 Okay, so today we complete the description of the power expansions of meromorphic functions, okay, as we started last time. So remember that we have a meromorphic function and we considered a singularity a and an annulus. The notation was like this, if I remember over a, r and capital R, right? So this is to raise, I'm considering, so we have an annulus around this a and we also considered two circles, gamma 1, gamma 2, this is gamma 1 and this is gamma 2. Then we took also this lambda, this was the notation I used last time. Gamma 1 is a circle centered rotate over radius r1 and r1 is smaller than the radius of gamma 2, another circle centered at a and the radius r2. So in a explicit way I can write this and t varies. So we already observed that the two circles can be one, one circle can be scratched, this can be deformed into the other, right? And then we consider the curve gamma which was gamma 2 plus lambda minus gamma 1 minus lambda or vice versa, I don't remember the correct, but this was the close curve we were considering and this close was of course, homologous to 0 because gamma 2 and gamma 1 are homologous, so they have more topic as functions, okay? And then we, well then we established this first relation which comes from Cauchy integral formula and this is for 1 over 2 pi i integral over gamma and this is the gamma we were considering of f c, c minus a, c is what, f of, sorry f of z, sorry f of z and z is such that, right? This is in between, right? So the z is in between, it is contained not being contained, it is contained in this small annulus, okay? Then I have to put the index, did I use this notation or I don't remember? This is in any case, this is 1, okay? Why? Well because we are taking z here which represents the, well a point in this, in this annulus so that it is in the bounded component of the complement of gamma 2 and in unbounded component of the complement of gamma 1. So the contribution calculating here is 0 for this is to cancel and only just this is to be more correct, this is n gamma z, gamma 2, right? Because gamma 2, okay? Only the contribution of gamma 2 is important, okay? And it is 1 because the number of winding times is 1, right? So this is 1, so we forget about this index. I normally forget also this, but well now I have just to take into consideration this this formula and if you remember from this integral formula we have obtained that any holomorphic function is in fact complex analytic. Now we are dealing with a meromorphic function in principle and we are dealing with a domain which is not a disk but an annulus. So we will see that there is a difference, okay? With the case we already seen, all right? If you remember we have considered the center to be a, right? In our expansion it was the center of the disk at that time. Now a is the center of the annulus and it is not contained, okay? We added and subtracted a and we wrote, okay? An expression which was then a series and then proved that the series, sorry the function involved in the series were uniformly convergent to something and then we used this fact to extract a sum from the integral and then express everything in terms of power series expansion. Now if we repeat the same game, okay? Now first we notice that this is 1 over 2 pi i, so remember that gamma is gamma 2 plus lambda minus gamma 1 minus lambda. So these two contribution cancel each other but this is important to remember. So we have integral over gamma 2 fx xc minus zdxc minus 1 over 2 pi i the integral over gamma 1 fx c xc. Why this minus appears here? Because well we have a minus gamma 1 so it means that we are considering the curve with the reversed parametrization so the reverse orientation and so this has the effect of changing this sign of the integration, right? So this is the minus which comes from this minus here. This minus here is formally means the same curve but with minus t instead of t, okay? So the index is negative if you want of the point a. Now we repeat the same game as we did before that is to say if we add a subtract a and the denominator in both integrals then we have this but this time if xc is in gamma 2 we are integrating, okay? The distance from a is r2 which is greater than the distance from a of z. When xc is in gamma 1 then xc minus a is r1 which is smaller than the distance from a of z. Remember that z has a modulus since z minus a is therefore we will do the following. In the first integral so the integral over gamma 2 we take out xc minus a and observe that this is 1 plus a minus z over xc minus a, right? Like we did before and this number here is such that the modulus is smaller than 1, correct? Because we are integrating over gamma 2 and on gamma 2 on the other side oh sorry I forgot this is a minus, right? Plus the integral over gamma 1 but the minus is in front, right? So minus then I repeat the same stuff but instead of collecting xc minus a I collect a minus z, right? So that I have 1 plus xc minus a over a minus z of course dc and dc because on gamma 1 I have that xc minus a is r1 which is smaller than z minus a and that's give you that this is smaller than 1, correct? Now so the first part can be regarded as something we have already done, right? Because then we express this as a summation of this number here, right? So I have that the first part is in fact the integral over gamma 2 of what? Of f of xc then I have xc minus a to the power n minus n