 So, last time we saw that a power series is uniformly convergent inside its radius of convergence. And the complementary theorem is that outside the radius of convergence it diverges. I am assuming here that this power series is expanded around 0, but this result will hold if it is expanded around any other point as well. For any point outside this disk f is absolutely divergent and the proof is actually almost directly follows from the definition of radius of convergence. What was that? The definition of radius of convergence was that it is a maximum or the limit of r's for which individual elements of this converge to 0. So, when you go outside that then this limit does not converge to 0. In fact, that means there are infinitely many elements here. In fact, let us be a little more formal. So, let us choose the modulus of z b r and s p a number between r and r and this sequence does not converge to 0. This is at least some non 0 number let us say some epsilon which is greater than 0 and this implies that now this is a number bigger than 1. So, this actually shoots up to infinity. So, inside the disk it uniformly converges outside the disk it diverges on the circle defining this disk. It could be either way it may converge may not converge that depends on the particular series that we are considering. Yes, it might be in peculiar cases it might still converge in a non absolute sense, but that those are bizarre convergences anyway. So, we will not get into there the minimal sensible convergence that we want is absolute convergence. So, which means that the behavior of a power series outside its disk of convergence is wild completely when it is essentially diverges everywhere. So, it is only make sense to study a power series inside its disk of convergence and which is very interesting because it actually defines this nice circle. So, it is not convergent inside some funny shapes convergent is precisely inside a circle unlike an analytic function which may be convergent inside strange boundaries good. So, now that we have the convergence issue settled for a power series we want to see when it is convergent what can we say about a power series and the natural thing we want to say about a power series is given by this theorem. So, again I am eating away that the point around which we are taking out power series as assuming it default to be 0. So, if f is a power series with radius of convergence r then f is analytic inside the disk and the proof essentially follows from the uniform convergence of f inside the disk and that is why we insisted on uniform convergence. So, what does it mean that to say that f is uniformly convergent? We say that let us define the sequence of polynomials P m z which is a truncation of the power series up to the first m plus 1 terms each one of this is a polynomial and this sequence of polynomial of naturally converges to the power series in a uniform way each one of this is analytic that we know every polynomial is analytic everywhere actually right. So, in particular each of this polynomial is analytic inside this disk and we will just use this fact and uniform convergence. So, let us see how do we use it to show that f is analytic we want to show one that f is continuous inside the disk and two that f is differentiable inside the disk or convert or alternately and that is again what we will use we prove Morera's theorem for f that f vanishes inside a rectangle. So, how do we show that f is continuous? What do we need to do to show f is continuous? Well if you look at let us say f z plus delta z we would like to show that as delta z goes to 0 the value converges to f z for every polynomial it is true that is that is the starting point yes. So, let us see so what let us see is this so for every let us write down for once this proof because this will run across this kind of arguments a lot of times and whenever there is a uniform convergence these arguments just go through. So, just to make sure or convince yourself that these arguments do go through we will do it for once firstly we observe this and this is should be less than equal to some epsilon m or let us call it delta m because I will be using epsilon somewhere else where of course delta m depends on the actual value of delta z all we will always assume the delta z is very close to 0. So, we can say this and we know that as delta z goes to 0 this goes to 0 and now what we want to show is that so let us look at f z plus delta z minus f z you want to show that this is also very close to 0 and the way we do it is we will let us first consider this quantity this is and this is less than equal to we can just write this as this plus this this by absolute convergence is less than some epsilon m this part right and that value of epsilon m is independent of the actual point. So, whether is z plus delta z or z it does not matter fine and again the same this quantity is also less. So, this is less than equal to 2 epsilon m and now this follows that this. So, this quantity is very small delta m. So, the maximum value of this quantity is 2 epsilon m plus delta m and as delta z goes to 0 that vanishes how it does not vanish does it. So, we have to send both delta z going to 0 and m going to infinity. So, then it is all vanishes this goes to 0. So, f is continuous and let us prove that now I have to