 Okay, last time we were dealing with some application of the residue theorems theorem. And now I want to enter in the details I left for this case, right? For the integration of a rational function over the reals times x alpha minus 1, where alpha is smaller than 1, alpha not real, okay? This is the interesting case. In general, you can take alpha to be real but not an integer, but then you reduce to a polynomial times something which is like this, okay? So the assumption we made is that this tends to 0 as z tends to 0 and z tends to infinity. And if you remember, I just made a slit in the plane because 0 is a branch point. Remember that in order to define this power, we have to introduce the logarithm and the logarithm is not defined in 0 but it means it is not defined in the neighborhood of 0. And we considered as a region G, the region which has contours as follows. We have two circles centered at the origin. One of radius R, I say CR1, CR2. I don't want to use capital R for the ray because capital R is also the rational function, okay? And then I also consider these two segments parallel to the real axis and I use this orientation, right? So what I am trying to do is to apply the regular theorem for this case, for this class of functions, all right? And we have this choice of contour. So the region G is this. So I will consider then the residues inside the region G and the boundary of the region G is CR1 where CR1 is this part of the circles of radius R1 plus L2 plus CR1 minus L1, okay? Plus L1, sorry, the orientation is this, okay? So this is the contour. And if I show that the contribution on the circles is vanishing as R1 times to 0 and R2 times to infinity and if I show that these two contributions on the segments which will become the offline zero infinity are in fact the ones you are considering, you can calculate this integral using the residue theorem, right? So let us start by considering this situation. Let us take mod of Z equal to rho and consider this function here which extends to the plane this which is defined only on the real axis. So Z is some rho times EI theta and theta bar is, okay, in general. What I can do is the following is substituting, substituting the Z with this, I have the DZ is I rho EI theta D theta and therefore the integral, this is plus, no, this rho is general. What I want to show you is that, okay, G is a region, G CR1 plus L1. Okay, okay, okay, thank you. So your question is about this symbol here, right? So if I say union, I mean a union of sets. When I say sum, I already use this symbol, sum for curves. It means that I consider the function without just the position of curves, okay? And this plus or minus means that I preserve or invert the orientation. So as a set, the boundary of G is in fact this set union the other sets. But as a function I use this notation because I can just consider one curve. So reparameterize the segment from 0 to 1, for instance, and stop at T1, then consider the other which begins from here where the other ends, continues up to here, then and so on and so forth. So this plus is in general adopted as a symbol when you consider in say in topology, okay? The curve which is obtained by just the positions of curves, okay? And the symbol plus, you can also take actually a combination of curves putting some coefficients, normally integers, which means that you consider, for instance, the same curve count several as many times where the orientation over the reverse orientation, okay? According to that, the sign is positive. So if I understood the question now, I adopt this notation, which is the one where we also use some, in some previous example probably. Well, I used this last time when I consider the Laurent series, remember? Okay, let us come back. We consider neighborhood of a singularity. Yes, and we consider two circles, n-side the annulus. Remember this? We consider this and this. These are two circles. We consider a segment connecting the two circles. And then I consider as a curve C1 plus lambda minus C2 minus lambda, right? I use this notation here. Check it. Check on their nodes. What I meant was the following. You say, move this way, go this way, then move this way and come back. And this becomes a closed curve. So the symbol I used here is exactly like the ones I'm using here, okay? So what I'm doing in this calculation is to show that in order to calculate... So when I calculate the integral over the boundary of G, the contribution along this portion of the circle is the zero. When the radius tends to zero and the small value tends to zero and the other tends to infinity. Okay? And so let us continue here. With this substitution, this becomes the integral between zero and two pi of what? Of I times rho EI theta square, rho EI theta. And then I have something, rho EI theta, right? Not square, sorry, alpha minus one, sorry, alpha minus one, the theta. Which means that... Notice that this is rho EI theta and this is rho EI theta alpha minus one, correct? So I have that the integral is in fact the integral between zero and two pi of rho EI theta to the power alpha r rho EI theta d theta. And this from our assumption tends to