plus 1 summation and greater or equal to 0 and here I have z minus a, right? The power n which leads to what we already know, right? Then I have a minus here and in fact sorry if I stop here but in fact if I had a point inside of this I wouldn't consider this gamma 1, right? I could just consider gamma 2 circle center a radius greater than the modulus of z minus a. Then I have this minus here and here I have the summation n greater or equal to 0 f of xc then I have here z minus a, right? n plus 1 and here xc minus a but with a minus in front you see because let me show you this. I have xc minus a plus a minus z then I have as I said a minus z times 1 plus c minus a 1 plus sorry over a minus z and this is minus z minus a or if you want minus here and z minus a here, right? So I need this, okay? So that I have this minus here which cancel this and xc here, right? So surprisingly then I obtain, oh sorry and n is of course here, right? So I obtain the following that when you have a singularity and then we'll see how to describe which kind of singularity according to the to what we obtain we have not just a simple power expansion like this but generalized power expansion with coefficients also negative sorry not coefficient but exponents negative, right? So this number here represents what this stuff, right? Because when n is 0 this is z minus a to the power minus 1, right? And going on we have all possible negative exponents as well the positive exponent coming from here, okay? So this generalized power expansion is known in complex analysis as Lorentz series and of course it applies not only to holomorphic functions in which case we have only n non-negative exponents but also to meromorphic functions. Now we'll get a bit more into the details of the description. So what are the cn for the different cases, right? So in general we have that cn so the coefficients with this notation so cn can be obtained as 1 over 2 pi i the integral of gamma of fxc c minus a n plus 1 d xa. This is the way to calculate the coefficients and in particular we observe that z equal to a is removable singularity f c n are 0 for n negative. It is to say if the function is in fact holomorphic, okay? Correct? So the Lorentz series with coefficient exposes so with parameters with indexes going from minus infinity to plus infinity instead of going from minus infinity starts from 0 to infinity. So the normal power series expansion so it is no it is analytics complex analytic. z equal to a is a pole and remember that pole is a certain order finite order f and only f okay cn is equal to 0 for any n smaller than minus m okay m order of the pole pardon me well m I write m m is in z right the order is m is okay minus m or if you write m yes we write minus m it's positive sorry sorry sorry yes it's m and is in z and m negative means that minus m is the order correct what I mean is that in this case you have a summation of just a finite number of summands with negative exponents and this expansion in the Lorentz function as it is suspected and finally the singularity is essential if none of the previous cases applies in a sense as then if it is not possible to follow a minimal negative index or positive or zero or non-positive say any such that all the coefficients are zero in the power in the Lorentz series right so all right so as I probably already said the interesting coefficient in the Lorentz series is this one and this is called the residue okay f f is meromorphic well sometimes well this is probably something I already I've already done sometimes there is an abuse and and the information and then in terminology sometimes you use meromorphic is to say not holomorphic but singularities are restricted to poles and that's what I also said probably last time in this case I'm saying well a is a singularity and well not necessarily a is a pole so probably in the first slides I used I just said f meromorph is a not holomorphic at a to be more precise okay but sometimes some books you can find meromorphic within singularities and then they precise meromorphic with poles meromorphic okay so just so if f is meromorphic so when there are poles in particular well c minus 1 is very interesting it's called residue of f and and and I already I think that I already told you why this c minus 1 the residue is important but residues of f for the pole a okay and this is somehow too long so I write it this okay residue function is f and the singular the pole is a this coefficient is important because as I showed you ff is meromorphic so if f as the power essential with the principal part which is the part of the Laurent series with negative coefficients the finite part then it has also the the the power city expansion with positive exponents then it has this expression in a neighborhood of ways like this plus summation remember that I use this notation right okay and I noticed that if I consider a curve gamma closed curve the integral over this closed curve gamma uh say around 0 around a sorry this is what is c minus 1 right so in order to cut well this is the only contribution to the integral if you know c minus 1 well and you know the index in case you have a very simple curve like we normally use a circle around the point a so to know c minus 1 is equivalent to know this in this integral that's why it is important so now the problem is maybe how I calculate this so since we have this expression here remember