pick up another symbol for a rectangle we know that by Cauchy that for any P m this integral is 0. So, now let us look at this this is equal to because this integral is 0 we can do this and this is. So, the absolute value of this integral less than equal to this we know is bounded by some epsilon m and then this integral around the rectangle is some finite quantity and as m goes to infinity this goes to 0 that makes f analytic inside the disk which is nice that now we know another whole class of functions we have shown to be analytic. So, the fact that e to the z or sin z cos z turned out to be analytic is no coincidence it is all power series are analytic. So, that is good to know now the next question is what else is analytic we had polynomials being analytic now power series are analytic what more kind of functions are analytic and the answer is quite nice nothing more this is it almost. So, let us start with an analytic function on some domain D. So, there is a domain and f is analytic on the let us pick any point inside this domain let us call it z naught f is analytic on z naught. Now to show that f is a power series we cannot do it point number 1 on the entire domain we may not be able to do it let us say on the entire domain because as we saw power series is precisely defined inside it circular disk. So, if f is not analytic outside this domain for example, and this domain is funny shaped and if you pick this point the maximum you can hope to show that f is analytic when you expand it as a power series around z naught is this the largest disk that fits in inside this domain where z naught is the center that is a maximum you can hope to do and that is what we will show with a k being. So, the theorem says that if f is analytic on D and we pick up a point z naught inside D then for any for the largest disk essentially that fits inside the domain with z naught as center f on that disk f can be expressed as a power series in z minus z naught with coefficients given by this integral again this constants I am not sure also as we derive we will figure out the exact value. So, that means there is this unique power series which expresses f in this disk you can almost guess the proof actually is yeah almost that and use uniform convergence essentially that is it. So, let us prove this the simplest way of doing it is like through Cauchy's integral formula this integral is w minus z naught that is what Cauchy integral formula tells that f of z is equal to this integral. So, z is inside the disk and you integrate along the circle and this is the quantity you integrate note that. So, I have to be a little careful here in fact in this specifying this I should write for what z is this valid and this is a valid for less than not less than equal to r that is a slight bit of distinction I am going to make I will need to make actually. So, since the absolute value of z is less than the absolute value of w because w runs over along the circle. So, not w but w minus z naught. So, I just rewrite this as now this ratio is absolute value is less than 1. So, I can expand this as a convergent power series which is the k plus 1 and now I am going to swap the sum with integral this is an infinite sum. So, I have to be careful again when swapping and again I invoke the uniform convergence because this series is uniformly convergent which series I am talking about. You can show that this series that the sums inside which I am running in is uniformly convergent inside the disk we are observing and therefore, we can swap the two and the proof is again I will not give the details the proof is very similar to what we saw in the last proof you truncate this sum go up to a constant amount to the 2 m and then you can surely swap and then you look at the difference as you send m to infinity you will see that difference goes to 0. So, this therefore, I can write as no 1 over k factorial and this is the a k good. So, we have now expanded I expressed f as a power series around a point inside a disk. Now, as I said the domain can be in a funny shape. So, which means that this power series which you described is valid only in this region to get hold of f around as a power series in a different region you pick another point and look at the circle around it and f is now can be expressed as a power series around this circle. And so, you can essentially cover this entire region by circles overlapping circles essentially and around each one of those on each one of those circles f is a power series a different power series, but a power series another point here to notice that it is a same function we are expressing. So, whenever 2 circles overlap for example, here the 2 power series must be equal have to be it is the same function after all. So, in this sum of piecewise sense any analytic function can be written as a power series. Now, this also provides interestingly a way to extend a given analytic function beyond its given domain. So, for example, suppose again this function f is given along with this domain on which it is analytic. So, we write f as a power series around a point inside. Now, this power series is valid around any circle inside the disk. So, we keep blowing up the disk. So, it is possible that if you blow up the disk further than the boundary of this that power series remains convergent. And