 Alright, so what we are going to discuss now is about dealing with the residue at infinity which will make sense since we have been studying about the point at infinity okay and so you know let me tell you at the outset something that you have to remember which distinguishes between the residue at infinity and the residue at a finite point in the complex plane. The point is that the residue at infinity for an function which is analytic at infinity can be nonzero whereas the residue at a finite point in the complex plane okay there is a point in the usual complex plane for an analytic function the residue is 0 okay so this is very very important okay. Of course if the residue is 0 at a point it does not of course mean that the function is analytic okay but the fact is that if you have an analytic function at a point then the residue at that point is 0 so long as that point is a point of the usual complex plane but if it is a point at infinity okay it may be analytic at infinity but yet the residue may not be 0 so we are going to talk about residues at infinity and we are going to talk about the version of the residue theorem at on the extended complex plane okay. So let me write this down residue at infinity and the residue theorem for the extended complex plane so this is what we are going to talk about. So you know let me begin by recalling what is the usual idea of residues okay so recall if Z0 is a point in the complex plane okay and f of Z is analytic in a deleted neighbourhood of Z0 okay so when I say this I am saying that Z0 is actually a singular point okay so it could be a removable singularity which means that it is not really a singular point it could be an analytic point but on the other hand it could be a honest singularity it may be a pole or an essential singularity and then the residue of f at Z0 is given by well residue of f of Z at Z equal to Z0 is equal to well 1 by 2 pi i so you integrate the function f Z dz around a simple close contour you a simple close contour which goes around the point in the positive sense okay so basically so Z0 is this point and of course there is a deleted neighbourhood of this point sufficiently small deleted neighbourhood of this point so in particular there is a disc of sufficiently small radius open disc surrounding that point where the function is analytic and then I am just taking a simple closed curve gamma okay of course it is simple closed contour so simple means of course that it does not intersect itself it has positive orientations which means it is going anticlockwise around that point so in other words the interior of the contour contains the point okay so the point lies to the left of the contour as you walk along the contour okay and the region to the left of the contour as you walk along the contour in the direction specified by the contour is called the interior of the contour okay and of course when I say contour you must remember that gamma has to be piecewise smooth so gamma is whenever you parameterize gamma mind you gamma basically is a continuous image of an interval okay that is what a path is but it can always it can also be a piecewise continuous image okay and well well in fact we always take gamma to be a continuous path so but the thing is that it is also continuously differentiable at least piecewise okay that is the condition for a contour and this much is required for this integral to be defined as a Riemann integral alright so this is the residue and well I am wondering so I am wondering if this if that is right so I was just wondering whether the factual 1 by 2 pi i is correct like this so for example if I plug in 1 by z minus z not if I take the function f of z to be 1 by z minus z not if I integrate 1 by z minus z not over a simple close contour sufficiently small contour then I am going to get 2 pi i so the residue is 1 and well see of course it is very important that I choose this gamma in such a way that there are no other singular points of f inside other than z not and that is of course true because I am choosing gamma inside a deleted neighbourhood of z not where f is analytic so there are no other points where f is singular other than z not okay. Now the point is so this is the definition of residue this is one definition of residue and as you can see the importance of this definition is that if you multiply out the 2 pi i on the right side if you multiply both sides by 2 pi i what you will get is 2 pi i times the residue is the integral of the function over gamma so what it tells you is that it tells you how to integrate a function around a singularity okay that is the important thing so the fact is that why is this definition interesting it is interesting because you get to know how to what the integral of the function is around a singularity okay but the important thing is that you should be able to compute the left side without having to do the integral okay and that is the method of residue that you would have learnt in the first course and so there is so you also can recall the usual definition of the other usual definition of residue in terms of Lorentz expansions. So the point is that since f of z is assumed to be analytic in a deleted neighbourhood of z not there is a Lorentz expansion for f centred at z not and then you write out there is a Lorentz expansion it will contain both positive and 0 and negative powers of z minus z not and the residue is precisely the coefficient of the of 1 by z minus z not namely the z minus z not to the minus 1 okay so that is another so let me write that down recall also that if f of z is equal to sigma n equal to minus infinity to infinity a n times a n times z minus z not to the power of n is the Lorentz expansion of f around z not or f centred at z not mind you that is a that exists because of Lorentz theorem Lorentz expansion exists because of Lorentz theorem and you know what is the big deal about Lorentz theorem it is the analog of Taylor's theorem so Taylor's theorem tells you that if a function is analytic at a point z not then you can expand it as a power series in z minus z not namely you can expand it