 Let us continue with whatever we were doing. So you know let me recall few things. So you see we started by asking what the image of an analytic mapping is okay and of course the if you we have already seen that you know because of the open mapping theorem the image of a non-constant holomorphic map is an open set and in fact it is an open mapping so it takes open sets to open sets so the image of a domain is again a domain okay and then I told you that we wanted to know what we wanted to know more about the image set okay namely the set of values that the analytic function takes and you know somehow the open mapping theorem tells you that for example you cannot expect the image of a non-constant holomorphic map to be to lie inside a curve for example okay because it is it does not have the property that it can accommodate as sufficiently a small disk okay. So then of course we where we stated the so called little Picard theorem or the small Picard theorem which says which deals with the case of an analytic function which is analytic on the whole complex plane so called entire function and the theorem says that the image will be either the whole plane or it will be the plane minus a single point okay and interestingly the proof of this theorem which is usually stated only stated in a first course in complex analysis involves interesting amount of analysis and that is what we will try to do as part of our this series of lectures and I told you the key to this proving this little Picard theorem is a so called big Picard theorem or great Picard theorem which has got to do with the image of a deleted neighborhood of an isolated essential singularity under an analytic mapping okay and the great Picard theorem what does it say it says the conclusion the great Picard theorem amazingly is the same as that of the little Picard theorem which says that you take a you take a deleted neighborhood of an isolated essential singularity of an analytic function then the image of that under the analytic function will be again the whole plane or it may be the plane minus at the worst one point or one value that is missed okay. So it is either the plane or it is a punctured plane and I told you that the proof of the little Picard theorem that we will give is going to be as a corollary of the big Picard theorem and to study the proof of the big Picard theorem or the great Picard theorem we need to study what are about Meromorphic functions okay and these are functions essentially these are functions which are these are analytic functions which have the only singularities as poles okay and so this led us to understand the and recall rather recall the notion of singularity so I told you that singularities of analytic functions are of two types namely the isolated and the non-isolated ones okay and of course the classic example of non-isolated singularities are the points on the negative real axis which have to be cut out if you want for example define an analytic branch of the logarithm of log z logarithm of z okay z being a complex variable. And I told you that these non-isolated singularities require much deeper techniques for example the study of Riemann surfaces to deal with them but then we are going to be worried only about isolated singularities and I told you the isolated singularities come in 3 categories of groups and these are mutually exclusive the first kind of isolated singularity is called removable because the idea is that the singularity can be removed in the sense that you can redefine the function so that it becomes analytic at that point and so it is not really a singularity then there are the poles which are thought of as 0s of the denominator okay and of course if the function does not have a denominator which is not writable as a numerator by a denominator then poles are just 0s of the reciprocal of the function okay and I told you that a function has a pole of a certain order if and only if the reciprocal function has a 0 of the same order at that given point and then what we are left with is are the singularities which are neither removable nor poles and they are cleverly called as essential singularities and they are very very important as the for example the great Picard theorem tells you. Now what I am going to do is first prove a certain weaker form of the great Picard theorem so called Casorati-Weierstrass theorem which says that you know if you take an isolated essential singularity then you take a small neighborhood of that deleted neighborhood of that no matter how small the image of that will be dense in the whole complex plane namely it will you can always find a sequence of complex numbers in any neighborhood such that the values approach any given complex value okay and that is a very deep that is already a very deep result but it can be you know reduced from the Riemann theorem on removable singularities so what I am going to do is now I am going to tell you something about the Riemann theorem on removable singularities because it involves a lot of nice concepts okay so here is Riemann's theorem theorem on removable singularities and it is a very very deep theorem because you know it already uses some of the most basic theorems in complex analysis in its proof okay so we will see that. So let me state that let Z not be an isolated singularity of the analytic function f of z then the following conditions are equivalent so number 1 Z not is a removable singularity that is a first condition