if it is convergent then that has allowed us to extend the definition of the function itself beyond its defined domain. Now, it may happen it may not happen, but if it does happen then it we have now again let us say a circle of this size. And this particular power series is convergent inside this then we have extended the definition of f beyond this domain original domain of definition. And we will see examples of this, but here the raises comes a fundamental question. This gives us one way of extending this definition. What if I picked up another point and maybe there was another power series around this with another circle and this gives another extension of the definition. And these two circles may have an intersection. If there were no intersection then I can say f extended on this side is this value f extended on this side is this value. But if there is an intersection we have a slight problem that this circle may say that in this region f has this particular value the other circle may say in this region f has another value. And then there is a problem because we do not know which one to choose. Fortunately power series are very well behaved or rather analytic functions are very well behaved and this never happens. If two analytic functions agree on a any dense set of points now density is defined as in points not being isolated then they agree everywhere. See that again it is a reasonably simple proof but we need to define or rather look at the zeros of a power series or again not power series I keep saying that zeros of an analytic function. Because the reason I want I am looking at zero is that if there are two definitions of f in this region if I want to show that the two definitions are equal which is that is equivalent to saying that the difference of the two is zero function right. So, if there is an analytic definition of f here and another analytic definition of f here I take the difference which is another analytic function and that is zero in this entire region. And I want to know what kind of analytic function be zero on an entire region like this. So, let us study zeros of an analytic function because definition is obvious I want to say is the order of zero let us say f be. So, suppose f is analytic on D and f of z naught is zero. So, we say that z naught is a zero of f of order n if the following two conditions whole this one is saying that if you divide f z. So, f z if we are looking at z around z naught take f z and divide it by z minus z naught to the power n minus one and take the limit as z goes to z naught we still get zero on the other end if you divide it by z minus z naught to the n and then take the limit then you get non-zero. So, just to take an example or trivial example suppose f z is z cube then this has a zero at zero that equals zero and that is a zero of order three because if you divide it by z square and take the limit as z goes to zero you still get a zero but if you divide it by z cube and then take the limit you get one. So, an alternative way of seeing this or visualizing this definition is to look at the power series expansion of f around z naught we know it exists around a small enough disc and that power series expansion around z naught will have terms of the kind z minus z naught to the power something times constant now the first n terms of that are zero and n plus first term is non-zero which is correspond to the z minus z naught to the power n then f has a zero of order n yet another way of looking at it is that if you look at the first n minus one derivatives of f at z naught they all vanish these are all equivalent ways of looking at. So, here is a very important theorem about zeros of an analytic function so if f is analytic on d then either f is zero completely on d or if there is even a single point on which f doesn't vanish on d then zeros of f are isolated on d and when I say isolated I mean every zero of f on d has a small enough neighborhood on which there is no other zero. So, two zeros are separated by a finite space so let us pick up a zero of and let us look at a power series on a disc around z naught so f we know can be written in this fashion now there are two cases since f of z naught is zero this zero is of some finite order or it is of infinite order that is all of these coefficients are zero so we will distinguish between these two cases. So, this is somewhat a bizarre case that how can all the coefficients of f vanish and we will show that if this is true then f actually is zero how do we show that. So, let us collect let u be the set of so let us collect all such points in the set u which is a subset of points in d such that for every point in this the power series expansion of f around z 1 is zero so the saying all a k is zero is equivalent to saying that power series expansion of f around z naught is zero now pick say any z 1 in u and pick a point which is very close to z 1 and let us look at a power series expansion around that point. So, what is the power series expansion of f around z 2 see if z 2 is within this disc which is what we will make sure that the this disc of convergence for power series around z 1 z 2 is within that disc then f of z 2 is zero by this formula what is the first derivative what how about the f prime of z 2 you can calculate the derivative of this derivative of this f