that we also have this fz times z minus a to the power m m is the order of the pole as a holomorphic function g of z right right so from this you obtain that this is a remark right put it here that the residue of f at a remember that m is the order as well c minus 1 is 1 over m minus 1 factorial times g the mth derivative m minus 1 sorry derivative of a of m of g at a that's the way to calculate if you want if you find this you calculate the residue in this way and this is number six I guess so let us now consider some applications of this calculus of the residues of a meromorphic function in particular let me consider this function here this is an exercise if you want and then we'll see a bit more in detail the general situation so consider this function here okay well this function is of course not holomorphic in the entire plane right well this is the ratio of two polynomials and the complex setting any polynomial of degree m and greater than one a greater than zero as m root so this polynomial here the denominator vanishes four points right so the f is holomorphic and c minus z1 z2 z3 and d4 where zj is one of the roots of one plus z the power four so this polynomial okay so if you want to solve this well I might say that this this is just an exercise but in case well this is the problem you have to solve and I think that in the first lessons we already solved similar examples and and we are looking for an application of the moir formula and these four roots which are all the strings are in fact on the unit circle and they are okay equally they are like this right this this this so they have this aspect a plus a a and a is plus or minus no possibilities okay so these are the four so the singularities are of course isolated as expected can they be essential singularities so these are singularities these are not removable singularities because well if you take z minus one of these roots okay take the limit as it and one of these roots of f of z z minus this root okay you obtain something which is not bounded so you cannot you cannot consider this as a as an removable singular any of these roots to be removable singular so it is not the case of removal singularity can it be that an essential singularity why not the answer is not of course sure this is the correct answer so thank you so for any of these roots you can always find a positive integer associated with this such that the z minus this root to this power times the function is in fact holomorphic so we are dealing with poles okay and these poles are okay finite number isolated as expected and not real because no real number can have this property to be z is false power to be minus one okay and it is obvious why all right so I write this one plus z square in this way so okay this is one plus not z square but z to the power four sorry hmm because I apply the fundamental theorem of algebra four times okay and I I know that I have four distinct roots so all these roots are simple roots right okay and here I have harmonic polynomial so that I have a z minus z j okay as as a factor correct good so I consider z one to be well this is the one this is the two this is the z three and this is the four according to the standard orientation okay the first positive real and imaginary part so it should be like this z two is minus and then the three is minus and the four is now the now the next step is calculate the residues associated to this poles four poles pole residues okay so we want to calculate the residues of f dj how can we do this suggestions no okay so theoretically in principle you are correct I agree with you but this is a long this is a long procedure I'm not going to okay all right all right so I prefer this okay okay all right so we are dealing with a simple pole and so that the function f of z is in fact z squared over z minus z one z minus z two z minus z three z minus z four okay to calculate the residue of f for the pole z one I simply calculate what the limit as it tends to z one of what the z minus z one f of z right so I have here z one squared because this is going to the z minus z one cancel and I have z one minus z two z one minus z three and z one minus z four okay and these are numbers which I can't calculate so in general the residue of f of the poles zj for f is zj to the power two over product okay zj minus zk k different from j k one two three four all right this is a general expression good so as I said just make this calculation in this particular case so remember that z one is this okay so z one squared is okay I have this squared minus z squared the same amount so I have only the double product so I have two square root two over two i I'm sorry times right yeah okay and then I invite you to to show that we have the following with z one minus z two z one minus z three z one minus z four which is the denominator in the calculus of the the calculation of the residue we are dealing is in fact something which can be okay we do it okay so z two remember z two is minus plus so I have z one okay from the difference okay and then I have nothing else then remember that the z three is minus z two over two minus i and z one is as I said before so I have okay minus so doubles the real part and doubles the imaginary part and final is d four is okay I have so that when summing together have i right correct so this is two i and this therefore the residue we are calculating is i over two i