and so here you can give several definitions of removable singularity but we will give the most natural one the most natural definition for removable singularity is that it can be removed namely that you can the analytic function can extend to an analytic function even at the point of singularity okay what it means is that you can find an analytic function which is also defined at the singular point and which equals the given analytic function outside that point okay that is what it that is what it means to say that the analytic function extends to an analytic function also at the singular point so that means you have by extending it you have actually been able to remove the singularity alright so of course you know the standard example you should think of is sine z by z at z equal to 0 you can you know the limited at z equal to 0 is 1 so you know you can take the function that takes the value sine z by z at z not equal to 0 and at z equal to 0 you can define it to be 1 and then this turns out to be an analytic extension okay so let me write that down that is there exists an analytic function function so g of z analytic at z not such that g of z is the same as f of z for z not equal to z not in a small neighbourhood of z not. So this is what it means to say that the singularity is removable I am able to extend the analytic function to the point to the singular point okay and to be able to extend the analytic function to the singular point means that I am able to find another I am able to find an analytic function which is also analytic at the singular point and restricts to the given function outside that point okay so let me rewrite it is very important in mathematics to be able to say things in different ways verbally without using symbols or notations as far as possible because that will help you to get a good understanding of the ideas okay so you should be able to state theorems at least as accurately as possible without using much notation and just using concepts okay this is something that you should strive to do you may be able to write technical mathematics namely you can write a theorem with all the technical symbols and so on but then it is very very important for the purpose of communication and understanding that you should also be able to say things in a way that does not involve any notation okay and so in that sense to when you say something is a removable singularity a point is a removable singularity what it means is that the analytic function extends to analytic function at the singular point okay so let me write that down in other words f extends to the to an analytic function g at z dot okay so this is the I mean so this is actually the definition of water removable singularities okay so this is the first condition okay so you can see that what this theorem does is that it gives you various equivalent conditions for a singularity to be removable so what is the second condition so you see in all these in all theorems connected with characterization of singularities there are usually at least three statements one is about the one is essentially the definition of that singularity the second one is the behaviour of the limit of the function as you approach the singularity okay and the third one is the behaviour of the Laurent series around that singularity okay so for example if you take the case of poles which I stated in the last lecture you see I stated that theorem in the last lecture and I wanted you to try to prove it and I do not know if you have done this exercise or not but essentially you see if you try to do that exercise at some point you might have to use Riemann's removable singularity theorem which I am going to actually prove now okay so but nevertheless the purpose of the exercise was to make you realize that you need to you might need to use this okay so you know in the case of a pole the condition for the first condition for a pole was that it is a pole namely which is the basic definition of a pole which is just that it is a 0 of the reciprocal okay the point is a 0 of the function which is the reciprocal of the given function okay that is the first condition the second condition is the behaviour of the limit of the function as you approach the pole and that condition turned out to be that the limit of the function turned out to be infinity okay and that is to be interpreted as the limit of the modulus of the function goes to plus infinity okay as you approach the singularity alright that is another characterization of a pole and what is the third condition the third condition involves the Laurent expansion and what is the condition based on the Laurent expansion the condition based on the Laurent expansion was that the Laurent expansion contained only finitely many negative powers of Z minus Z naught okay so you know a Taylor expansion is something that contains only positive powers and zero powers and this of course zero power corresponds to the constant term okay and the Laurent expansion is something that will also contain negative powers of Z minus Z naught okay where Z naught is the singular point and if you have a Laurent expansion which has only finitely many negative terms I mean terms with negative powers of Z minus Z naught that is an indication that Z naught is a pole okay now that these three conditions are equivalent was a theorem I stated last time so in the same way for removable singularities the first condition I have stated is what removable