 Okay, so what we need to discuss now is analytic continuation okay, so the first thing that I am going to talk about is what is called as analytic extension okay, so let me explain this analytic extension. So what is this notion of analytic extension, see the idea is that to give an analytic function okay, there are many ways alright, of course I am considering an analytic function on a domain which is an open connected set of the complex plane alright and an analytic function can be given by a formula or it can be given by a power series alright or it can even be given as the integral of a function okay, there are so many ways alright. Now but the problem is that if an analytic function is given in a certain way, say it is given by a power series okay, centered at a point, of course you know that the power series will converge in the disc of convergence okay but outside the disc of convergence what happens you do not know alright. Similarly I may give an analytic function by a formula or by some properties in a domain, I do not know whether that formula will hold outside the domain okay or whether it will define an analytic function outside the domain okay, so the question of trying to see how far this you can find an analytic function, how far means on a largest possible open set on which you can define this analytic function largest possible open connected set okay, that is the question that we first need to understand okay. So you know I will give you an example, so you see so let me say this, so the analytic extension the other word I would like to use is direct analytic continuation, this is the other word I would like to use and what is this, so you see start with a pair D, F where D in C is a domain which means it is an open connected set of course non-MT and F from F is a function that is defined on D it is a complex value function and it is analytic on D okay, analytic or holomorphic okay start with the start with the pair like this okay, we say that another such pair D prime, F prime okay is a direct analytic continuation, analytic extension of the original pair D, F if F restricted to if D and D prime intersect D and D prime intersect and F restricted to F restricted to D intersection D prime is equal to F prime restricted to D intersection D okay. So the roughly one picture that you can think of but of course it is not the best picture because I am in this picture I am only considering bounded domains which is simply connected so you know this may be a domain D on which you have function F with values in complex plane and well and this side of course is a complex plane and you may have another domain D prime and I would have an analytic function F prime defined on D prime D prime and the condition is that where D and D prime do meet and in the intersection which is also an open set F restricted to D intersection D prime is the same as F prime restricted to D intersection D prime we say that the pair D prime, F prime is a direct analytic continuation of the pair D, F okay and of course you can see that it is also the same as saying that D, F is a direct analytic continuation of D prime, F prime and why what is so special about this this the thing that is special about this is that these two functions glued together to give an analytic function on the union okay. So what you can do is you can define G so in this case in such a case define G from the union to C by G restricted to D is equal to F and G restricted to D prime is equal to F prime you define it like this this definition makes sense because if you go to D intersection D prime G restricted to D intersection D prime will be F restricted to D intersection D prime and G restricted to D intersection D prime will also be equal to F prime restricted to D intersection D prime but they are one and the same because of this condition okay. So this is just the gluing condition it tells you that the function F and the function F prime they give one and the same analytic function on the intersection okay and the intersection is non-empty right and what we say that G is obtained by gluing F and F prime okay. We say we see that G is analytic and we say that G is obtained by gluing F and F prime along D intersection D prime okay. So you have just glued the functions together to give you have two sir you have got two open sets and you have got two functions defined respectively on those two open sets and what you have done is you have put those functions together to get a function on the union of the two open sets and it makes sense because they coincide on the intersection alright and of course analyticity is not an issue because analyticity is locally defined this function G is on this set if you take a point in the union the point has to either lie on in this set or in this set and if it lies in this set then G is the same as F and F is analytic so G is analytic if the point is on this set then G is F prime so again it is analytic so G is analytic because analyticity is a local property okay so this G that you have defined is analytic right. Now the point is that so this is a very simple concept alright but the point is that the way in which F is defined may be very different from the way in which F prime is defined okay. So in other words I am saying whenever you want to specify a pair D, F or D prime, F prime you are trying to specify an analytic function on an open set okay but