right which can be also written as one minus i over four square root of two okay after multiplying times the conjugate of one plus i okay so I invite you to verify that the residue of f of z two for this f is just minus one minus i over four square root of two this is an exercise okay an exercise for you but it's not difficult to to understand why okay repeat the calculation just stupid calculation and now okay this is number nine sorry let us go to an application or what we have found because this is just trivial calculation I would say trivial calculation now let us consider this situation we have the four poles on the unit circle this is the one this is the two this is the three and this is the four right let us take the half circle this is the origin right of radius raw and raw is greater than one and we consider as a curve gamma the half circle from here and this segment so the segment so raw is half circle of radius raw and segment real segment minus raw raw okay this is minus raw and this is raw with this orientation this is a contour of what half disk diameter and half circle and I want to calculate the integral over gamma of the function z squared over one plus z to the power four d z this from our consideration is simply well I put one over two pi i in front because I normally forget it okay this is simply what the sum of the residues but which residues the residues of the poles which are contained in this curve gamma so the residue of z one and residue of z two so it is the residue the sum of these two residues okay and we already know what it is it is one minus i over four plus one minus one minus i over four square root of two which means that this number here is sorry it's minus two i over four should be right good that is to say that this integral d z is okay four pi minus one cancel this minus in front over four root of two is pi over two what is this useful for well it can be applied for real integration of of rational functions and over the entire real axis in fact the integral of a gamma of z squared over one plus zero is the integral of the the circle of half circle okay to produce raw plus the integral between minus raw and raw so I use the additive property of the line integral all right so when I restrict my consideration on the real segment in fact instead of considering z I consider the real numbers in these segments so this reduces to what to the integral which is normally very odd to calculate right especially when you want to be sure that this number exists or something like this and raw can be very large and become actually infinite and we show that when raw becomes infinite this is zero so that from the residue we know the value of this undefined integral real integral we conclude this once we prove that making raw okay big enough this number here is smaller than epsilon all right so it's all it's easy to show this to conclude because we have the following okay so let us consider the integral over c raw of d squared one plus z power four d z okay remember that over cr you have that z is raw e i theta we are moving okay and theta varies between zero and pi right so this z is i raw e i theta d theta good so after substituting I have that the integral there is equivalent to the integral of what of raw squared e i theta squared times i raw e i theta d theta over one plus raw the power four e i theta and that's it correct so I probably better write it this way I collect everything say i in front then I have e i three theta now let us consider for a while this picture here this is the origin and this is the circle of radius raw one plus raw to the power four e i four theta moves okay remember the raw is greater than one moves on so sorry this is the modulus points this circle here the center is one and the radius power four all right you say this this is a circle centered one of radius raw to the power four when theta varies probably see this better if you consider well theta varies raw is fixed so if you consider this number here as say w raw so raw is fixed it depends only on theta right raw is given so w theta minus one is modulus constant equally right so it is a circle it well it's not a circle well points with this property lies on a circle geometrically it means that we are in this situation so this this is the description of the point of the set and see where these points can stay okay varying theta and raw is greater than one so this is zero right because if raw is greater raw to the power four is of course greater than one okay so I have that from this picture I obtained one plus raw to the power four e i four theta is greater than one minus sorry raw to the power four minus one which is distance here raw to the power four minus one so this is the distance this small part of the signal and this can be seen easily geometrically okay so each point here as a distance from the origin which is of course larger than this this is the minimum right so with this inequality we actually conclude we conclude because we have that well remember that we have the integral we have to estimate this integral right this was the integral of sorry c r c raw of f z d z or if you want the integral over c raw of z squared over one plus okay so one extra raw appears so here the exponents are two and four four and three here is because when calculated the d z raw comes into okay into our calculations but then if I take this integral actually the