singularity is essentially and I will give you two more the two other conditions so the second condition is going to be a condition that is got to do with the limit of the function so the second condition says that the limit of the function exists as you approach the singularity okay just the existence of the limit is the second condition okay and why it is significant is because if the limit exists what you are saying is that the function extends to a continuous function at that point okay so mind you it is certainly weaker than the first condition because the first condition namely the definition of removable singularity is that it extends to an analytic function at that point whereas the second condition is the condition on the limit that the limit exists at that point only tells you that it extends to only a continuous function at that point okay so it is now so what you are saying is that the fact that you can continuously extend the function to the point already makes it analytic at that point this is a characteristic of removable singularity okay so let me write the second condition so here is the second condition the limit of f of z as z tends to z0 exists this is the condition okay so in other words in other words f of z extends to a continuous function at z0 namely we simply define f of z0 to be the limit at as z tends to z0 of f of z okay you define you redefine f at z0 and the resulting function becomes also continuous at z0 but what you do not have immediately is that it is also analytic at z0 okay and that is the serious that is the serious consequence that is the that is the Riemann removable singularity theorem okay just continuity at that point is good enough for analyticity okay that is the crux of the Riemann removable singularity theorem so this is the second condition and here is the third condition so I told you the third condition is something that has got to do with the Laurent expansion and you know that so I want to give you some background on this you know if you have a function which is analytic at a point then you have a so called Taylor expansion at that point okay if the point is z0 then you have power series in z minus z0 which involves 0 and positive powers of z minus z0 with some coefficients and of course these coefficients are you know they are just the related to the nth derivatives the derivatives of the function at z0 okay and so this is the Taylor's theorem that you can have a Taylor expansion for the function at a point z0 where it is analytic and an extension of the Taylor's theorem is the Laurent theorem it is an amazing extension because it deals with the case when z0 is not a point of analyticity but it is a point it is an isolated singularity what Laurent's theorem says is that you can still get a series but now this time you will also have to allow negative powers of z minus z0 that is the Laurent series okay and Laurent's theorem says that you can get that series and the coefficients are again given by integrals okay the see in the case of Taylor series the coefficients are given by integrals and these integrals are essentially connected to the derivatives by the general Cauchy integral formulas okay and in the Laurent expansion the coefficients of the negative terms the coefficients are all anyway given by the integrals okay there is no question of derivatives at the point z0 because at z0 is not a point where the function is analytic so the coefficients of Laurent series are given in terms of integrals okay. Now so you see the Taylor series is a very special case it is a special case of Laurent series okay and the fact that so this is the point if you have an analytic function at a point okay if you try to write the Taylor series at the point you will get the same result as if you try to write a Laurent series at that point okay see the fact is that if you have an analytic function we know that it is given by a Taylor expansion at that point but if you throw that point away okay then you get a deleted neighborhood of the point and in a deleted neighborhood of a point you always have a Laurent expansion for any function regardless of whether the function is analytic at that point or not okay and the point is that if you write out the Laurent expansion for an analytic function at a point okay you will get only the Taylor expansion you will not get the negative terms in the Laurent expansion the negative terms in the Laurent expansion constitute what is called the singular part or the principal part of the expansion okay and the principal part will not exist that is a sign of the fact that the function that you are actually expanding into a Laurent series is actually analytic at that point okay so that is the third condition there is a third condition in terms of Laurent series for a removable singularity is that when you write the Laurent series for the function at the removable singularity centred at the removable singularity you will see that the negative terms the principal part does not exist it is 0 okay so that is the third condition which is based on the Laurent expansion so here is the third condition the Laurent expansion of f of z at z not has no negative powers of z minus z that is it has 0 principal part principal or singular part okay this is the condition in terms of the Laurent expansion and then so these are the three conditions that you will always have as equivalent in any theorem on singularities characterization of singularities now