that can be given in many different ways you can do for example analytic function can be given as a power series because you know power series is an analytic function okay it can be given by a formula alright involving say some standard functions like polynomials, the standard trigonometric inverse trigonometric exponential logarithmic functions okay or it can be also given by an integral of an analytic some other analytic function alright with the variable being the upper the end point of the path of integration which starts from a fixed point in the domain provided the integral is well defined okay. So there are different ways of defining the analytic function now the problem is that from the way in which analytic function is defined it is not it is not at all sometimes it is not very easy at all to guess that it really extends to a region beyond which it is defined okay. So I will give you an example so the point is that you know the first question is if you give me a pair can you find an analytic extension a direct analytic continuation for that in a domain which is which contains points different from the original domain okay. So the problem is suppose I start with the pair D, F namely analytic function on a domain D my question is can I find a can I extend it to an analytic function on a bigger set. So in this case you know what has happened is both D, F and D prime, F prime have been extended to G on D union D prime okay. So the question is therefore the idea is therefore to try to see if at all there is a largest open set on which you can extend it okay that is the first question. The second question is on the largest open set the new function that you get can you describe it in some way okay that is the question. So just to tell you the kind of things the myriad things that happen so I will give you an example so you know the first example is rather very interesting so take D to be a set of all Z such that mod Z less than 1 is unit disc okay and take F of Z to be 1 plus Z plus Z squared plus Z cube the geometric series okay you take the function and take geometric series. Now you know that F of Z this is a power series centred at Z okay so if I draw a diagram my situation is like this so here is unit disc so I have unit disc so this is my D and here is this function is given by power series you know the radius of convergence of that power series is 1 alright and therefore inside the disc of convergence okay the disc of convergence in this case is unit disc okay the region inside this unit circle okay there you know the power inside the disc of convergence the power series always represents an analytic function and what is the analytic function that it represents it is that analytic function for which if you write out the Taylor series expansion at the centre of that disc you will get back the power series. So this is an analytic function okay for this analytic function if you again try to write the Taylor expansion at 0 you will get back the series so I am just saying that if you start with a power series okay centred at a point then inside the disc of convergence around about that point the power series represents an analytic function for that analytic function if you write the Taylor series about that point you will again get back the power series because the power series this is the statement that you often see in a first course in complex analysis which says that power series are analytic functions okay and the proof is essentially the fact that the power series if you take a disc which is if you take a disc close disc which is contained inside the disc of convergence the power series converges uniformly and absolutely and in fact therefore what will happen is that it can be it can be differentiated infinitely many times and the differentiate and every time you differentiate it the new power series will also have the same radius of convergence the fact that you can differentiate it tells you that is analytic that is why power series are analytic okay. So here is an example so here is my power series and now the point is so here is my pair d comma f alright now you know you can easily guess that there is a you know there is a way to extend this okay there is a way to extend this because you know that this if you consider now you consider d prime to be the plane minus the point 1 and you consider f prime I think f prime is a very bad notation so let me use d1 f1 is well 1 by 1 minus z okay you take this. Now this function f1 is certainly analytic on though see this is a this is just reciprocal of a polynomial and you know reciprocal of a polynomial is of course an analytic function everywhere it is an entire function and the reciprocal of an analytic function will also continue to be analytic so long as the denominator does not vanish and the denominator vanishes only at z equal to 1 alright. So this so far as so long as z is not equal to 1 this is an analytic function so here is another pair and the fact is that these two this pair is a direct analytic extension of this because you know for 1 by 1 minus z if you take this function and try to write out the power series at the origin you will get exactly geometric series which is which you know as you write 1 by 1 minus z as 1 minus z to the minus 1 and then you use binomial theorem and expand it you get the geometric series 1 minus z to the minus 1 is 1 plus z plus z square that