the modulus of this integral this is smaller or equal to what to the integral zero pi of low to the power three of one sorry raw to the power four minus one okay v theta because I have that one plus raw power four e i four theta because of this inequality so this number here is pi or cube over raw four minus one which tends to zero and also raw tends to plus infinity because of the denominator then the degree of the denominator this is a rational function in raw the degree of the numerator is larger okay so that this number can be made as small as I want and so that I conclude that the integral of this rational function over the entire layer axis is this as an application of the residue theorem the residue theorem is the result we applied this in a sense calculates the value of an integral along a curve as the sum of the the residues of times 2 pi i of the residues of the poles in the in the bounded in the bounded component of the complement of the curve gamma okay so this is not by accident what so it's just an example of course but this is not one example this can be generalized okay so let me just catch the ideas for the general situations assume that we have a rational function and you want to calculate the integral of a rational function that is to say pq polynomials and we have to well to calculate this but this has to have to exist first okay so we have to be sure that this is something meaningful so first of all we have to assume that no singularities can be found on the real axis so first of all q of x is different from zero for any x okay or to be more precise or r has no poles on r which is more correct so q can be a zero q can have a zero but ps to have the same zero okay furthermore this is not enough because if I consider the integral also in a smaller domain of something like this this is not converging okay right so I'm assuming assume that the degree of the denominator is greater or equal to the degree of the numerator plus two which guarantees that convergence exists okay because then we can compare this ratio to one where x squared at least and one over x squared is converging okay which is interesting and important to know so in fact in the previous example let us go back to the example we started from we have a polynomial with no poles on the axis so we have a sorry a rational function so two polynomials one of the numerator one of the denominator the degree of the denominator is exactly two plus the degree of the numerator which is good and no poles of f can be found on the real axis so the general situation is like this with this assumption we can show that this number can be calculated using the residues and it goes like this or we need some extra so we repeat the same argument as we did before we take this to be the curve so minus raw and raw so actually it's suffice to consider the residues of the poles which lie in the upper half plane so the imaginary part of zj has to be positive and this suffices okay if I will consider or if you prefer you can repeat analogous consideration for so but so it's better to to to reduce our calculation to what we are interested in okay and the previous case you could repeat everything with the other two poles and they have circle it was exactly the symmetric problem okay now r of z can be written as you know and according to our assumption a0 plus a1z plus a n zn and q of z is like this where m is and at least m plus two in other words if I collect here the leading term this so I have a0 well a n plus a n minus one over z plus plus a a0 over z to the power n here I have z to the power m and then I have bm plus bm minus one over z plus b0 over z to the m which a is also one over zm minus n and then this is like before which tells us that the important part to be considering the integral is this right the others are somehow can be neglected and the fact that m is at least n plus two means that m minus n is at least two which is good because we can compare then okay the the integrand with one over z to the square z square one over z square right but and our explicit example we also showed that the contribution of the integral along this half circle was infinitesimal when rho tends to infinity more rho tends to infinity well is it normally the case well it is because with our assumption it is because well since we have that well actually I probably have to use the same we have that r of z s one over z m minus n and remember that m minus n is greater or equal to two and then I have here something which is a n which means that when I consider z times r of z because of this assumption and because of this expression this can be made smaller than epsilon if z is greater than rho star so if z is taken far away from the zero this number this product can be made smaller than epsilon and this is what we need in fact the integral over c rho of r of z this z is well rho c rho is what the set of points of rho e i theta right and theta varies and this segment so I substitute and I obtain that this is the integral over and then again i rho times probably e i theta right in here that is to say that when I consider the integral the modules of this integral this is smaller or equal to the integral of rho and this can be made smaller than epsilon because of this so we can control this and this explains why in the previous consideration we obtained the final result as an application of the residue theorem well and