there is one more condition is pretty interesting and it is a rather remarkable condition the fourth condition is the following if you have a removable singularity okay at a point you see the let us believe that is a removable singularity just to understand this condition then it becomes analytic at that point okay and if it is analytic at that point it is also continuous at that point okay and you know a continuous function on a bounded set okay if you take a continuous function on a close and bounded set it is going to be bounded in modulus okay because if you take a close and bounded set it is compact okay and if you take a real valued continuous function on a compact set the image is going to be also a compact set topologically the continuous image of a compact set is a compact set and a compact subset of the real line is going to be bounded okay so the point is that if you really believe that z naught is an essential singular is a removable singularity for the function it should the function should extend to an analytic function at that point in any case it is going to extend to a continuous function at that point for example that is what condition 2 says then in a neighbourhood of that point the function should be bounded in modulus of course see whenever we say bounded for a complex valued function we mean bounded in modulus okay that goes without saying okay so that is the next condition so it is an amazing condition you just assume that there is a small deleted neighbourhood of the point z naught where your function in modulus is bounded by a positive constant okay that is also as strong as saying that it can analytically extend to that point that is the really amazing hypothesis because it is a very weak hypothesis you see the first hypothesis the first condition is that the function is the it has a removable singularity namely which by our definition is that the function extends to an analytic function at that point okay. The second condition is slightly weaker it says that it does not extend to an analytic function at that point but it extends to a continuous function at that point okay that is a slightly weaker condition and of course the third condition has got to do with the Laurent series which says that essentially the Laurent series is a Taylor series but the condition I am going to state now it is a very weak condition it just says that there is a deleted neighbourhood of the singular point where the function is bounded in modulus you see the bounded function is a very weak condition okay but that is strong enough to make it analytic at the point. So that is the amazing power of the Rayman removable singularity okay so let me write that down f is bounded and of course bounded means in modulus in a deleted neighbourhood so I am using n B D for neighbourhood as an abbreviation of Z0 so this is the fourth condition which is by far the weakest condition on a removable singularity okay so you know why it is weak if you want if I might say a little loosely is that you know continuous function is always bounded but there are bounded functions which could be very highly discontinuous okay so the moral of the story is that boundedness giving rise to continuity is already something that is very hard to expect you should not expect that and in this case boundedness is giving me analyticity so you can imagine analyticity is a terrific condition because you know analyticity at a point means that you know not only that it is differentiable in the neighbourhood of the point including that point but it is also infinitely differentiable there so it is a terrific condition okay that you are able to get this from boundedness is an amazing thing that is what you should appreciate okay. So alright so now what I am going to do is I am going to you know prove that these various conditions are equivalent okay and in the process help you revise some basic complex analysis okay fine so let us look at these conditions so you know let us look at condition 1 and 2 you see it is very clear that 1 implies 2 okay if the analytic function extends to an analytic function at that point then certainly it extends to a continuous function at that point because an analytic function is continuous okay differentiability implies continuity so 1 implies 2 is very trivial alright and now so you can see so I am just trying to see which of these are very easy to deduce which of the equivalences are easy to deduce so 1 implies 2 is pretty easy okay and well I think if you look at 2 and 3 2 and 3 are equivalent okay 2 and 3 are equivalent that is very easy to see because you know you take the so let me so let me first tell you in words suppose you assume 2 suppose the limit of the function at as z tends to z0 exists okay then I can take the limit of limit as z tends to z0 in the Lorentz expansion and as limit z tends to z0 in the Lorentz expansion exists means that cannot be any negative powers of z-z0 in the Lorentz expansion because if you have a negative power of z-z0 in the Lorentz expansion it will be a term of the form An by z-z0 power n and that as z tends to z0 will go to infinity okay and you cannot get a finite limit. So the moral of the story is that 2 implies 3 is obvious and 3 implies 2 is also obvious because you know if the Lorentz expansion does not have any negative terms then I can take limit as z tends to z0 in the Lorentz expansion and essentially what I will get is a constant term okay as I if I take limit z