prime of z 2 is also zero f double prime of z 2 is also zero essentially all derivatives of f on z 2 or of zero which is equivalent to saying that power series around z 2 also vanishes because all these coefficients see we saw that the previous theorem that is gone each of this coefficient was that Cauchy's integral which I although I did not mention that we have seen that that is equal to a high order derivative around that point z naught. So, if fact that a k is all a k's are zero is equivalent to saying that all derivatives of f at z naught vanish and what we have we are now observing is that all derivatives of f at z 2 also vanish and therefore, z 2 is also in u which means that u is an open set because any point for any point in u an open neighborhood around that point also belongs to you. So, just to make it obvious I will remove the equal to sign here it does not make any difference, but it just clearly shows that this is an open set. Now, let us pick a point in the complement of u what do we know about this point z 3. So, I can write f as a power series around z 3 also. So, let us say it has of course, different set of coefficients which are derivatives of f around z 3 on z 3 fine. So, again let us pick up a pick up a neighboring point to z 3 for z 3 we know that one of these coefficient is non-zero. So, fix the smallest k for which b k is non-zero and pick up a point which is very close to z 3 now what is f of. So, let choose. So, look at the value of z 4 as given by this power series again this point is close enough to z 3 is that it is within the disc of convergence of this series. Now, f of z 3 f of z sorry around z 3 is b k the first b k which is non-zero z minus z 3 to the power k plus higher order terms. Now, z 4 is very close to z 3. So, when you plug for z 4 for z 3 the first term is totally non-zero right the higher terms may or may not be 0, but if it is close enough to z 3 let us say z 4 minus z 3 that most epsilon then this value will dominate the sum of all other values. If epsilon is small enough because those are epsilon to the power k plus 1 k plus 2 and so on. So, those terms can never cancel this which means that f of z 4 is non-zero and this shows that this is also an open set. So, D is a domain which consists of two disjoint open sets which is union of two disjoint open sets and this is the only place we are going to use the fact that D is connected. So, if D is connected then it is easy to show that one of these two splits of D must be empty. All you need to do is to pick up a point in U a point in D minus U since connected there is a path within D connecting these two. As you traverse along that path at some point you have to move from U to D minus U, but at the point there is no point because these are both are open sets. So, there is no particular point where this change over occurs or other limits of those points will be in either of the two sets. So, either way if that is if both are non-empty so that is not possible. So, this implies that either U is empty. So, if U is D that is the first case f is 0 and everywhere on D and that is again you can show by looking at power series expansion of f around on any point inside D because it is all the derivatives vanish around every point. So, f is 0 everywhere and on the other hand if U is empty so that I will still have to show. So, if U is D f is 0 on D if U is empty then we can directly conclude that 0s of f are isolated and the proof is right here. Z 3 was a 0 of let us say 0 of f in this set. Now, we know that some B k will be non-zero because U is empty and here is a small disc around Z 3 on which we show that f of Z 4 is non-zero and this theorem is really powerful. The corollary of this is one of the corollaries is that the previous discussion that if extension of f overlap on two regions they must agree on that region because otherwise in that region the 0s are not isolated. Another very interesting corollary is that the relationships that hold over reals can be lifted to relationship over complex numbers. For example, we know that sin square x plus cos square x is one over reals. So, when you extend sin x to sin z over complex numbers cos x to cos z over complex numbers then this relationship is also lifted extended because if not consider this as one function consider this as another function. Again take the difference these two the difference is 0 over the real line. Since it is 0 over the real line it must be yeah or I should not say take this difference take this minus this. So, look at this minus this. So, over real line this is 0. So, that is a non contiguous set of points on which this vanish. So, it must vanish over the entire complex plane because this function is defined is analytic over the entire complex. So, all such relationships over real just lift naturally into complex planes once you extend the function definitions to complex plane. And this also allows us to what I have been intermittently calling analytic continuation of function that is when you given analytic function defined over a domain we can extend this definition to beyond the domain by using this power series idea or maybe some other idea. And this extension by this theorem is unique. So, we can freely extend it we need not ever worry about whether there is alternative definition of f beyond the domain.