as a series which involves only 0 and positive powers of z minus z not Lorentz expansion tells you that you can do something similarly something as good if even if z not was a bad point if z not was a singular point and the only singular point in the in that neighbourhood in its neighbourhood then you can expand f in both positive and negative powers of z minus z not okay so the fact is that if you are dealing with a singular point you must allow negative powers of z minus z not also in the in the series expansion and of course if z not is a removable singularity you know roughly I mean this is exactly what remains removable singularity theorem is that if you write out the Lorentz expansion you will only get a Taylor expansion if z not is actually a removable singularity okay which is equivalent to saying that f is analytic at that point so anyway so the point is that if f is has this Lorentz expansion then the residue of f of z at z equal to z not is none other than a minus 1 and so you know so if I write this down this is also equal to as I wrote earlier it is 1 by 2 pi i integral over gamma fz dz okay and so the advantage of this formula is that you know I in many cases if I can write out the Lorentz expansion at z not then I can explicitly find out what this a minus 1 is and that a minus 1 will give you that multiplied by 2 pi i is going to give me give you the integral of the function around that point okay so that is the advantage of this and you know you have used this in a first course in complex analysis to evaluate real integrals and so on and so forth so it is a very useful thing. Now the point is that you know and then of course what is the generalization of this the generalization of this is that suppose you are going to integrate a function around the simple closed contour and assume that inside that region the function has only isolated singularities okay and you know they are going to be only finitely many isolated singularities and then the formula reads that if the formula that you get is essentially the residue theorem which says that if you integrate the function around a bunch of isolated singularities what you are going to get is 2 pi i times the sum of the residue of the function at each of those singular points okay and that is the residue theorem and what we have written down here is that the integral of the function is 2 pi i times the residue at z not because z not is the only singularity and the point is that this extends to 2 pi i times sum of the residues at various points which are the isolated singularities of f inside the contour okay so that is the residue theorem essentially. So the point is that you know it is you know in a way it is very easy the residue theorem is very easy if you look at it like this okay it is rather not a theorem it is more part of the definition except that if you want to really prove it you will actually be applying Cauchy's theorem okay by surrounding each of the singular points by a sufficiently small disc and noting that outside this disc and inside your contour the function is actually analytic okay and you will have to apply the version of Cauchy's theorem for multiple connected regions and you have to use this definition of residue to get the residue theorem. So let me write that down more generally if f of z has only isolated singularities inside simple closed contour gamma and of course I am assuming that on the contour there are no singularities okay and is analytic on gamma so this means that it is analytic on a small neighbourhood an open set which contains the contour gamma so in particular I am avoiding the situation that there are singular points of the function on the contour okay. Of course you know if there are singular points of the function on the contour then you know at those singular points the function will fail to be continuous if they are honest singular points and once the function fails to be continuous it is very difficult to define the Riemann integral of the function over the contour. So you must understand that the moment you define the Riemann integral more or less you are assuming that the function has nothing wrong going on the contour it is only what happens inside that matters okay. So well then there are only finitely many and integral over gamma f of z dz is equal to 2 pi i times the sum of the residues the residues of f at the singular points. So this is essentially the residue here okay of course that there are only finitely many follows from the fact that if you take the contour along with the of course whenever I say contour mind you it is always an oriented contour and without if nothing is mentioned always the contour is oriented and it is given the positive orientation which means that you go around the point in the anticlockwise sense okay and that is always that is always taken for granted unless something else is mentioned. So whenever integral over a contour is mentioned you must understand the contour is already oriented and the orientation is positive okay. So well so this is essentially the residue theorem so let me write that down so this is just the residue theorem and well as I told you it is very important it is very useful to evaluate integrals which integrals of functions around a contour which have only isolated singularities inside the contour okay. So this is it now the point is that what we want to do is that we want to do this for the point at infinity okay and we want the residue theorem for a domain in the extended complex plane so how are you going to do it? So the idea is very very simple we start off by using this by adapting by literally using the same philosophy namely the residue should be you know you integrate the function okay around the point okay and then divide by 2 pi i okay and then that should give you the residue. So if you want to get the residue of a function at the point infinity first of all it should be analytic in a deleted neighbourhood of the point infinity okay and then I must take a contour that goes around infinity in the positive sense okay and I have to integrate the function around that contour