the topology point of view it's like an annulus so that for instance this loop can be evidently shrink into a point but this cannot remaining in the domain so there is an obstruction topological obstruction well this will be probably shown in the course of topology of algebraic topology I don't know and is well related to the fundamental group to homology and so on which is another topic but for the case we're interested in it suffices to know that the index is the index i so we define is homotopic invariant okay and that's that's a result which I don't want to prove you so fact assume that gamma and gamma one are two closed curves which are homotopic then the integral of a gamma of f of z d z is the integral of a gamma one of f of z d z say in domain d and f holomorphic in d and well turns out that actually if these are closed curves this number is here it's zero so any other deformation if you something which does not provide any special contribution for the integral in particular as we know then consider a simply connected domain which is a domain with the proper you know what I mean what we mean when I say when we say simply connected what do we mean so any loop is in fact homotopic equivalent to a constant so in the previous sketchy examples I gave you well a domain in which this property apparently works fine and another one the the domain homomorphic to the annulus where it doesn't apply good so but luckily enough in the case of playing domains this property can be simply characterized and it be kind of the rest in the following way well simply domain d in c this is equivalent to say that c minus d is connected luckily enough so that which any closed curve is homotopic to a constant but this is just in c this characterization is true not in for instance in an r3 remove a one point to sphere well this is simply connected to okay take a sphere minus of one point the complement is of course not connected but the sphere minus one point in r3 is in fact simply connected right okay so the last thing I wanted to show you today since time is running right as the following okay as a consequence of this consequences so if f is homomorphic in a simply connected domain then as I said this is zero or there exists a holomorphic function capital F such that f prime of z is f of z potential primitive this is because whenever I have this property for closed curve gamma closed curve curve in d I can define the potential of primitive which turns out to be analytic as an example of application let me consider this one I have no is the following f holomorphic and such that f of z as differ from zero for any z so f holomorphic you say in d I'm not assuming that d is simply connected but I'm assuming that the function f is well I'm assuming actually that d is simply connected but also I'm assuming that f of z is different from that what I'm not assuming is f of d is simply connected okay so since f is not zero for any z in d I can consider the functions g of z to be f prime of z over f of z and g is well defined and holomorphic remember that f holomorphic guarantees that this function is defined and turns out to be holomorphic at the f of z is also different from zero so I can consider the ratio of two holomorphic functions and they are this ratio is still holomorphic I can consider the derivative and the derivative is well defined for any z in d now since d is simply connected then this implies that there exists a holomorphic function g such that g prime of z is equal to g of z so there exists a holomorphic primitive and I define h of z to be the exponential of g of z this is also well defined and holomorphic now I consider finally the ratio f of z over h of z now remember g of z is whatever you want can be also zero but x of g of z is not zero again so that this ratio here is holomorphic and d let us calculate the derivative okay h of z oh sorry this one can call it p of z okay has a derivative the derivative of f times h h prime z f of z over h of z square now h of z is x of g of z and g prime of z is g of z which is remember f prime of z over f of z when I substitute here so h prime of z is in fact x of g of z times g prime of z so it is h of z right times g prime of z this so p prime of z is f prime of z over h of z because I cancel one h okay from this summand here minus and h prime is h f prime over f time f so f prime h h cancel h so it is f prime of z over h of z h is zero and this is true for any z in other words the function p of z is constant because we have already shown that holomorphic function whose derivative is constantly zero on an open well I didn't say this but when I'm when we are talking about the simply connectedness we are talking about connected open sets with this property because otherwise we have to talk about each component each connected component should be simply connected right um so that let me conclude so p of z as constant in other words h of z is times c is f of z and this is c times x of g of z since c are constants from the properties of the exponential this can be also written as x of g of z plus c prime okay called this function m of z which depends on c prime so you have freedom of choice c prime but then what you have here is that m of z is the logarithm of f and if you want assume that f of z naught which is different from zero can be written as x of w naught so that we can take c prime in such a way that well well there are several but when you fix the value to one point so if you if you fix that at for z equal to z naught m of z is w naught then there is only one logarithm so you have the choice there's the opportunity to choose whatever you won't see here see one c prime here but when you fix the value of the function then the function is also and this sense it means that whenever you have a function holomorphic and not zero you can always define its logarithm in a simply connected domain which explains why we have we're in troubles in the plane because as i said well the exponential complex or real exponential is never zero so we have to remove zero from the plane okay on the other hand when i remove zero the complement is not simply connected in the extended in the extended Riemann sphere two points the common sorry the common is not connected sorry the common is not connected so this setting fact is not simply connected so if we're restrict and we want to make it simply connected restrict well we have to take a slit on half line connect zero to infinity or another curve and then we are done well there is no way to define a logarithm the function f of z to be considered in this case is simply z which is not zero only for z different from zero right whether you move zero and then from topological properties of the set you also to remove an upline starting from zero and going to infinity okay that's why in any also from from topology there are some obstruction to extending the complex logarithm to the entire infinite annulus or if you want to the puncture plane all right this property will be used several times so remember the please simply connectedness if you don't know anything about how motopi well it's nothing very difficult in a plane because you simply consider the complement of the set improve it to be or not connected unfortunately simply connected simply connected can be also meant as simply pause connected but it's not a domain which is connected and nothing else so simply means simply connected it's a unique unique definition which is bad translate bad calc from from the french because this was invented about punk array okay all right so in case you are interested in how motopi and the staff please study it and which is an interesting images which is an interesting subject which interplays topology and algebra but this will not be the main the main so we will not use very much of except for this definition all right which in particular for the plane case is somehow trivial because well you have this very simple characterization of all simply connected domain however simply connected domains turn out to be incompletely very important and as we will see at the end of the course or maybe just measure if we change our there are essentially three fundamental simply connected domains which are studied more and which represent a good to give you all the interesting example since any simply connected domain can be put in in relation with these three models which are the disk you know this the plane and the Riemann sphere they are all simply connected they are not equivalent in the sense of holomorphic functions but they represent the the unique examples interesting example which you can deal with for the very generic study of all the domains in the plane okay thank you i stop here