 Whereas, for the that Riemann surface for log, but I did not get time therefore, you get a home assignment. So, all you need to do is to make stack them together and then make a branch cut to the center together. So, that you get perfectly aligned branch cut and then stick them together. One more home assignment which something which I sort of hand not sort of completely hand waved last time and none of you caught me there. So, let us refresh your memory. What I said last time was that you take a and I think I started from here take a point z naught here which is at this point and write the power series of log around z naught. So, look at the disk around z naught and then I said that this disk actually intersects goes beyond or goes above the real axis and that and then I also said that once it the moment goes above the real axis this power series coincides with the log 1 z definition and I did not prove it I just said it does and none of you pointed out also. So, as a punishment for that you get this home assignment which is to take a point look at the disk take a point z naught oh sorry z 1 what will that be and or let us make it simple the two main negative signs here. So, it is just flipping this you approach from here this is at pi minus epsilon and this is at as you traverse further down it is at pi plus epsilon and let log z around z naught b of course, you know what it is it is i pi minus epsilon and we saw this last time all of this k greater than equal to 1 minus 1 to the k minus 1 divide by k z minus z naught to the k. Now, this disk hits z 1 as well and then what I want to say that look at the disk disk around z 1 on that disk this power series will coincide with the power series around z 1 and the way to prove it would be to show that this is equal to i pi plus epsilon plus k greater than equal to 1 minus 1 to the k minus 1 over k that is the power series around z 1. So, it need to show that these two are equal of course, this is equality holds for certain z also only which are reasonably close to both z naught and z 1 for assuming all of this. So, this clearly establishes the fact that the power series around z naught coincides with the power series around z 1 and therefore, which and that is certainly not the power series given by the log 0 z at z 1 because for log 0 z at z 1 will be a will have a minus epsilon plus pi here. So, now let us continue we will discuss a little bit more on this. So, this Riemann surface that you get for log z it is again a mental construct to allow us to nicely visualize mapping log set right. You can instead of looking at infinitely many log function this allow us to look at a single log function sending this Riemann surface to the complex plane. Now, one could argue that this is as you know arbitrary as that cutting of infinitely many log functions you could start at any point make any branch cut at any angle and then you get different set of infinite classes. But in very in one very interesting way the Riemann surface associated with log z is not arbitrary it is a it is really extremely useful. Besides allowing you to imagine the mapping log z in a nice fashion it is actually useful in a very more or I say much more concrete sense. So, let us see that what is now this I am going to invoke Cauchy's theorem what is this this is integral around not less equal to 1 this integral around unit circle of 1 by z what is Cauchy's theorem tell us 2 pi i right because Cauchy's theorem in general says f z over z the integral around a disk where f is analytic is 2 pi i times a f our likely z minus z naught is f z naught. So, this is z minus 0 that is f 0 I can view this integral in another way 1 by z is differentiation of log z and we establish that it holds over all complex so 1 by z dz equals d log z it is d log z by dz times dz and d log z by dz is 1 by z. So, integral of d log z is what log z and then there are this is circle of course now circle there are no limits. So, one can see, but there is a bit of an issue about here because about how do you exactly define this business of limits as one thing. Secondly, log z is not properly defined over this entire circle anyway again thinking of log z over complex plane and here let us say thinking about the differential another differential the log 0 z by the way differential of any variant of log function log 1 z log 2 z log 3 z log 0 z the differential is always 1 by z because of the difference between any of these variants is only in the constant part and that gets wiped out by differentiating. So, this differential is not defined at the branch cut. So, there is one line of in case of order is a negative real line is not defined. So, you cannot really integrate over this both of this problem can be resolved in the following way we can say we can integrate from z equals e to the i minus pi plus epsilon to e to the i pi minus epsilon now in this sort of this is starting point this is ending point and in this region this is very well defined and this also gives me a starting point and definition point. Now, by Cauchy if the well actually you do not you would need Cauchy, but there is a fact use the analyticity of log z function to say that this integral is this log of this minus log of this which is 2 pi i minus 2 i epsilon and as epsilon goes to 0 this becomes and that matches perfectly with this Cauchy's formula. So, we could do this round about way of integrating and get the same value, but that is not the end of story what about or let us say in general c by z let us say t by z t z what about this this is clearly by Cauchy is 2 pi i t. So, I just write 2 pi i t is this, but if I write it in the same fashion this is equal to d log z to the power t is simple to verify is differentiate z power t you get t by z here fine and now if you try to do the same thing we will fail why here the z moves around this unit circle in the counter clockwise, but we are taking log of not z but z power t. So, as z moves around a circle counter clockwise what happened to z to the d z to the t will move to t times the speed of z. So, when z will complete one circle z to power t will complete t circles and now we are taking log of that function which is actually the completing t circles. So, now there is really we cannot really do any limit because it is there is no to end point as such