this is one step then let me also say that the application of the residue theorem works fine also in another odd situation for real integration it is to say the following 17 so assume that we have to calculate this integral it's another exercise okay 1 over a plus b cos theta d theta right this for a b real and assume that a is greater than b and I think it well we probably have to consider a b positive this is not a situation because it's not a rational function it is trigonometric polynomial the okay so it's a rational trigonometric polynomial which is normally odd so how can we transform this integral in order to apply the residue theorem so here apparently we are dealing with something which is real everything is real right here but remember that passing from real to the complex the cosine of theta is 1 over 2 e i theta plus e minus theta theta is 1 over 2 i of this right now consider z to lie on the unit disk not unit circle sorry in the complex plane so z is e i theta and theta varies sorry 2 pi correct modulus 1 e i theta good so that cos theta is one half c plus 1 over z or z minus 1 z to the power minus 1 right similarly sin theta is one half 1 over 2 i sorry z minus 1 over z but now I have also that from z equal to e i theta and theta varies i is fixed right the z d z is i e i is theta d theta or if you want d z is i z d theta right in other words I can substitute d theta with d z over 1 i z that is to say the in this the expression the integrant can be the can be substituted by this so the integral is d t d theta is okay d z over i z and then I have 1 over a plus b z plus 1 over z and then half okay so I substitute it cos theta and d theta when considering z to be e i theta so that this becomes the integral over this unit circle of what 1 over 2 a plus b z plus b over z times i z d z then I have to put a 2 in front right which comes from this denominator all right so I have I have 2 over I put this way 2 i b okay z and z comes here b is here and have 1 over 2 a z plus z squared plus z 2 over b 2 a over b z plus z squared plus z and then I have b z finally so that after this change of variables I have again a rational function the degree of the denominator is greater or equal to 2 plus a degree of the numerator in this case the degree of the denominator is 2 and the numerator is constant so that we can apply okay something which is well in this case we are actually integrating over a circle not over the entire real axis the last time is sure thank you this is 1 b s is here z and z cancel correct yes so let me just conclude if I can so I have the this number here is continue 1 i b integral over c squared plus 2 a over b z plus 1 b z this is a polynomial in c and it has two roots right okay so now up to now we haven't used the assumption about a and b they are real but what about the the discriminant we have this is a squared over b squared minus 4 right minus 1 sorry right and this is well if a as I am assuming is greater than b this number here is this is positive right so we have two real and we can also say that if alpha and beta are the two roots the modules of the two roots is one you see this the constant term right okay in other words if modules of alpha is greater than one modules of beta is to be smaller than one and they cannot have p it cannot have both modules equal to one which means the one is inside the unit circle and unit disc and the other is outside so that we have to consider just the residue for this calculation of the root which lies in the unit disc because we are integrating over the unit circle in other words the this integral is the integral 2 1 b and this is 2 i b times 2 pi i the residue of this function here alpha well of beta okay beta because beta is modules smaller one well these two roots as I said are simple roots so that we have this right or plus 2 a over bz plus 1 is in fact z minus alpha times z minus beta so that the residue of f f beta where f of z remember is 1 over the square plus 2 a over bz plus 1 okay is limit as it tends to beta or what 1 over z minus alpha and then I have to put something in front I guess right no it is 1 over beta minus alpha okay but then remember that alpha was some well can be calculated is minus 2 a over b plus what a square over b square minus 1 and over 2 and beta is minus 2 a over b minus where b square minus 1 right so that the residue is or should be something like these two parts cancel right no these two parts cancel and I have what is left here is so we are finished about 20 so putting together we have the integral we were started from it is this as okay let me probe 2 pi i i over 2b times what 1 the residue was calculated somewhere else so I said this a square over b square minus 1 if I'm not mistaken okay yes I hope so the residue times 2 pi i so this is well 2 and 2 cancel minus b over maybe there is a mistake somewhere because in my calculation should be different but anyway please follow the calculation repeat the calculation and check where the mistake well some somewhere I probably wrote something wrong here because there is a b in bed okay just check okay the idea is that if you know the residue theorem you can solve this problem but you have to be probably more careful about the calculation of the real roots alpha and beta last but not least let me just point out something which maybe has to be reminded okay check it we come back to this yes we have dealt only