zero as z tends to zero or infinity. Which means as rho tends to zero or infinity. And this applies either for r1 or for r2, okay? Good. Now, so the integral over the boundary of G is in fact, is the integral over CR1 over the same amount. But these two integrals are infinitesimal as rho tends to infinity or as rho tends to zero. Okay? And then I have the two integrals over L1 and L2. Well, L1 and L2 are the two segments, different opposite orientation parallel to the real axis at a certain distance, positive distance from the real axis. So they are symmetric but with inverse orientation. Okay? So we are moving like this, then like this, like this, and then like this, okay? Now what is the problem now? It's how to write z alpha minus one in the two segments, the problem. So now we have to be careful about the notation. Remember that z alpha minus one is a definition, the exponential of log z. And this is the complex logarithm times alpha minus one. So that this becomes, right, x of alpha minus one log, this is the real logarithm. And here, when I write this way, I mean that I've made a choice of the argument, all right? And if this is true in L1, using the same notation in L2, we have that z alpha minus one becomes x of alpha minus one logarithm. And this is the same, you see, because the distance from the origin is the same, points on L1 and on L2. So it is independent of the fact that they are symmetric. But theta becomes 2 pi minus theta, right, on L2. So L1 is here, this is the real axis, this is L2, this is L1, and this is L2. So if this is the angle theta, sorry, theta, this is the angle 2 pi minus theta. In order to use the same principle argument in the definition of the functional log, okay? So when I have to calculate, so I just put, therefore, when I calculate the integral, the integrals over the two segments, I have, well, in fact, this is the integral, all right, the z as x plus i delta. Delta is the distance from the previous picture, I could put this delta, okay? This is the distance from the origin. So z is x plus or minus i delta, that is constant, right? And x varies, so we are integrating over the reals, essentially. And the two extremal values of the integration, for the integration, is R1 and R2, precisely the center, the radius of the small disc, small circle and of the big circle, right? Remember that these two L1, L2, where, this is a zoom of the previous picture, right? Where in G, the two segments cutting parallel lines, the annulus, okay, center at the origin of radius R1 and R2. So x is moving between R1 and R2 here and here. Therefore, the integrals reduces to the integral over the reals sometimes, because delta is fixed. And I have, so using this notation, I have that this can be written as z alpha minus 1 Rz dx, say, okay? And plus, this is the integral of what? Well, remember that, as I said, well, probably I said, but I didn't write it explicitly. So z alpha minus 1, I write it here again. z alpha minus 1 is xp of alpha minus 1 times log mod z plus i theta in L1. And this is xp of alpha minus 1 log mod z plus i to pi minus theta in L2. Therefore, we have that this is, okay, like the other one with a minus here, but then I have a coefficient 2 pi i alpha minus 1, which come out. So we have the same amount with a minus in front, but I have also, I have xp of 2 pi i times, sorry, yes, times xp. You see this? 2 pi minus alpha. And I have to multiply this entire value of the complex logarithm times alpha minus 1, okay? So there is an extra term, 2 pi i alpha minus 1 as part of the exponent. So here I have, then the integral, this is the integral over, this is L1, say, this is L2. And here I have 1 minus e 2 pi i alpha minus 1 of the same, z alpha minus 1 r z dx. And I inverted R2 to R1 because I'm integrating over L2, probably vice versa, right? This is R2, R1, vice versa, I don't remember, sorry, the orientation, okay? But I have to revert because in one case I'm moving from the right to the left and the other from the left to the right. So when R1 tends to 0, so probably this says yes, okay? When R1 tends to 0, no, what I had here is something different. Sorry, sorry, sorry. This is R2 integral e 2 pi i. This is what I finally have, the integral in between R1 and R2, sorry. 2 pi i alpha z alpha minus 1 r z dx, where this is the integral. Now, changing the sign and putting together, I have this coefficient coming out which does not depend on x and on z in general, this is constant. In other words, when I then consider the case that R1 tends to 0 and R2 tends to infinity, on the right-hand side I have what I'm looking for times a constant. So the integral between 0 and plus infinity of this integrand function times this number, which is... and on the left-hand side, I have the sum of the residual sum 2 pi i included in the plane with the slit from 0 to infinity because the contribution on the portions of circles are vanishing when R1 tends to 0 and R2 tends to infinity. So to conclude, this is an application of the theorem. I have the integral 0 plus infinity times 1 minus c 2 pi i alpha is 2 pi i sum of the residue of the function f j alpha j in G. And G is the general situation. G is in fact... So this is the application. So not simply the sum of the residues, but to be divided by this constant, which comes out from the calculation. So the two contributions do not cancel each other, but give you something which appears here. So let us probably see in one of the examples I wanted to consider. This is the one we consider, right? 