okay and divide by 2 pi i and I must get the residue at infinity okay. So this is so the definition for residue remains the same namely you just integrate the function around by going over a contour that goes around the point in the positive sense as far as the point is concerned and then you divide by 2 pi i okay that is the definition alright and let us see what that brings up. So incidentally I wanted to point out something here which I just remembered so let me tell you so for a moment in a way you know once you believe Laurent's theorem once you believe Laurent's theorem that this is exactly I mean this is that this form was correct is something that you can more or less see because you know so you know I have this expression there I have this expression of Laurent series okay and what I need to compute is that I need to compute integral of f over gamma okay and that is the integral that is the same as integral over gamma of the series on the right side okay and now the point is that I can push the integral inside the sum because the fact is that the Laurent series like a Taylor series you know wherever it converges it converges normally that means it converges uniformly on compact subsets and since I am going to integrate over gamma is of course mind you any simple closed contour is a compact set because it is both closed and bounded. So therefore the integral of the sum is same as sum of the integral so I can push the integral across the sum and when I push the integral across the sum what is going to happen is that you know I you will see that if I take positive powers of z minus z not the integrals are going to vanish because the positive powers of z minus z not are analytic functions and Cauchy's theorem is going to tell you that whenever you integrate an analytic function around a simple closed contour you are going to get 0 in fact the positive parts are all entire functions okay they are just polynomials and the same thing is going to happen to the constant term okay and then if you take the if you take the negative powers of z minus z not greater than 2 okay so I mean what I mean by that is if you take terms involving 1 by z minus z not the whole squared 1 by z minus z not the whole cube and so on the integrals of those terms will also vanish that is because they all have antiderivatives. So you know it is something very basic that I want you to understand the fact that an integral vanishes is you know basically it is equivalent to saying that the integral is independent of the path and especially when the integral when the integral has an antiderivative then you know that the integral is actually the final value minus the initial value okay of the antiderivative and if the final value is equal to the initial value which is what will happen if you go around a closed path you are going to get the integral to be 0 okay and so you know all the all the integrals which involves involves z minus z not to the power of n where n is minus 2, minus 3, minus 4 and so on they are all going to vanish so the only thing that is going to survive is going to be integral over gamma A minus 1 z minus z not power minus 1 dz okay and you know the integral over gamma z minus z not to the minus 1 dz is just going to be 2 pi i and that is just because of Cauchy's theorem because the integral over gamma is not going to really depend on the shape of gamma you can replace gamma by a small circle okay going around z not and then you actually parameterize the circle as usual write it as z equal to rho e to the i theta varying from 0 to 2 pi and compute the integral you get 2 pi i okay. So therefore what I want to tell you is that once you know Laurent's theorem okay once you believe Laurent's theorem this formula for the residue that A minus 1 gives you the residue is more or less direct okay so that is something that I want you to recall now let me go back and define the residue at infinity. So suppose f of z is defined in neighbourhood of infinity okay so mind you that what this means is that f of z is defined on a circle on the exterior of a circle of sufficiently large radius that is what it means a neighbourhood of infinity is by definition you know by the geographic projection the same as the exterior of a sufficiently large circle okay and so at this point what we do is that we define the residue of f at infinity to be just the integral over gamma 1 by 2 pi i integral over gamma fz dz where gamma is a simple close contour that goes around infinity in the positive sense okay this is exactly the way we define it as we defined it for a point in the usual complex plane okay so let me write that down. Then residue of f of z at z equal to infinity is defined to be so I put a colon and equal to 1 by 2 pi i integral over gamma so you know let me let me use let me let me change notation to fw because I will need to appeal to something else namely I need I need to appeal to changing the variable to 1 by z so let me do that oops so fw w equal to infinity 1 by fw dw okay 1 by 2 pi i integral over gamma so where so let me write that this is very very important where gamma is a simple close contour going around the point at infinity in the positive sense. So you know this is the this is the this is the definition and well you know so this is exactly the definition that you would have made for a point in the complex plane and this is the same definition I am using for a point for the point at infinity but then there are 2 or 3 things that one has to be careful about the first thing is that what do you mean by a contour that is going around infinity in the positive sense so the fact is that you know you must think of a contour in so here is where you again appeal to the stereographic projection okay you know that sufficiently small neighbourhood of infinity is given by the exterior of a sufficiently large circle centered at the origin okay so in some sense a sufficiently large circle centered at the origin should be a contour which goes around the point at infinity okay and you can you can see this you can see this so more generally if you take a sufficiently large