with simple poles so far and with simple poles it is so in a sense it is simple also to calculate residue so let me just go to more genetic situation assume that we have this which is of course a function which is a meromorphic function but not rational function so the poles are a and b so the residue of f a is simply e a a minus b and similarly okay b minus a right now the question is what if I consider g of z to be e to the power z over z minus a squared in this case the pole a is not simple so what is the residue of g at a how we calculate this well we cannot apply this simple factor we we can do like this we can do something like this okay well first substitute z with z minus a plus a so that we have z minus a as up to a constant we have e z minus a as in the denominator as a numerator the same factor which appears in the denominator so then we expand we have e a times one plus z minus a plus z minus a squared n over n factorial plus something and here z minus a squared which means f of z sorry g of z is in fact e a times one over z minus a squared plus one over z minus a plus one over two factorial plus something else which is one over three factorial z minus a plus blah blah blah and hence the residue at a is so the general procedure requires to write the Lorentz series that in practical use you have to simply find a way a simple way very simple way to to determine this coefficients okay good or if you are in big troubles and well in this case it's quite easy to see also that the order is too right but if you are in big troubles the only solution is to calculate the rebuttives and so on and so forth all right so let me now almost conclude what I have to say about the application of the residue but I just want to point out that in some books there is also this interesting case and it is like this so you have rational function over the real integrated with the positive reals times something which is like this and r r a has no poles and no no rear poles and we also assume so we substitute x with z and we consider r to be the extension over the complex of a rational real function okay real rational function and we have this assumptions here so the novelty in this situation is that we cannot apply for example we have to calculate something like this x to the minus and c say this is standard example but here apparently there is one pole for sure x equal to minus one and so positive real okay that's what I'm assuming so it has a real pole but there is also the zero appearing as a singularity but zero is the branch point so it's not point where the function can be extended so what I'm saying is that remember that we are considering here so it cannot be that this function here is in fact defined at zero so just just a sketch idea of how things go on but we consider zero is in fact like for the for the square root and so on it's a point where you cannot define a whole multitude function when you have powers okay with non non integer non positive integers powers and we have this pole at minus one for the case we are considering we normally take two disks two yes and that's the curve gamma this well this distance is delta this this is the origin so sorry maybe the origin is here right so this is r this is raw and this is in the complement of the plane with the infinite slit from zero to infinity here we can define also the lower rhythm which is because it is a simply connected domain whose complement contains zero now it is that gamma can be modified the form to a point this gamma is homologous to zero but if we apply the residue theorem passing through the complex station we just sorry just take and minus one is here right we just take the residue in our example we had okay just take this show that what you are considering on the contour depends only on this part here but not so the contribution on the portion of circles is irrelevant for the integral and then we can conclude something okay I just wanted to mention this because this is maybe interesting to at least know that it can be possible to because it is an example where you can you do not take the same contour okay circle half circle and okay so what is left to show is well well what I said is correct okay so I will just come back to this see again the calculation and the previous exercise and come back to this but I promise that this is the last time I'm talking about residue theorem and application because otherwise it becomes just a course an entire course in how to apply the residue theorem according to several different situations okay so each time you have to know which contour which function what are the manipulation you have to deal with so I'm not I'm not going to give you all the say details in this very complicated stuff however it is important to know that in some cases this is the only way to calculate some integrals and it is it has also many applications in in engineering calculations so for instance people from other faculties like engineering do not know anything about all of architecture but they do know how to apply residue theorem okay but in case there are books there are there are okay so these are somehow just basic examples in particular the first one was a reasonable example the second one was somehow an application of a previous result and this is a more complicated example which we'll just mention and quite quickly show you how how to be faced okay thank you for your attention