tends to z0 I am just setting z-z0 equal to 0 z equal to z0 and then since it is already a power series in z-z0 if I put z equal to z0 I will get the constant term and that will be the limit okay. Now so 2 and 3 are clearly equivalent alright now there is only one technical point which I want you to notice because this is an advanced this is a course in advanced complex analysis the technical point is that you know how can you take a limit in the Lorentz series okay how can you take limit z to z0 z tends to z0 of a Lorentz series see so basically the we are arguing in the following way we are arguing as if we can take the taking the limit as z to z0 z tends to z0 in the Lorentz series is the same as taking the limit in each term and then summing it up okay that is the way we are arguing okay and why is that correct that is because the fact is because of the fact that you know the you see the Lorentz series as it is you know that is its convergence is uniform okay and whenever the convergence is uniform okay you can take a limit okay and so that is used okay you can take a term wise you take a series or you take a functional series okay and you take limit z tends to z0 of the functional series okay that is the same as taking limit z tends to z0 of each of the terms of the functional series and then taking the limit of the resulting numerical series this is allowed provided the functional series converges uniformly and basically I am just using the fact that if you have a convergent uniformly convergent series of continuous functions then the limit is also continuous okay that is all I am using so that is a little bit of technicality that is used when you want to prove 2 and 3 are equivalent okay and of course you know 1, 2 and 3 all the 3 will imply 4 because you know continuous function is bounded, continuous function on a compact set is bounded. So the difficult part is to go from 4 to 1 okay 4 is the weakest condition the condition 4 is the weakest condition is the condition that says that you just tell me that near z0 near that singularity the function is bounded and low behold it becomes analytic at z0 that is the most tremendous observation okay. So 4 implies 1 is the most difficult part that is the crux of the theorem okay which we will try to prove and essentially 1 will again essentially use Laurent expansions okay so let me write down all this so clearly 1 implies 2 which is equivalent to 3 and they all imply 4 okay so this is what we have seen the non-trivial part is 4 implies 1 okay so you know if you want 1 implies 2 basically uses analytic implies continuous okay and 2 implies 3 is going to use well uniform limit of continuous functions is continuous okay and of course 1, 2 and 3 imply 4 all these basically use the fact that the continuous function on a compact set is bounded okay continuous function on a compact set is bounded okay so the non-trivial thing to prove is 4 implies 1 namely that boundedness in a neighbourhood of the point gives you the analyticity at that point which is an amazing thing which is and interestingly the way you prove it is again using Laurent expansions okay you just use the Laurent expansion. So here is a proof recall that so let me go to a different colour recall that the Laurent expansion of f and z0 is f of z is equal to sigma n equal to minus infinity to infinity An z minus z0 to the power of n where An is 1 by 2 pi i integral over gamma fw dw minus z0 to the power of n plus 1 this is the Laurent expansion okay and where gamma is a simple closed contour which goes once around the point the singular point z0 in the anticlockwise or positive sense okay we can very well take gamma to be a circle centred at z0 sufficiently small radius okay. So now the point is that the point is to you know let me explain the idea of the proof see what are you trying to show you are trying to show that the function is analytic one way to show that the function is analytic is that its Laurent series actually a Taylor series because you know a Taylor series always represents an analytic function okay a convergent power series within its disc of convergence always represents an analytic function okay. So incidentally that is also that is also probably used in one of the earlier equivalences okay that is something that you have to remember if you have a convergent power series okay then it can any power series converges in a disc okay the disc could have possibly infinite radius in which case it is the whole plane as happens in the case of a polynomial or an exponential function and in that disc the convergence of the power series is always normal namely it is uniform on compact subsets okay and it is also absolute okay. So this is something that you should have come across in the first course in complex analysis where essentially you make use of the Weistra's M test okay so and a convergent power series by Abel's theorem is that it can be differentiated term by term and what you get is again a power series is the same radius of convergence and that is the derivative of the original power series and this is one way of showing that the derivative of a power series exist and it is gotten by differentiating it term by term the term by term differentiation is justified because of the uniform convergence okay and if you apply this ad infinitum what you get is that a power series is infinitely differentiable it is actually analytic and it is infinitely differentiable. So a power series