1 plus x. So in this case, R of z is 1 over 1 plus z. And so we have that z to the alpha times R of z and alpha is in between 0 and 1, right? So this is in fact infinitesimal as z tends to 0 and as z tends to infinite. Because this is a rational function. The order is 1. The denominator is degree 1. The numerator is degree 0. When you multiply this rational function times z to the power alpha, alpha is positive. So you have a numerator which is infinitesimal as z tends to 0. The denominator tends to 1, so no problem. But then when z tends to infinity, the degree of z is 1, which is greater than the degree of the numerator, which is alpha, so that it goes faster to infinity. So the denominator goes faster to infinity and so the ratio tends to 0. So we have the hypothesis 5 to apply the theorem. And there is only one residue for R of z. And 1 pole is minus 1 for R. So it's not only on the positive relaxing, which is good because we're integrating between 0 and 2. And so it's quite easy. This sum reduces to 1 summand. And what is the residue of the function f at minus 1? The residue is minus 1 to the power alpha minus 1, y. Well, the function f of z we are considering is z to the power alpha minus 1 over 1 plus z, which is a pole in minus 1, sorry. So that if I consider the limit as it tends to minus 1 of 1 plus z times f of z, in this case I have to write 1 plus z. And then this is what? Minus 1 to alpha minus 1, the power alpha minus 1, right? So that this number here is to be calculated. Is it normally a real number? No, it is not. We are calculating some root of a negative number. So in general this can be odd in the reals. But in the complex number we can do this because we can write minus 1 as the logarithm, sorry, that's the exponential of the logarithm of minus 1, right? And this is the exponential of logarithm of modulus of 1, which is 0, right? And what is the argument of minus 1 pi, right? Plus i pi, this is the principle argument. So minus 1 is in fact at e i pi. So this number here is e i pi alpha minus 1. Or this is e i pi alpha times e i pi with a minus in front. What is this number? Minus 1 again. So this is minus e i pi alpha. So this is the residue. So if we apply the previous general consideration to this particular case, we can say that the integral we are dealing with is 2 pi i times the residue of f at minus 1 over 1 minus e 2 pi i alpha, correct? But in this case, this is 2 pi i. The residue is minus e pi alpha and then I have 1 minus e 2 pi i alpha. Then I notice that this expression can be somehow simplified. And the following way I write e pi i alpha to be u. Therefore the integral which was 2 pi i minus u over 1 minus u squared, right? With this substitution, now becomes minus 2 pi i u over u u minus 1 minus u. This is never 0, right? For any choice of alpha, e 2 pi alpha is never 0. And here I recognize up to the sign. And so I change again. I have 2 i over u minus u minus 1 times pi. And this is the reciprocal of what? The sign of alpha. Because remember that u is this. This is 2 i, so I write this way. E pi alpha minus pi alpha pi. So I have that this is pi over sin pi alpha. Complex sign. And this is the final expression in our calculation. So which can appear to be difficult but it's much easier than to try to use standard integration properties from the real. This is an improper integral which is normally not very affordable. As I promised, I don't want to make you become such, to annoy you with integrals. And let me continue by saying something else. That is to say, let me add some comments on closed curve. So again, I will use the notation and I explain once again why I use plus minus and so on instead of using. So in fact, you can consider the curves and to put some algebra structures on curves on the set of curves which is quite important to make chains of curves. Then you apply for instance homology and homotopy techniques using this notation. I will not go into the details. However, let me say that this is quite standard to think of. To put a coefficient in front means to consider this curve re-parameterized and counted with positive or negative orientation according to the sign of the coefficient you put in front. The sign is always an integer. The coefficient is always an integer. So let me just give you a geometric application of what we have known about the complex line integration. So you probably all know but I guess none of you have seen a complete proof that closed-injective curve, continuous-injective curve, is a Jordan curve. So no self-intersection are allowed and the curve is closed and continuous. That is essential, continuous and not self-intersecting. Then in the complement as two components, one is bounded and the other is unbounded. This is the famous Jordan theorem which states for the circle