is what I have written. So you can see that d1,f1 is an analytic extension or direct analytic continuation of d,f and the point is that it is rather what is really mysterious is the following. See if you consider the functional form as given by the power series in this form the function does not live at any point on the boundary okay because you know if you take any point on the boundary it will be of the form e power i theta it will be a complex number of modulus 1. So it will be of the form e power i theta for theta real okay where of course theta is this angle if you want alright and if I plug in e power i theta into this series then the nth term is e power i n theta it is modulus is 1. So if I evaluate this power series at any boundary point of the circle the nth term always has mod 1 therefore the nth term does not go to 0 and if you know for a numerical series if the nth term does not go to 0 it cannot converge. A necessary condition for a power series to converge is necessary condition for a numerical series to converge is that the nth term should go to 0 okay. So this tells you that this power series does not live even on the boundary it lives strictly inside okay it is life is only inside even on the boundary it does not make sense okay notice that this is equal to this which lives everywhere except the point 1 okay 1 by 1 minus z of course lives everywhere except for z equal to 1 because z equal to 1 is a pole of order 1 it is a simple pole it is a 0 of order 1 of the denominator of the function right. So the moral of the story is that if I give you a if you give you if I give you an analytic function in a certain way the way I have defined the analytic function may restrict it from extending you know in the form I have given it to you beyond the region that I have given. So I cannot expect this power series to extend even to the to any point on the unit circle okay but that does not mean that the analytic function does not extend. So the analytic function as a function is actually 1 by 1 minus z it extends the problem is with trying to only look at it as a power series it is a the power series has limited life only on the interior of the unit this but it does so what you must understand is so what is it that is happening I have this power series that certainly gives me an analytic function but just because the power series has a life only sense makes sense only inside the unit disc does not mean that the analytic function it represents lives only there. The analytic function which actually it represents may live on a much bigger open set and the whole theory of analytic extension analytic continuation is to try to find whether given a function on a domain which may be given by a power series or whatever it is you have to find out whether it really extends okay and it is a pretty involved kind of problem alright I will give you I will give you another example. So here is which is very similar to this and that is the example of zeta function so you know incidentally before I do that I want to tell you that you know if I try to make an analytic extension of a pair the analytic extension that I will get if I take an analytic extension of a given pair okay then that extension is unique and that is just because of identity theory okay see so here is a uniqueness the uniqueness is well if D prime, so even here I think it was bad notation you have used F prime because it would confuse you with the derivative of F so maybe I will better late than ever I will fix to D1 F1 and change everything to I will change all the super primes to sub primes so let me do this so of course when I wrote F prime I did not mean the derivative of F okay so let me change that sorry for the bad notation so let so the point I want to make is if D if D1, F1 and D2 and D1, F2 are analytic extensions of D, F then F1 is equal to F2 okay this is just a consequence of the identity theorem because what you will get is you see why is that true because you know if you take F1 and restrict it to D1 intersection D or that is D intersection D1 I will get F restricted to D intersection D1 and that is supposed to also be equal to F2 restricted to D intersection D1 okay this equality is because F1 is analytic extension of F and this equality is because F2 is analytic extension of F so what will happen is you have two analytic functions F1 and F2 defined on the same domain D1 and on a smaller open set D intersection D1 which is non-empty on a non-empty smaller open set they are equal therefore they are the identity theorem will tell you that they are throughout equal so let me recall what the identity theorem says. The identity theorem says if you have two analytic functions defined on a domain and suppose their values are equal even on a convergent sequence of points okay and with the limit of the convergent sequence in the domain of analyticity okay then they are always equal their value at every point is equal so to check that two analytic functions are equal on a domain all you have to do is to find just one sequence of convergent sequence of points in the domain which converges to a point in the domain and verify that for each point of the sequence the two analytic functions take the same value okay. So and of course if you give me a non-empty open set I can find so many convergent sequences there okay because it will always contain a disk and I can always take a convergent