 So, this is the theorem we wanted to prove last time right. So, the idea is pretty simple this is let us say suppose this is the point z naught and we create an analysis which should entirely lie oh enter into 0 then analysis must lie entirely inside z naught otherwise it will not make any sense to have because f may not even be defined there. So, let us pick up an analysis which is this one. So, this is the domain D and this analysis that I am picking up this is centered at z naught inside D and this I will cut by making a single slice here. So, that it becomes a single curve the boundary of this analysis and I traverse this boundary counter clockwise starting from outside going here then coming here then going down here. So, this is clockwise then going back on this and then continuing counter clockwise completing this circuit and we use the standard trick that at this point instead of running over the same cut twice you just separate them by a very tiny amount. So, that it is this traversal is very well defined everywhere and let us call this region. So, that is the region which is encompassed by this curve R. Now, f is completely analytic inside this region forget about completely f is analytic inside this region. Therefore, by Cauchy integral formula if there is a point let us say any point z here I can write f of z as the integration along the boundary of R of f w divide by w minus z. Now, let us look at the boundary. Boundary is this, this, this and this. Now, as this two parts tend towards each other their integral around those parts will cancel each other out. So, what will be left with the limit is integral counter clockwise on this and integral clockwise on this inside circle. So, let us say this these are both circles centered at z naught. Let us say this is beta this distance and this distance is alpha. So, then I can rewrite this as now the first part is familiar here the integral is on w's which are on the outside circle and z is inside that is z is closer to z naught then these w's are. So, let us just rewrite this a bit because I want everything in terms of or expand around z naught. Now, w minus z naught as w moves around this circle is larger in absolute value than z minus z naught. So, this is standard or familiar thing that we did for power series also. So, therefore, I can rewrite this as and in this integral w is running over the inside circle. So, absolute value of w minus z naught is smaller than z minus z naught. So, I can expand this, but taking out z minus z naught and then doing it and now we can use the uniform convergence of this power series in the region of interest to swap the integral with summation and we get summation k greater than equal to 0 integral of I got I forgot 1 over 2 pi i here. So, there is 1 over 2 pi i everywhere and if you just change this summation or by replacing the index k with minus 1 minus k. So, just redefine k to be minus 1 minus k. So, this becomes k this becomes k plus 1 and then you get there is bit of a problem here because these 2 integrals are over different circles. So, I cannot write it exactly like this. So, let me still go with a split definition. So, we are almost there well actually we are there because now this is a Lorentz series. We also have the value of the coefficients of the Lorentz series that it is for the on the positive side it is these integrals on the negative side it is these integral. So, that completes the proof of the theorem, but actually we can do one step better and make these 2 integrals over the same circle. Do you see the reason? In fact, let me state it this way let me just write it as equal to sum k going to minus infinity to infinity 1 over 2 pi i integral. I can replace this integral over circles of radius alpha or beta by a circle of radius r as long as that circle lies inside the analysis not inside the region the domain D. Exactly, so this function that you are integrating f w over w minus z naught to infinity k plus 1 is analytic inside that analysis because f is analytic this is analytic why is this analytic? Because it never vanishes and it is inverse of an analytic function w minus z naught absolute value is always non-zero. So, this is this is the product of 2 analytic so this is an analytic and since it is analytic inside that analysis is integral over any circle inside the analysis it is going to be the same that is why caution and so we have a uniform definition of the constants of this expansion circle around it not yes. Yes, so it is defined see it is we are only see we are if you integrate around this circle on this circle the function is well defined and inside this this and this this another look at this analysis inside this analysis function is analytic the complete not at f w what f w divided by w minus z naught to the k plus 1 is analytic. So, if you cut make a cut here and look at that region the integral around the boundary of this by caution the integral is 0 and therefore, the 2 integrals are same of course, the same thing does not apply here because w minus z naught can become 0 in this integral not this in not in this integral, but inside the region inside this disk where we are doing the integral it can become 0 or here