contour which goes around the origin okay a simple close contour which goes around the origin sufficiently large means that it encloses a sufficiently large area okay then for that for that matter I mean it lies in the exterior of I should rather say it lies in the exterior of a circle of sufficiently large radius okay then that itself is an example of a simple close contour that goes around the point at infinity and that is because of the stereographic projection and the only thing is that what is this business of the positive orientation of that contour with respect to infinity mind you the positive orientation means that infinity should lie in the interior of that contour okay so you should orient the contour in such a way such that the interior of the contour contains a point at infinity now you know if I take if I take a circle sufficiently large circle centered at the origin and you orient it the usual way I did which we do namely give it the anticlockwise orientation then the origin becomes comes into the interior the interior just is the interior of that circle okay and the exterior will be the exterior of the circle and that will contain the point at infinity so you can see from this argument that you will have to orient it clockwise okay so the so gamma must be a simple close contour lying in a sufficiently small neighbourhood of infinity so it should be lying in the exterior of a circle of sufficiently large radius and it should be given the clockwise orientation that is what it means okay. So let me write that down this means that gamma should be a simple close contour in the complex plane lying outside a circle of sufficiently large radius but given the clockwise or negative orientation so here you see you have to be careful I am saying that gamma should have negative orientation here but in the statement before that I am saying it should have positive sense so mind you in the statement preceding it was positive sense with respect to the point at infinity and positive sense with respect to the point at infinity is negative sense with respect to the origin okay so these are not contradictory statements okay you have to understand this subtlety and well of course you can also see this pictorially more or less see you know if you draw the stereographic projection so here is the x y plane which is a complex plane or the z plane and then I write then you know we draw this third axis which we call it as u and then you take the unit sphere surface of the unit sphere in 3 space I am going to get something like this this is the stereographic projection so you know if I take now you know if I take sufficiently if I take a circle on the complex plane of sufficiently large radius okay and then you know well you know the point at infinity corresponds to this point here on the stereographic on the Riemann sphere which corresponds to the north pole so this is the this corresponds to the point at infinity okay so I put this triple line to tell you that it corresponds to the point at infinity under the stereographic projection and well you know what happens to the well you know the exterior of this circle the exterior of the circle in the complex plane that is going to correspond to well on the Riemann sphere is going to correspond to this small disc like region though it is curved surface it is a small disc like region when I say disc like I mean topologically you can flatten it to look like a disc topologically and it is a disc like neighbourhood of the north pole on the Riemann sphere okay and that is how the shaded regions on the plane namely the exterior of the circle and the small cap on surrounding the it is like a polar cap okay if you imagine the earth it is the north at the north pole so now the point is that you see what you must understand is that if I now give these if I now give this so in particular you know if I now take sufficiently if I take a contour which lies in outside the circle okay if I take a contour like this now that contour is going to correspond to a contour here on the Riemann sphere that goes around the point at infinity okay and as I make this circle bigger and bigger that is going to give a smaller and smaller contour simple close contour that goes around the point at infinity the only thing that you have to worry about is orientation the orientation the way I have drawn it the orientation should be like this mind you it is the it is actually it is actually clockwise about the origin and the reason for that is that if you define it like that which is what you should do if you are dealing with the point at infinity then the interior of that contour is the exterior so this region this thing outside this is the interior of that contour okay and well and it is actually the region that lies to the left of the contour as you walk along the contour okay and you can see that this region corresponds to well this region on the Riemann sphere and that of course contains the point at infinity so the point at infinity is an interior point for the contour oriented in the clockwise direction okay so you can see this diagramatically okay fine. So very well now that we have defined what the residue at infinity is so of course this is gamma right so let us compute it let us compute what the residue at infinity is so f of z is well f of w well let me write f of w f of w is going to be sigma n equal to minus infinity to infinity n w to the n this is the Laurent series of f this is the Laurent series of f at in a in a neighbourhood of infinity okay and mind you in principle it is actually also the Laurent series of f at the origin in some sense centre at the origin but in a domain that is a neighbourhood of infinity and the only thing that you distinguish is that when you say it is a Laurent series at infinity you see the singular part is the one that involves the positive powers of w and the analytic part is the one that involves 0 and negative powers of w because it is the negative powers of w that behave well at infinity okay so that is the only difference but what you are actually looking at is actually the Laurent series of f at