sequence of points going to the center of the disk if you want okay along the radius. So the identity theorem will tell you that F1 and F2 are one and the same so direct analytic continuation to a given domain is it has to be unique okay. So this is because of this and the identity theorem so the analytic the direct analytic extension is unique alright. Now what I want you to understand is that of course there are situations where if you give me a domain and you give me another domain which intersects it analytic extensions are not possible. So for example you know if I if you in this example suppose I continue with this example suppose I take D2 to be D2 is equal to any domain in C containing the point 1 okay. Then there is no direct there is no analytic extension of this function to the point 1 okay there is no analytic extension of this function to the point 1 because of the simple reason that the point 1 is a singularity see the maximal analytic extension of this function is the function 1 by 1 minus z that is the maximal analytic extension that is the largest possible open set on which this the function representing this power series the analytic function that represents this power series can be extended okay. So the function is 1 by 1 minus z the largest set on which it may which is this analytic is leaves out the point 1 because the point 1 is a pole it is not a removable singularity if a point is a removable singularity then you can correct it by defining the value of the function at that point to be the limit that you get as you go to that point but this is not a removable singularity it is a pole it cannot be corrected okay. So if you take any domain which contains that pole there is certainly no analytic extension of this function to that okay. So then there is no analytic extension d,f or d1,f1 to d2 there is no analytic extension okay. So of course this fact I made about uniqueness is true for any analytic extension and here I am returning back to this example right. So there are so the idea is that you cannot expect an analytic extension across a singularity right. Now so the other example that I want to talk about is the Riemann zeta function okay. So that is a function which is so you know you define the so let me give this example d so this is another example let me call this example 2 so this is example 1. So let us go to example 2 define zeta of z to be let me take it from n equal to 1 to infinity okay. So I will start with n equal to 1 to infinity and so this is just if you write it it will be 1 by 1 per z plus 1 by 2 per z plus 1 by 3 per z and so on right. So I will define this okay and by that I mean of course sigma n equal to 1 to infinity as I just explained it is 1 by e power z ln n okay and the point you must understand is that each of these functions ln n is a constant okay ln n is a real logarithm of n that is a constant. And z into ln n is of course analytic function because it is a polynomial of degree 1 okay it is just the analytic function z multiplied by a constant and e power that is also an analytic function. So each of these functions is an entire function each so each each function 1 by e power z ln n is entire each of these is an entire function okay and mind you since I have put an exponential the denominator can never vanish so I can put it in the denominator okay. So whenever I put something in the denominator I should always be worried about whether it vanishes but what I have put in the denominator is an exponential and you know exponential never vanishes so it is always defined and it is always analytic. So these are all entire functions so if you take this sum this is again an entire function okay but where so there is a there is an issue there so what you do is you take the so here is my so let me write D let me define D to be the right half plane okay to the right of the point to the right of the vertical line passing through z equal to 1 okay. So you define it like this so you take so here is a diagram so here is the complex plane okay and here is the point 1 and I draw this line this is the point real part of z equal to 1 okay this is that line the real part of z is x so this is the line x equal to 1 line parallel to the y axis this is the imaginary axis and I am taking this region so this is my D it is a right half plane okay so D is a set of all z in C complex numbers such that real part of z is greater than 1 imaginary part can be anything okay. So this is the right half plane it is a right half plane that is to the right of the line the vertical line passing through z equal to 1 which is given by the equation real part of z equal to 1 okay. Now the fact is that zeta is actually an analytic function on D so fact is zeta is analytic on D okay function zeta is analytic on D. So in fact probably you have seen this in a first course in complex analysis first of all this is a series okay it is a functional series with the nth function given by 1 by e per z ln n okay first of all functional series need not converge. So first of all for this to make sense as a function it has to converge okay and the fact is it will converge okay and then the next fact is that this convergent function once it converges it defines a function zeta it is called the Riemann zeta function okay and the fact is that this Riemann zeta function is actually analytic okay. So the so what is the so let me briefly recall