here which we can take a limit and get around this. So, we cannot sensibly discuss this integral over the standard view that log is infinitely there are infinitely many log maps one each from a complex plane with a branch cut to a complex plane. So, if you adopt that view we cannot really do this integration. However, if you adopt the Riemann surface view then t circles are fine because every time you complete a circle you move up. So, you are spiraling t times up and as you spiral the definition of log changes right you jump to the next log function. So, as you do t times you will make t jumps and here. So, this is same as integral z power t starting from and now we do not need to start from here just to make life simple we can start at one start from here and do t circles. And as you do one circle where does z power t goes e to the 2 pi i and you do it twice twice t times reach here. Now of course, this is all same there is no difference between these two values, but when we talk of taking log of this and because as every time we circle around we change the value of log function itself. The first log function we start with this the log 0 which is. So, log 0 of 1 is 0 log 1 of 1 is 2 pi i log 2 of 1 is 4 pi i. So, as you do a t time circle. So, log this would therefore be 2 pi this is log t of 1 minus log 0 of 1 which is 2 pi and that is perfectly matching with the Cauchy integral. So, this allows us to view the integral in this different fashion and still make complete sense out of it. Of course, in this simple case because we can apply Cauchy's theorem directly here we get the value of integral without any problems. But when there are more complex functions sitting here instead of z to power t let us say there is some f z sitting here. Then integrating so if you do this and you come here you get f prime z over f z d z and integrating that over a circle may not be very easy. On the other hand if you take this view and you look at d log f z then all we need to do is that as z varies we need to count or see how many times does f z wind up or wind down and that will give us a the value of the integral without worrying too much about see all we need to do is to count how many times does it wound up. I do not need to worry about exactly what path does it follow while winding up because by Cauchy's formula all paths will lead to the same integral because it is an analytic function. So, just need to see that as z moves around a close contour and again this need not be a circle it could be any close contour as z moves around this close contour how does f z move and how many you know levels does it go up. The moment we calculate the levels we got the integral. So, it provides a very convenient way of integrating function integrals of this kind f prime z over f z d z and when we look at zeta function this will occur very regularly. So, we need to understand this value yes this function. So, typically what we will have is that the function will be f z here. So, let us look at this there will be some close surface let us say some bolted in delta r d log. So, that is a kind of integral we will see. So, here say the z moves around the boundary of r this is z is on complex plane it just moves around the boundary of the complex plane. Now, z is first sent by f to f z now of course, f z sends f z to just the complex plane. But, instead of viewing f z as lying on the complex plane we think of this as on the Riemann surface because there is a log sitting after this. And we calculate how many times does f z as z moves around this how many times does f z wind up around it around some or say 0.0 let say yeah around 0 is a key thing. How many times does f z wind up around 0 and that number. So, it is convenient to view f z as lying on the Riemann surface because winding means going up in a circle fashion right. And that number is the value of this integral that numbers times 2 pi i of course. So, it is still a mental construct you can still view as f z being on the complex plane. You can still do this integral by as Orko pointed out by taking the limits cutting out pieces where log is not defined. But, it is far more convenient to think of this lying on the Riemann surface. So, all you need to do is to study this map z to f z to evaluate this integral. And as you see there as z moves in a circle how does f z move around 0 not clear. You can also think of if you do not want to think of Riemann surface you can just think of f f mapping complex numbers to complex number. And in the complex plane itself you see how f z bounce around 0 it will on does it bounce 1 time does it bounce twice does it bounce 3 times like z to the t winds t times around 0 as z moves once around fine. So, that is perfectly fine view and you can calculate the immediately conclude the value of this integral. But, to justify this to be the value of the integral we have to say that f z is actually lying on this Riemann surface. Because, then we can say that log of f z is completely defined over this curve. So, the Riemann surface provides a way of justifying the value of the integral. So, now we have I will study the analytic functions we have also that the power series expansion of analytic functions. And we know now that every analytic function around every point expands as a power series inside a disk. So, analytic function can be defined in a strange domain. But, they will be sort of if this is the domain of analytic function then around any point there will be disks over which a power series represents the analytic function. And the size of this disk can be inferred easily you start at the point at the center and just blow up the circle as much as you can until you hit bound one of the boundaries of the domain. And that is the size of the disk on which this is well defined on this on which this uniformly converges this power series and is equal to and in this way fashion you can keep on defining in this console. So, one thing that is clear is if it is a finite domain no it is not clear that you will have that is not necessary that you will only have a finitely many disks here it depends on the boundary the boundaries messy then you may end up having infinitely many disks required to capture all of this. Now, this is nice, but this is still little little unsatisfactory and the reason is that first is