whenever you are looking at a power series inside its disc of convergence you are actually looking at an analytic function and what has the analytic function to which it converges got to do with the power series this power series is nothing but the Taylor expansion of that limit that limiting function. So you start with the power series it converges within its disc of convergence to a certain function that function is an analytic function and if you expand it as a Taylor expansion at that point you will get back the power series. So the moral of the story is that whenever you are looking at a convergent power series you are actually looking at the Taylor series of an analytic function okay and what is that analytic function it is exactly the function to which this power series converges okay. So what we are trying to show is that we are trying to show that this point Z0 is basically a point where the function is analytic so what do you expect is that you expect the Laurent series should actually be a Taylor series that means all the coefficients of the negative powers of Z minus Z0 in the Laurent expansion should be 0. So you try to show that all those coefficients are 0 then you are done okay and how do you show those coefficients are 0 the coefficients are given by integrals and integrals can always be estimated by the so called ML inequality okay. So what we will do is we will show that all the negative coefficients in the Laurent expansion they are all 0 and we are done okay that is exactly what I am going to do. So what is An? An is 1 by 2 pi i integral over gamma fw dw by w minus Z0 to the power of n plus 1 where you know the pictures like this you have Z0 and you have gamma a simple closed curve going around Z0 sufficiently close to Z0 going around once okay. Now you take for take gamma to be the circle okay you could the shape of gamma really does not matter because of course it is theorem actually take gamma to be the circle mod Z minus Z0 is equal to epsilon for epsilon positive small enough so that the circle on the circle and inside the circle except at the point Z0 the function is analytic okay and well you know then you can so the equation Z mod Z minus Z0 is equal to epsilon you can use that write it as a parametric equation and do an integration okay. So this is the same as writing it as Z mod Z is equal to Z0 plus epsilon e to the i theta where theta varies from 0 to 2 pi okay and so this integral becomes so this integral so if I calculate the modulus of An mind you I am trying to show that the modulus of An is 0 I am trying to show that the An are 0 for negative N this formula is valid for all values of N okay I am trying to show An is 0 for all negative N that is good enough to say that the Lorentz is the Taylor series and that will tell me that the function is actually analytic at the point okay. So I have to calculate mod An mod An is going to be modulus of this integral okay and then you would have come across this estimation formula in a first course in complex analysis which is used all the time which says that the modulus of the integral is less than or equal to the integral of the modulus so this is less than or equal to integral over gamma of if I take the modulus I am going to get 1 by 2 pi mod Fw mod Dw by mod W minus Z0 to the power of N plus 1 okay and this integral becomes therefore just integral 0 to 2 pi because now I changed the variable of integration from W to theta mind you you should always remember that whenever you write an integral then you have an integrand and you have a variable of integration okay and the integrand is a function of the variable of integration and the variable of integration should always vary on the area on the region of integration okay in this case the region of integration is the curve gamma which you have taken to be a circle so your W is actually varying on the circle okay so W should be written as Z0 plus epsilon e power i theta okay. So what I will get is I will let me put this mod F Z0 plus e to the i theta and then I will have to write out so I should change this Z to W so it will become W equal to Z0 plus e to the i theta Dw will be i epsilon e to the i theta D theta and if I take mod Dw I am going to get epsilon D theta okay and then I am going to get here I am going to get epsilon mod W minus Z0 is epsilon e to the i theta its modulus is epsilon so I will get epsilon to the n plus 1 okay so this is what I am going to get alright and you know what I am going to do next you see whatever I assumed I assumed condition 4 condition 4 is that the function is bounded in modulus in a sufficiently small neighbourhood so you know this term mod F of Z0 plus e power i theta I am going to remove that and put an M because mod of F Z is going to be less than or equal to M in a sufficiently small disc and I am assuming that epsilon is small enough so that this circle lies in that disc in that deleted neighbourhood of Z0 okay. So what I am going to the next step is I am going to get rid of this mod F Z0 plus e to the i theta I am going to pull that out and instead of that I am going to put an M I am going to get it I am going to get oh I forgot 2 pi this is 1 by 2 pi outside and then so you know what I will get is I will get basically I will get M times epsilon times 2 pi divided by 2 pi times epsilon M plus 1 okay mind you when I integrate 0 to 2 pi D theta I am going to get 2 pi okay and that 2 pi is going to cancel the 2 pi outside so basically