it is easy to see. So any deformation of the circle in some sense, this topological stuff, any deformation of the circle but without overlapping. Imagine that you have an elastic band and you move it. So you can make different shapes but in a sense from a topological point of view you are always dealing with the circle. Until you overlap the bands then this changes a lot in topology. You are dealing with the circle. So it is quite natural. It is obvious in the circle I have something which is bounded and the other component is so not in the circle. In the complement of a circle you have something bounded. This is not very easy to prove in general. This is the famous Jordan theorem about Jordan curves. However what we can do now, we have enough instruments to say that in the complement of the Jordan curve we have at least two components which is a weak reformulation of the theorem. In order to prove this we apply the fact that we know something about the index of a point. Let me just remind you the basic notation for Jordan curves. Jordan curve is a continuous closed curve. In fact you can define also a Jordan path and then consider a closed path to be Jordan if it is a closed curve which is also Jordan. Jordan means which is injective. To be more precise I have the curve gamma to be defined in A, B into C, the plane curve. Then I have the gamma of A is gamma of B so it is not injective in A, B in the open A, B. Gamma of T is different from gamma of S for any T different from S and T and S are in A, B. It is not injective in A, B because gamma of A is gamma of B because it has to be closed. No other points are allowed to have the same value in the segment which is quite natural to consider. You can have different pictures. This is one example. This is another example. Now assume that we start from a Jordan curve closed Jordan curve not passing the origin. We also assume that there exists Z1 and Z2 gamma such that this is our gamma. We call it this is Z2 and this is the gamma such that I have that the two points Z1 and Z2 divide the curve gamma into two arcs called gamma 1 and gamma 2. For this properties gamma 1 is not intersecting the negative real axis and gamma 2 is not intersecting the positive real axis such that call this part of gamma from Z1 to Z2. According to orientation which is implicit with the definition of the Jordan curve gamma. Consider this path gamma 1 and the other gamma 2. Assume that gamma 1 has no intersection with the negative real axis and gamma 2 has no intersection with the positive real axis. This can be always done because since 0 is not contained you can always move this curve or if you want the frame in such a way you put the origin in the middle and separating the two arcs the two paths gamma 1 and gamma 2. So, just a matter of how to then if this is the case and then I have the index of the origin with respect to gamma is 1 and I show you how. And first I indicate by L1 this segment so L1 as the this is the statement okay this is kind of proof. So, call L1 that segment from Z1 to the origin of course it is oriented I choose this direction and L2 is the segment from Z2 to the origin right. Then the origin stays away from the curve the curve is continuous so that I can consider a small circle center such that doesn't intersect the curve gamma okay. Now but it intersects for obvious reasons this segment segment L1 in C1 and the segment L2 in C2 so Cj is Cj that's obvious because the segment Lj connects point outside okay but in any case it connects the origin with something else which stays away okay. Now I restrict my consideration on this is the C but then I put well C1 and C2 the two portions of circles which are with endpoints C1 and C2 so respectively this is C1 and this is C2 okay I use this notation and the part of the segment in L2 from Z2 to C2 I call it delta 2 and this from Z1 to C1 delta 1 alright just names okay C is C1 plus C2 okay so if you want to see C to be a curve then I prefer this notation if you want to see it as a subset of C I use the other okay good. So then I define so keeping in mind this picture I cannot write everything on one slide but keeping in mind this picture so define sigma 1 and sigma 2 the two paths are the following so I show you first what I'm doing I'm just considering this and this okay so if you follow my if you follow the orientation we have given to the curves so this is a positive orientation okay then if I use this orientation I have gamma 1 then delta 2 then minus C1 minus delta 1 to define sigma 1 for instance okay. So gamma 1 plus delta 2 minus C1 minus delta 1 and similarly I define sigma 2 as well I have gamma 2 then delta 1 minus C2 and minus delta 1 okay and I have you know I said gamma 2 plus delta 1 minus C2 plus delta 2 so and this probably also gives you an explanation why we prefer this notation instead of union okay because when I have well I notice that sigma 1 plus sigma 2 is what yes gamma 1 you can it's gamma 1 plus gamma 2 which is gamma sorry minus C1 minus C2 this minus C probably I put it wrong it is minus delta 2 right but let us check okay gamma 2 delta 1 minus C2 minus delta 1 minus delta 2 sorry okay