the origin okay now but anyway the fact is for the same reasons that I told you earlier if you compute the if now if you if you compute integral over gamma f w d w if I do this what I am going to get is and you know of course I think as a 1 by 2 pi i this is what it is well you know mind you if I compute this integral the way I normally would compute the integral on the plane what I will first do is I always compute integrals with my contour being positively oriented with respect to the usual plane that is with respect to the origin so what I will do is I will first write this is minus 1 by 2 pi i integral over minus gamma where minus gamma is the gamma oriented in the in the anticlockwise sense and that is the positive sense for the plane okay with respect to the origin if you want okay and then I will get this and you know now if I plug in the series here okay and and remember that I can do integration term by term because of the same reasons I told you earlier because the the laura series always converges normally it converges uniformly on compact sets and gamma or minus gamma for that matter they are as sets they are compact sets so if I do that again what is going to happen is that what is going to survive is only the coefficient of 1 by w okay and that is going to give me that is that is the coefficient when n is minus 1 so I am going to get a minus 1 the only thing is that I am going to get I am going to get a minus a minus 1 okay. So what you what you must see is that I mean what you must see what you will see is that I will get 1 by 2 pi i times 2 pi i times a minus 1 this is as before and I get minus a minus 1 okay so so you see so the so the moral is the moral of the story is that when you are looking at the residue at infinity okay what you do is you literally get minus of a minus 1 okay which is with an extra minus sign added to it okay whereas if you look at the usual Laurent series of a point around a point in the complex plane then the residue is actually a minus 1 which is just the coefficient of the first negative power of the of the variable okay whereas in this case it is the minus of that and so you know you can now believe that you can you can see from right from here you can see something happening this suppose this function f had only singularities at the origin and at infinity okay which means that this Laurent series has infinite radius of convergence okay so in by that I mean the Laurent series is valid for all w not 0 and not infinity okay so it is valid on c star c minus the origin okay if that is the case then you see this function if you calculate the residue at the origin you are going to get plus a minus 1 okay which is usual definition of residue if I if I take this function suppose it is also analytic in the name in a neighbourhood of the origin okay and suppose the only singularities are at the origin and at infinity okay then you see you notice that the residue of the function at the origin is plus a minus 1 and the residue of the function at infinity is minus a minus 1 what is the sum of residues it is 0 and that is exactly what the residue theorem is I say is for the for any function which has only isolated singularities in the in the extended plane okay so you know so the statement is that you take a function which has only isolated singularities in the extended plane okay which means that you know there are only mind you it means whenever you say isolated singularities in the extended plane there are to be only finitely many that is because the extended plane is compact and any subset of a compact isolated set subset of a compact set is finite in this case okay. So therefore you know what the residue theorem will say is that if you take the residues at all the points at the points in the finite complex plane and you take this residue at infinity and you add them up you will get 0 and here that is exactly what happens if f were having a residue if f was having a singularity at the origin only and a singularity at infinity okay then you see that from this computation the residue at 0 is plus a minus 1 the residue at infinity is minus a minus 1 the sum is 0 okay. So anyway so the moral of the story is that you know it gives you a very easy way of computing residue at infinity it is very very simple what you do is that you simply write the Laurent expansion of the function about the origin but write it so that it is valid in the exterior of a sufficiently large circle okay mind you you must always remember this that so this is probably maybe I should spend a few minutes on this see in a first course in complex analysis when you study about Laurent expansion at a point you must remember that there are several Laurent expansions there could be several Laurent expansions at a point that is because the Laurent expansions the domains of the Laurent expansions are actually annulite which are whose boundaries contain the singular points okay. So if I say if I talk about the Laurent expansion of a function at a singular at a point okay it could be even a analytic point does not matter but the problem is that because there are singularities there are other singularities the Laurent expansions will be different expansions because you will get different annulite okay and therefore when you write the Laurent expansion at the origin you should not look at the Laurent expansion at the origin that may be valid in a deleted neighbourhood of the origin which at whose boundary there is a finite singularity you should not look at that you should rather look at a Laurent expansion that is valid outside a circle of sufficiently large radius okay write that Laurent expansion that is the Laurent expansion that you need to work with infinity okay and in that Laurent expansion look at the again look at the coefficient of 1 by the variable and take it to the minus sign that is the residue at infinity okay and that is how you can very easily write out you can compute what the residue at infinity is and that combined with the residue theorem is another powerful tool for computing lot of integrals okay as we will see in the next lecture.