inside this region the delta r it can become 0. So, you cannot this is not 0 this is not 0. So, that is the Lorentz series expansion of an analytic function around a singularity and it expansion makes sense only around a singularity because if the point z naught is not a singularity then is expansion is simply a power series and this expansion is defined on the analysis which lies inside the domain and this the inner circle we can shrink to as small as possible and outer circle we can expand to as large as possible that does not matter. So, that will be remain the same. So, this gives us a more general way of expressing an analytic function specially it allows us to study the singularities in a much nicer way the power series always avoided singularities whereas, this actually uses a singularity to expand the function around it. So, we can study the behavior of an analytic function around a singularity using Lorentz series we cannot do that using power series and that is something that is going to be very very important for us because singularities of an analytic function are going to play a very important role. In fact, you already see this is an analytic function with a one singularity at w equals z and you already note that it has a very interesting role to play that it actually gives me a value of f at z. So, we will see as you move along more and more usages of analytic function around a singularity, but first let us address the singularity themselves a bit. So, we saw zeros of an analytic function we the most important thing about them was that zeros are isolated and every zero has an associated order with it. That is the order of the first few term that first number of term that is zero in power series expansion that is the order of the zero. Can we say same thing about singularities? Can we say that singularities of an analytic function are isolated make you guess and there is a very simple you mean to say they are isolated that is what you have to answer yes they are actually not okay. So, analytic function of course, if there is a domain associated where it is analytic the rest of it. So, presumably there are it is it may or may not be defined see it is possible that that analytic function can be a continued more in a larger domain. So, whatever it can be continued more and more and more whatever maximum possible continuation is there after that what is left is places where it cannot be analytically continued which means that these are singularities. So, are those singularities isolated well okay let us take an example analytic function this is analytic inside unit disk what about the singularities they are everywhere after that there is completely singular behavior there is nothing outside there. So, this immediately shows that singularities of an analytic function need not be isolated. Now, when the singularities are not isolated those are not very interesting cases because that is a complete mess we are typically not even going to be interested in those regions. So, we will only be interested in those singularities which are isolated because those are the interesting singularities. So, from henceforth we will restrict our attention to isolated singularities you understand what an isolated singularities a singularity such that a small enough neighbor it around it has no other singularity. Now, isolated singularities also have a very nice property that we can write a nice Lorentz series expansion of the function around that singularity in a small region and the analysis is just the punctured disk because there is only one singularity which is right now sitting there. So, we just puncture that and that analysis is the Lorentz series expansion of the function which means that we can study the behavior of through that Lorentz series we can study the behavior of the analytic function around that singularity completely we know the exactly how its value behaves around that singularity. So, we can let me just write it down because it is important we can write and we will use this observation to study the behavior of the not only analytic function, but also because we can study it around the singularity we will be able to say something about that singularity itself. So, let F be analytic on D with isolated singularity now using the previous observation let us write F as say some disk punctured at z naught of radius r. Now, when I say inside you take it to mean that it is punctured at z naught. So, what can we say about the singularities? Well it turns out there are going to be three types of singularities which I am going to describe now and these singularities characterized by behavior of F around z naught type one is called removable singularity this is actually not a singularity this is a phony singularity this occurs when the in the Lorentz series expansion the all the negative coefficients are 0. So, actually Lorentz series is a power series. So, actually there is no singularity there. So, then why do we even call it a singularity? Well technically in the we can a singularity is defined if the function appears to be undefined there. So, to take a most silly example of this if you look at the function z square by z at z equals 0 if you