so the contribution of the integral of this sorry of the of the segments cancel each other okay. Now I have that this number here is what and gamma 0 minus n gamma C but I can also say I can also say that n gamma 0 is n C1 0 plus n C2 0 plus n C 0 now you understand why we prefer this sum and minus in front of the curves okay because well this is a close this is the sum of two close curves sigma 1 is a close curve sigma 2 is a close curve right and we calculate this remember that the index was linear using this this definition of sums of curves right close curves so on the left hand side I have this and then this and this on the right hand side which means that if I calculate the integral of gamma so what I said is wrong here sigma 1 plus sigma 2 right this is what I have on the other hand this means precisely this because of the additive property of the integral of a curve if you have a sum of two curves and you have the integrals summed so it's essentially okay the two curves are the contours you are in where you are integrating and they are two split curves right but what can we say about this value here this is 0 and this is 0 0 no sorry this is 1 and this is 0 as well why can we say so well first sigma 1 has 0 by construction and the unbounded it stays away okay from it is 0 in the unbounded component and similar for sigma 2 the index is 0 because it is in unbounded complement in unbounded component of the complement right in unbounded component of the complement sorry whereas for the circle the index is 1 good so with this in mind let us go to the second part which is the interesting part and this time we consider this is Jordan lemma but say weak version which stays there for the complement closed Jordan curve has and then this is the weak version at least two connected components so we start from a closed Jordan curve with some hypothesis about the frame we are referring it so such that gamma as a set is contained in this half plane I write it this way because probably use it some other time later h plus is the set of complex number whose real part is positive so the right half plane and when I use this with plus on the top of h I mean the set of C and C such that in C is positive this is another half plane just the rotated half plane okay so I also assume and this is not this is without losing any generality for the curve is there so which means that in case I move the frame on the left right and I can also shift vertically the frame in such a way that there are two points on the curve one in the upper half plane h plus and the other in the lower so and such that there exists Z1 and gamma and C2 in gamma and M Z1 negative M Z2 positive okay so the situation is the following so for instance if I had a curve like the other one had to move the frame this way okay and this is always possible so you simply shift everything in order to have a picture with at the end of this okay I'm assuming that entire okay let me let me make you a picture so these are the assumption let me make a picture here so we are essentially considering a curve which is close and in this right half plane h plus with plus whose plus means real part positive but that the curve could be also in the this first quadrant for instance so what I'm assuming is that well now this is not the case I'm interested in and I want the curve like this so there are points here this is the one and this is the two for instance okay this can be done because as I said you can shift the frame as you want okay good now well let us start from this okay so here is the origin so the origin is not is in in the unbounded component of the complement of okay what I can do is well I can join these two points sorry this is the one is here right and Z2 is here Z1 is here and what I have is that it is two segments eventually intersect the curves into other points here in two points and here also in two points okay so I can over the sticks my consideration to what to two points Z1 and Z2 in the respectively in the lower half plane and in the upper half plane we call lower plane the the half plane of the gas plane whose elements have imaginary part negative so lower upper means imaginary part positive then they say right half plane the plane where the curve is that is the plane the half plane where whose elements have real part positive and then the left half plane okay so up low right and left this is just by convention okay but what I'm saying is that you can always assume that in fact the Z1 and Z2 can be chosen in this curve in such a way that there are these two points the Z1 and Z2 okay and this picture what I mean is that the entire segment 0L and Zj connecting 0 to Zj does not intersect any point of gamma okay right so we are say in a situation like this so we can assume I put this this way we can assume that the segments 0Zj j are such that Lj and gamma has no intersection okay it's not realistic at all now so let us go to an enlarged picture like they have here this this and then I put maybe this is another example so I put Z1 here and Z2 here okay just to be to have a larger picture because I have to add some extra extra rays and so on okay I could