how you prove such a thing so what you do is well you take you take you take a line to the right of this given by real part of z equal to 1 plus epsilon where 1 plus epsilon where epsilon is positive and you can take epsilon as small as you want but you take a vertical line to the right of this line okay and what you do is you use the Weierstrass m test to show that this function converges uniformly and absolutely on the right half plane starting from this line and including that line. So the so the Weierstrass m test shows that zeta z converges in the set of all z such that real part of z greater than or equal to 1 plus epsilon both uniformly and absolutely that is the Weierstrass m test okay. And so you know one can easily check that it is pretty easy to check that so let me let me continue here in in this in this right half plane close right half plane given by real part of z greater than or equal to 1 plus epsilon what you have is z is x plus i y and x is greater than or equal to 1 plus epsilon okay so what you will get is it is a it is an estimation so if you calculate e power if you take the nth term of that of that functional series it is 1 by e power z ln n okay you see so it is equal to 1 by x plus i y ln n ln n which is that is right it is 1 by because it is multiplicative so it is x to the ln x ln n d e power i y ln n and as you rightly pointed out the mode of this is 1 so I end up with 1 by e to the x ln n this is what I get that is good and now I think more or less than because x is greater than or equal to 1 plus epsilon x ln n is also greater than or equal to 1 plus epsilon ln n and e to that will also be greater than that and the reciprocal will go the other direction so you see x greater than or equal to 1 plus epsilon will tell you that x ln n is greater than or equal to 1 plus epsilon ln n and that will also tell you that e to the x this ln n is greater than or equal to e to the 1 plus epsilon ln n that is because ln n is, n is greater than or equal to 1 so ln is positive that is why this inequality holds and then the exponential function for T L's is an increasing function okay so this holds so this will be the same as e to the x ln n because it is a real number okay and that is going to be less than or equal to 1 by e to the 1 plus epsilon ln n and that is 1 by n to the 1 plus epsilon this is what it is okay. So if you take the nth term of the, if you take the nth term of this series then in modulus it is dominated by this numerical term okay and if you take the corresponding numerical series for this what you will get is you will get sigma n equal to 1 to infinity 1 by n to the power of 1 plus epsilon okay if you take this, this is convergent this is a fact that you would have learnt in any first course in analysis if you take sigma 1 by n power alpha okay that will always be convergent for any alpha greater than 1 alright. So this is convergent so the moral of the story is your whole if you so you know if I start with the, if I start with this functional series for the zeta function if I take the absolute series that means I take the series that you get by taking absolute values for each term then the absolute series is dominated by this numerical series which is convergent okay and this is exactly the situation of the Weierstrass M test which says that whenever a functional series is dominated by which is a absolutely dominated by a numerical series which is convergent then the original functional series converges uniformly and absolutely. This is the of course since the absolute series itself so first of all there is a dominated convergence theorem which says that if a series is dominated by another series and the dominating series converges then the original series converges. So what this will tell you is that if I take sigma if I take the absolute series for the series corresponding to the zeta function that will converge and you know it is also a fact that absolute convergence implies convergence okay therefore that will also tell you that the original functional series for the zeta will converge. So the moral of the story is that the original series that was used to define the zeta function that converges both absolutely and uniformly and that happens for every epsilon greater than 0 and therefore if I make epsilon small enough I can cover every point in the right half plane right to the right of the point to the right of the line x equal to 1. By taking epsilon small enough I can cover every point therefore this gives you the fact that the zeta function the series for the zeta function actually converges for every point lying in that right half plane to the right of the point to the right of the line vertical line passing through z equal to 1 okay. So this implies that zeta z converges absolutely and uniformly on in the set of all in D which is the set of all z such that real part of z is greater than 1 greater than or equal to 1 plus epsilon this is not D 1 plus epsilon for every epsilon greater than 0 okay. Now from this there are 2 facts that you can get the first fact is that zeta of z converges in D that is the first fact because any point of D I can make it lie in a region like this by taking epsilon small enough okay so zeta converges in D okay. The second fact is not only does it converge I now claim it is analytic I am claiming that the zeta is analytic why is that so that is so because of a fact that