are treated completely dumbly z square goes to 0 and z goes to 0. So, it is 0 by 0 therefore, undefined there are not. So, silly examples also I mean there are for example, sin z by z that is also 0 by 0, but of course, you all know that we can remove it, but why do we know that we can remove it only through this actually even this an argument that this is not really a singularity at z equals 0 is by plug in the make the Lorentz series expansion of this and show that all these are 0 all the negative coefficients are 0 and therefore, it is non singular it is removable. So, this is a just a formalization of this whole thing clear. There is a nice character I said that these are characterized by behavior of f around z not. So, how are these removable singularities characterized as? So, f has a removable singularity at z not if and only if f z is bounded around z not and when I say around z not is taken to mean a small enough disk around z not and the proof is fairly straight forward. Let us tackle this let us get the proof out of you guys which way do we start suppose f is f has a removable singularity then how do you prove that f is bounded around z not. So, let us assume less than equal to epsilon z minus z not the absolute value that is a very tiny disk you need to show that absolute value of z is bounded that is pretty straight forward use the definition f is at this Lorentz series expansion all the negative a k's are 0. So, only the positive a k's survive. So, a 0 will be a 0 and the other ones a 1 will be multiplied with an absolute value if you look at the absolute value of this epsilon, epsilon square epsilon cube the higher coefficient and since epsilon is tiny is all converges to some finite value this is less than equal to sigma k greater than equal to 0 a k epsilon and this therefore converges to something well of course here one is assuming that f is analytic on d. So, there is a non trivial analysis on which this is convergent if it is not convergent anywhere then of course there is no point talking here of anything at all or if it is convergent inside an analysis which is this it is epsilon is much smaller than r. So, it is certainly convergent and it converges to some finite value and therefore this is bounded by some value the converse if f is bounded around z naught that is that is another way instead of using the counter positive if f does not have removal singularity then that is not going to work directly because what might happen is that infinitely many of a k's are non 0 negative a k's are non 0 and then somehow those sums converge to something well one can argue that if they do not but you need to measure on with infinite sums one would like a more direct and simpler proof here which fortunately exist you can definition of exactly you want to show that a k's are 0 for all k less than 0 what is the definition of a k a k we saw was 1 over 2 pi i integral of w minus z naught equals an hour which is can be any circle as long as within this. So, I can choose it to be epsilon as long as 0 f w divide by w minus z naught to the k plus 1 d w if you look at the absolute value of this is going to be less than equal to forget about constants this is bounded f w. So, there is some c constant c such that absolute value f w is bounded by c this is in absolute value epsilon to the k plus 1 right. So, this is and this as you integrate once you take this out and integrate you get 2 pi epsilon. So, basically what you will get is a k you will get to be order 1 over epsilon to the k epsilon to the k plus 1 here 2 pi epsilon comes after integration the 1 epsilon cancels out forget about 2 pi's they are all constants and this is also a constant. So, all gets absorbed in this now as you send epsilon to 0 what happens for negative case they all go to 0 neglecting z 0 of course, because z 0 is singularity. So, you do not know what it is happening, but that is good enough because you do not actually have to take this go to epsilon equal to 0 we are showing that a k cannot have a non-zero value because we can make choose any small element epsilon to make it smaller than that. So, that is pretty straight forward. So, these are sort of silly singularities as I said they do not even exist what is more interesting is and actually the most interesting kind of singularities are type 2 these are called poles a k is not equal to 0 when a k is non-zero for a finitely many k less than 0 and that has to be non-zero value. So, the examples of this 1 by z is the simplest possible example here a minus 1 is non-zero. So, what with as with zeros we have an associated order with a pole it is equal to the largest value of n such that a of minus n is non-zero. So, this is almost dual of the order of a 0 and how is a pole defined by the behavior of f around z naught it has a pole around z naught if and only if as you send z goes to z naught the absolute value of z goes to infinity. Now, this is a bit of a loaded statement because when I say z goes to z naught I can take z to z naught along any path that is very important not only one path. So, it does not matter which path I take it to z naught f z always diverges and the proof of this is again fairly straight forward if