continue on the other but it would be too difficult to then to see the differences okay so this is what I denoted by L2 and this is what I denoted by L1 and I consider the segment sorry the segment the portion of the curve from Z1 to Z2 to be called let me say gamma 2 and from Z2 to Z1 gamma 1 these two points are distinct and they are the endpoints of two paths Jordan path gamma 2 and gamma 1 assume that we have also this orientation and that we choose this orientation for the segments as I said connecting the origin to the four point Zj so this way so gamma is gamma 1 plus gamma 2 once again what I mean is move and continue okay this plus means the fall and I define this time sigma 1 to be what I have L1 this this okay L1 and then minus gamma 1 minus how 2 and I have sigma 2 to be L1 plus gamma 2 minus these are two closed alright so sigma 2 minus sigma 1 is gamma in this case but in general it can be plus or minus if I have the difference of that is these two closed curve have plus or minus gamma okay so I have in this case with this notation of this other possibility because I've chosen a in any case the difference of these two closed curves is plus or minus gamma good so now let me let me make some observations which depends deeply on the fact that we are dealing with continuous functions but these two points have imaginary part as you want to which are one is positive and the other is negative the imaginary part of a continuous function if you think that this is function from an interval a to c sorry this is gamma right this is a continuous function so it can be viewed as a pair of functions which are continuous and in the complex planes that the pair correspond respectively that the first element is the real part and the second is the imaginary part right so if you say that gamma t is x t y t and this is also this x t plus i y t that is z of t since this is positive and this is negative and y is continuous then necessary it is a zero so that I can say that gamma 1 intersects the real axis and similarly gamma 2 intersects the real axis as you can see here gamma 2 in this this picture this is just an example okay intersects in more than one in fact this is the this is a consequence of the statement okay any continuous function in an interval this is a real real valid continuous function in an interval and assume that on the end points the values are such that their product is negative okay so one is positive and other is negative right this is only possible then there is at least one zero this is the existence of a zero but nothing is said about where it is and how many there are right in fact in this case there are more than one here there is only one okay for this path we have one zero but for here for this path we have for this function we have more than one actually to apply the zero theorem existence of a zero for a continuous you have to see it as the projection okay just to see it on the imaginary axis however what I'm saying is that well define x2 to be an element to be d element in gamma 2 intersected there with with maximal distance from the origin so I choose this okay what can I say now with this picture in mind I continue and I say well what about the index of x2 with respect to sigma 1 let's go this way x2 is defined as the element in gamma 2 such that it is a real number so the imaginary part is 0 maximal distance from from the origin sigma 1 is this closed curve so I can definitely say since the curve gamma is Jordan so it is injective that you cannot have self-intersection x2 is in the unbounded component of the complement of sigma 1 it cannot be in here because otherwise the curve would have self-intersections so I can actually say that this number is 0 remember that we have proved that for any point in the unbounded component in the unbounded component of the complement of the curve the index is 0 correct so what I say now is the following this is 13 right in particular this implies that n sigma 1 of z is 0 for any z in gamma 2 because remember that we have proved that this is true for any element in the in the component so it's property of the component over the point the index is constant on the component right and since gamma 2 is continuous path any point stays in the same component cannot be in different components so alright now using the same picture and the previous lemma fx is here and so very close to the origin on the real axis positive relax but very close to the origin we have that the curve sigma 1 and sigma 2 are surrounding okay the region where x lies okay it can be considered as the 0 of the previous lemma so if x is very close to the origin the origin is one of the points of the curve sigma 1 or sigma 2 but x plays the role of point okay of the pointed origin in the previous lemma and therefore we can say that if x is a real number such that modules of x is smaller than epsilon for a suitable choice then this is true for lemma 1 now we are very close to the end so let us go back to this picture so consider as we did here x1 to be the first of the intersections of gamma 1 with the real axis the first in the sense that we have segment AB to consider so if x1 