I had stated in several lectures ago namely whenever you have a sequence of analytic functions if it converges normally okay whenever you have a series of analytic functions or if you have a sequence of analytic functions if you have a sequence of analytic functions if the sequence converges normally then the limit function is also analytic okay this is a fact that I stated in an earlier lecture and the proof was essentially that the limit function the normal convergence will mean that it is convergent on it converges uniformly on compact subsets and this uniform convergence will ensure that the limit function is continuous and then what will happen is that analyticity will follow from Morera's theorem and you use also the fact that because of uniform convergence you can interchange limit and integral okay. So you can go back to that earlier lecture where I proved in detail that you know if you have a normal limit okay if you have a normal limit of analytic functions and the limit function is also analytic so what is happening in this case is that you can consider this function zeta as this series but what is a series a series is just limit of partial sums and the partial sums are analytic because they are sums of analytic functions in fact they are the partial sums are entire functions okay so you the partial sums of this series are entire functions they are analytic functions and the series converges uniformly on any such closed right half plane. Therefore on an interior of such a closed right half plane it will represent an analytic function because the only condition that is required is you must have normal convergence you must have convergence which is uniform on compact subsets but you here you have uniform convergence on the whole closed half plane that is the right half plane including the boundary which is aligned to the right of z equal to 1. So that theorem will tell you that zeta is actually analytic that is how you get analyticity okay so and is analytic indeed okay so this is the zeta function this is the famous zeta function and so here is my here is my here is my pair which consists of the famous zeta function and this domain which is the open right half plane okay which consists of all points to the right of this line okay and the question is can this be analytically continue can this be analytically continue. So this question is very similar to the simple first example that we saw the simple first example also consisted of a function which is defined by a series on a domain in this case a series was a power series and the domain was a unit disk and one could see by introspection by inspection that well one could see that it is just 1 by 1 minus z and therefore one was able to guess that it could be extended to the whole complex plane except the point z equal to 1 which is a pole okay. Now the amazing fact is the following the amazing fact is the zeta function also admits an analytic extension to the whole plane except the point z equal to 1 exactly like the geometric series okay when you try to extend the geometric series to the whole plane it admits an extension and the extension has a trouble only at z equal to 1 and at z equal to 1 what kind of trouble does it have it has a pole of order 1 okay it has a simple pole of it is a simple pole at z equal to 1 amazingly enough zeta also extends to the whole plane except the point z equal to 1 and at z equal to 1 it has exactly a simple pole okay so this is what happens but to prove this is not trivial it needs theory of analytic continuation okay. So the point I want to make is that trying to find whether a function has an analytic extension is not a easy problem okay but it happens that in the case of zeta function it can be extended to the whole plane minus this point what is the problem with this point at this point you know what is going to happen if the series becomes a harmonic series okay and the harmonic series is it does not converge okay so that is the only troublesome point okay it is hard to believe that it can extend to all other points on this line except the point z equal to 1 where all the other points on the line also seem to be very difficult points but the truth is it extends everywhere the only point where it cannot be extended is z equal to 1 where it becomes a harmonic series and that for the extended zeta function the zeta function which is a full extension of this function to the whole plane minus the point z equal to 1 that point z equal to 1 is only a simple pole okay so this is a beautiful fact okay and we will have to prove it I will try to prove it okay. So I just want to summarise by saying that you the problem of finding an analytic extension can be very simple or it can be pretty complicated okay theorem zeta extends to an analytic function that is it admits an analytic extension on c-1 which has a simple pole at z equal to 1. So this is a so this needs proof you will try to prove this at some point but so this is the so this is the story about trying to extend an analytic function. Now I want to tell you that it is purely a matter of abstract mathematics to ensure that if you start with a pair consisting of an analytic function on a domain okay a pair consisting of a domain and an analytic function on it namely a pair of the form d,f then there exists a maximal extension of that okay so we will continue with this discussion in the next lecture.