you if f has a pole at z naught then just pulling out the definition f as a Lorentz series expansion with let us say pole of order n. So, which means that there is if f has a pole z naught then I write f z as a minus n by n plus i degree terms and now if you see that if you send z naught z to z naught all of this in particular this diverges these will also diverge, but the rate of divergence here will be faster than here. You can make this argument more precise by considering if you let because sometimes this argument are subtle one has to be very careful when making this arguments. So, just to emphasize that I define g z to be z minus z naught to the n times f z what is g of z naught that is not 0. So, then I can write f z as g of z by z minus z naught to the n now you see that clearly as that goes to z naught this is non-zero and this goes to 0. So, clearly diverges it does not matter what path I take and conversely if you assume that f z diverges around z naught everywhere I mean it stands towards infinity then how do I show that f has a pole at z naught this is a little trickier not straight forward although pretty simple. So, again define g is of course likely differently to be 1 over f z then what can we say about g z g is analytic around z naught because f z does not take value 0 around z naught and is bounded why is it bounded because absolute value f z around z naught is very large. So, therefore g z is bounded around z naught. So, it is analytic and bounded around z naught of course g z naught I am saying it is not clear what is the difference I do not I am not even defining it right now, but the fact that it is bounded and analytic around z naught means that I can work the type this is a type if at all it has a singularity at z naught is a type 1 singularity which means it is removable and if it has a removal singularity at z naught it means it is g z naught can be defined in a unique way and how is it defined in a unique way well just take one path and send this this is going to diverge the g of z naught has to be 0. Now, g is analytic function in this disk it vanishes at z naught. So, it has a 0 at z naught now every 0 is isolated of course we know and there is an order associated with this. So, let g have a 0 of order n at z then we can write g z as z minus z naught to the n times h z with h z naught not 0 that is by definition of the order of 0 and now I can write f z as 1 over now here h z is analytic around z naught it is not 0 around z naught. So, 1 over h z is also analytic around z naught and n here and this shows that f has a pole of order n because 1 over h z I can write as a power series around z naught and which does not vanish at z naught. So, this is going to be a Laurent series with a minus n being the largest or smallest non-zero coefficient why is g bounded because of this in a small disk because of this property you have z naught and you look at a small disk here look at this region as this size shrinks absolute value of z goes to infinity. So, at some point let us say absolute value of z exceeds 10 everywhere it will be because no matter where you approach from any point no matter how you approach z naught you are going to diverge. So, for every point around this in a small enough disk the absolute value of z is say at least 10 and if absolute value of z is at least 10 on that disk g z is at most 1 by 10 mean the absolute value. So, it is bounded no at what z naught z naught equals 0 you are saying summation k greater than equal to 0 1 over z to the k. This is a Laurent infinite Laurent series yes and as it goes to z naught there is absolute value go to infinity does it that is type 3 singularity these are called essential singularities when a k's are non-zero for infinitely many k less than 0 of this kind the example is as just law you gave or e to the 1 by z scope you want to write is that there is a sequence of points tending towards z naught. So, each of this point is closer and closer to z naught that is what I mean here I am not writing it very precisely. So, essential singularities is a bit completely bizarre behavior they diverge they do not diverge as you go from z to z naught depending on what path you follow you can get any result that you want we prove it in the next class, but just to convince you of this let us look at this one. If you approach it from the positive real axis what happens approach towards 0 from the positive real axis goes to infinity approach towards 0 from the positive imaginary axis minus infinity positive imaginary axis. So, z is i y so you get e to the 1 over i y y tending towards 0 what is the absolute value of this e to the 1 over i y what is the absolute value of this is always 1 does not matter what y is. So, if you approach it from the positive imaginary axis towards 0 the absolute value stays at 1 and this is not the end of the story as this theorem says you can get any value that you desire as the limit depending on what path you chose to approach what same story absolutely same story this is true for every essential singularity and essential singularity is defined as infinitely many a k is being non zero which is the case here. So, we will continue this in the next class