as parameter T1 associated to it so the other points which lie and gamma 1 intersect to R are such that their parameter associated is greater than T1 okay so it can be only later if you give the role of time to the parameter T that gamma 1 of T intersects the real axis so I consider this point x1 in this picture there is only one so there is no no extra choice but so in general take x1 to be let x1 be the first in terms of the parameter point of gamma 1 and intersect with R which is not empty remember this is not empty so there is at least one if there are many choose the one with parameter associated the smaller okay it's not necessarily the closest to the origin but is for instance in this case assume that there is this is the closest but it's not the first the first is here okay if you use this parameterization this is the first this is the second third but this is the closest not the first in the sensitive the first from the left the first according to the parameter chosen okay good now what can we say about this the index of this point x1 with respect to sigma 2 let's go back to this this is sigma 2 and this is the point x1 okay it plays the role of the originals before okay in the previous lemma and this number is one okay good but this means then n sigma 2 of x is one for any z and gamma 1 as we said before any points on gamma 1 lie in the same each point lies in the same in the same component of the complement right so that the index is the same now between x1 and x2 these two points are separated the curve is Jordan all right so there is at least one x not actually there is an interval okay this is the first this is the last but I can say more you can find minimal distance between the two points in gamma 1 and gamma 2 this is an interval this defines an interval on the real axis and any point in this interval at this property there exists an interval real axis such that call it I okay such as if x is in I so the interval is clear this one this interval is with one and point in gamma 1 and the other gamma 2 that such that if x is in I we have that the index of this x with respect to the curve gamma is one put it together the previous result so the origin is outside the index is zero and for at least one x different from zero the index is one so there are at least two components that's what we can say okay as you can see it's not easy to even to prove this minimal so not it's not very direct the very direct it's an application of this in terms okay now this ends the part of consideration about integrals and application of the integrals let me go for a why just an introduction to what we will be the next topic this is up to my choice of course so we have seen that there are several topological instructions to to have sets to be homeomorphic right the mug the torus the donut one of the exercise in topology similarly there is no hope to have homeomorphism which extend to something which comes holomorphic between the annulus and the disc okay assume that we are looking for a function F which is holomorphic in the annulus such and with inverse G define on a disc there is no hope for topological reasons okay because any homomorphic function in fact continues it would then we would have a continuous function with the inverse continuous from non-simply connected domain into a simply connected domain we also know that there is no possibility to have in general a homomorphic yes we cannot have a homomorphic function with inverse between any pair of simply connected remains and Louville theorems in fact is an obstacle to this take the plane which is simply connected take the disc which is simply connected domain of radius whatever you like any function which is holomorphic in the complex plane takes its values and the disc so it is bounded and entire then as it is constant so there is no possibility to have a one-to-one correspondence right so the next topic is about well about the definition and so the description of possible sets which are in fact by holomorph which means that there exists a function from one set to the other holomorphic with embers which is also holomorphic Louville theorem is somehow special result because we can prove that in some cases even unbounded simply connected domains and bound the simply connected domains can be by holomorphic and one classical example is the unit disc and the upper half plane or the right half plane which is just the rotation of the other so we will investigate in general what happens for biomorphic functions and sets which are biomorphic set which are biomorphic so we give this definition definition to sets omega 1 and omega 2 and c are set to be by holomorphic or sometimes you can find by holomorphically holomorphically equivalent if there exists f from omega 1 to omega 2 holomorphic with embers holomorphic okay so of course there are some as I said topological restriction for two sets to be holomorphic to be biomorphic and such a function f or g f is said to be by holomorphism of course also g okay in particular we may wonder whether there is some easy example we start from two sets this is the general definition but what if I restrict my consideration