what I am going to get is I will get mod An is less than or equal to M times M by epsilon power M this is all I am going to get assuming that by 4 mod F Z is less than M for all Z close to Z0 where B assume also gamma lies okay. So I am going to get this now watch see epsilon is a small quantity okay epsilon can be made as small as I want I can make epsilon smaller I can make epsilon 10 to 0 okay. Now if N is negative if N is negative mind you I am trying to show that the An's for N negative are 0 I am trying to show that all the negative terms in the Lorentz expansion do not exist okay so all the negative Lorentz coefficients are all 0 I am trying to show so if N is negative that is the case I have to look at then this epsilon power N will go to the numerator so I will get a small quantity to a positive power okay and if I now let the small quantity go to 0 okay its positive power will go faster to 0 so the numerator will go to 0 and since this is valid for all epsilon greater than 0 mod An has to be less than or equal to 0 and that will force mod An is 0 and that will force at An is 0 okay and we are done okay and that is the end of the proof okay. So let me write that down now if N is negative then epsilon power N tends to 0 as epsilon tends to 0 and so mod An equal to 0 implies An equal to 0 okay so this implies that the principal part in the Lorentz expansion is 0 that is what we have actually proved is that we have proved that you know 4 implies 3 in fact you have actually proved 4 implies 3 and so 4 implies 3 alright and of course 3 mind you is equivalent to 3 implies 1 because if the Taylor see if the Lorentz expansion is a Taylor expansion okay namely if it has no principal part then it is a power series it will converge to a function and that function is going to be equal to the given function outside that point and therefore what happens is that you extended that function analytically to that point also okay. So let me write that down there is a little bit of there is a little bit of technicality so let me write this down we observe that 3 implies actually 1 for if 3 holds we have sigma N equal to 0 to infinity An Z minus Z naught power N converges to G of Z and converges to G of Z in mod Z minus Z naught lesser than epsilon and G is equal to F for Z naught equal to Z naught okay so see so let me repeat this see if you assume 3 what does 3 say it says that F of Z has a certain Lorentz expansion in which there are no negative terms if F of Z what does it mean it means that you have first of all a Lorentz expansion which converges to a function that function to which it converges is none other than F of Z okay and this is valid whenever Z is not equal to Z naught but what is the Lorentz expansion that converges to F of Z it is actually a Taylor expansion but you know namely it is just a convergent power series and you know the convergent power series is actually Taylor expansion of the analytic function to which it converges so there so you take only the Lorentz expansion which has 0 principle part it will converge to an analytic function call that function as G of Z now that function is going to coincide with F outside Z naught by definition because you already know that the Lorentz expansion also converges to F outside Z naught so in principle what has happened is you have found an analytic function G of Z which is analytic at Z naught and outside Z naught it coincides with F okay so finally this proves 4 implies 1 alright and that completes the proof of the Riemann removal singularity theorem that is the end of the proof which I will signify by putting a usually in books you see that people put a shaded square I will put something like this to indicate end of proof okay but there is a remark that I want to make so the remark is to fix some loose ends in the statement of the theorem in the so I am going back to the first condition okay the first condition namely the definition of a removal singularity what is the condition the condition is that the singularity is removable namely that there is a continuous function there is an analytic function to which this function converges what I want to say is that you see the condition only says that the function converges to an analytic function at that point but it does not say that this analytic function is unique okay so the first condition which is just the definition our definition of a removal singularity is that your function can be extended to an analytic function at that point okay but I say it can be extended to an analytic function I am not saying it can be extended to a unique analytic function and the fact is that it can be extended to a unique analytic function and the reason is that there is a deeper theorem behind us suppose it extends to 2 analytic functions at that point okay then you use the identity theorem which you should have studied in a first course in complex analysis which says that if 2 analytic functions coincide on an open set non empty open set or for example even if they coincide on a sequence of points which has a limit point at which both of them are analytic then they have to be identically equal. So that identity theorem will tell you that the if the singularity is removable then the function to which the analytic function to which the given function extends at the singular point is actually a unique function okay so there is the identity theorem there okay that you should remember okay so I will stop with that.