to one set and I say well I want to consider one set and see if there is a function from this set into itself which is one to one do we have examples of biomorphism that find on that time of that kind of course the identity is a good example the identity is well the exponential is not not any not any depends on the set so oh yes oh yes sure sure sure sure not I'm saying but I'm restricted to an open set for instance and I consider which if I wonder if I can find I'm able to find a function from this set into itself which is holomorphic with an embers I'm not saying that the entire plane okay well just to be on the safe side let us say that there is at least one example the identity which is always useful to remember okay to know that we have examples okay but then as your colleagues said of course if you take a linear transform a linear function which is a function like this this is an example of course a b r complex numbers and a is not zero right this is an example of biomorphism of course in this class there is also the identity when I a is equal to 1 and b is equal to 0 exponential is not even injective so I'm sorry this time no depends okay you can restrict and but in general it's not okay good so may ask you since you came you brought our attention to this linear function this is linear this linear it's not linear technically it's not linear it's an affine function anyway this is normally called linear function okay because it reminds the line okay in the real case anyway is this just an example of biomorphism of C or the exam so let me tell you that well as an notation if omega 1 and omega 2 are sets of C we can indicate by all omega 1 omega 2 the sets of functions from omega 1 and 2 omega 2 which are holomorphic this is a notation okay and and my holomorphic I don't know if this is the notation used f from omega 1 into omega 2 f by holomorphism office but I'm sure that if you restrict to the case which omega 2 and omega 1 coincide and you normally use automorphism as notation okay this is terminology stuff nothing more can use this notation automorphism so my question is now is the class of the automorph is the set of automorph is for the domain C completely describe of this so by these two part this this linear functions so just to leave you with some yes let me just point out that we can say what we are dealing with an entire function so no poles can exist right so take f holomorphic and C entire right half cannot have poles and C no other singularities okay not even poles right question what about the point of infinity so assume that f is injective and we are dealing with automorphism function which are invertible so they are for sure injective and take D any bounded open set in C for instance a circle sorry a disk but this is D and then consider the closure of the and its complement in C which is open because its complement is closed okay so what I'm doing here is the world this is D and this is C minus the closure of D which incidentally is unbounded what I can say is the following since f is injective f of D cannot intersect the image of the complement of the closure of D are you with me these two sets are separated and distinct this joint the image cannot have intersection which means that this set here is not dense because this is open because remember f is holomorphic D is open open set and f is open so f of D is an open set which means there is an interior okay can be whatever you like but it has an interior so there are points with some interior and this is the image of what of a neighborhood as I already observed is a neighborhood of infinity this is unbounded and it is any neighborhood of infinity because I chose I've chosen D to be any general open set but bounded which means that if you go very close to the origin or very far from the origin you take an open neighborhood of the not the origin sorry although the infinity its image according to f is not dense so it cannot be an essential singularity it can only be eventually a pole infinity but so just to okay I run out my time but just to leave you some ideas okay so it can be a pole most it cannot be it cannot be a removal of singularity because otherwise we would have a holomorphic function over a composite this would be constant so and it's not invertible so in any case so the only possibility is it's a pole but if it were a pole of order say m and m is greater than 1 then infinity would have m inverse images right condense at infinity and then the function wouldn't be 1 to 1 so the only possibility is that this pole at infinity is of order 1 which means that in fact if you write down the Laurent series okay it reduces to a polynomial okay but the polynomial degree at most one because it has to have it has to be injective otherwise if the if the degree is greater than one then necessarily there would be several images of zero so the example you gave us is precisely the set of all automorphism of the plane linear functions so I invite you to think about the fact that automorphism of a set first include the identity form a group with respect to compositional functions okay and we'll continue on Friday thank you for your attention