that this power series expansion what it says is that outside the power the disk everything diverges on every point outside the disk it is a it is a certainly not absolutely convergent. So, it is generally diverges absolutely. So, it does not quite allow us to study properly I should say the singularities of an analytic function well analytic function by definition cannot be singular in its domain. But there are certain points which are outside the domain on which the analytic function is singular. Now, behavior of analytic function around those points is not that properly expressed by the power series expansion. Let us take an example is let us take just this function the simplest possible such function 1 by z it has just one singularity at z equals 0. And if you so is domain is the entire complex plane except this point there is this puncture made at z equals 0 and that is a domain on which it is analytic. Now, if I try to write this same function as a power series around this point z equals 0 well what I can do is I can write write put a circle here which converges everywhere inside this power series expansion. But which touches z equals 0 this is not defined then I have to cover the other regions I have to define another circle then another circle another circle. So, I have to define this multiple circles each capturing part of the behavior of the function around the singularity. So, there are times when this is good enough, but there are times when one would like a you know simple or one single representation of the behavior around the power around the singularity. And for example, for this function this representation itself completely defines the function around the singularity and anywhere else, but of course, this is not power series. So, that is the problem, but this also says the solution there instead of just taking to power series if we relax our requirement a bit and allow inverse polynomials also to occur in the power in not the power series by in the more general series. Then at least certainly for this function we can just write the entire function as a single such series. And that give rise to the notion of Lorentz series. So, this is the last concept before we dive into zeta function. So, this Lorentz series and its discussion for singularities will complete the basics of complex analysis. So, Lorentz series is defined as this is Lorentz series is an infinite sum of this form. The only difference with power series is that we allow negative powers of z also to occur. So, infinitely many negative powers, infinitely many positive powers. This this is clearly a Lorentz series. Now, as with power series we have to understand where is this analytic this clearly this is more general than power series. So, every powers every analytic function we can write as Lorentz series around a certain region, certain region where it is analytic. So, that is that is clear. So, we need to understand exactly where is a Lorentz series convergent. So, let us start with this as a Lorentz series and let us split it into two parts. This part is a power series and this part is a pure negative power series. And let us give it certain names this is a f minus z plus for f plus z we know precisely where this is convergent. There is a disc associated with this where this is convergent. What about f minus z? So, f let me just write it. f z is convergent inside and whenever I write convergent I mean uniformly convergent because that is the only notion of convergence for power series that we will be interested inside say some disc z equals alpha. Oh by the way I have again eaten up the more general form here it could be it should be z minus z naught to the k. So, you are expanding the series around z naught. So, let us assume that that is 0 here. So, this we know what about this? Well we do not know its negative powers of z but it is purely negative powers. So, if we flip z and replace this is 1 minus z we get a power series. So, we know precisely where this is convergent is convergent or let us instead of z let us write it as 1 over w this is convergent inside some disc w equals beta. So, this implies just taking z to be 1 over w this is convergent outside the disc mod z equals 1 over beta w is 1 by z. So, just replace 1 by z here the by mod w we replace by 1 over mod z and then you get z equals 1 over beta and inside gets flipped to outside. You can see this clearly by we is writing this replacing equal to by less than equal to and that is it we know precisely the convergence of these two parts. So, therefore, f z is convergent in this annulus annulus is this name for the disc with the smaller disc taken out from inside. Yes it is possible and then it is nowhere convergent then it is really not a very nice definition. So, it will make sense only when 1 over beta is less than alpha only then there is some region of convergence for the Lorentz series. Coming back to this example what is the annulus of convergence for this 0 to infinity. So, it is basically convergent everywhere except this single point 0. So, that is punctured out and that is becomes an annulus. Now, it is outside this annulus is this convergent can it be convergent anyway no why well if you look at outside this annulus go beyond alpha going beyond alpha this part is convergent that is not a problem. So, this is some finite value take any point z which is with modulus bigger than alpha this is absolutely convergent some finite value, but this diverges. So, some diverges similarly inside below 1 minus less than 1 more beta this diverges this converges. So, some again diverges. So, again exactly like power series it is convergent precisely inside the annulus. So, again this is limited in that fashion, but less limited than a power series because this is a living example here you can capture this function just by 1 single Lorentz series as opposed to no many power series. In fact, to capture this function as power series in the entire complex plane you will need infinitely many power series because any power series start you know at any point in the complex plane and will only be able to take out a finite region from the complex plane. So, it is a more general and more powerful form of representation of an analytic function and it is a pretty straight forward to show that any Lorentz series inside the annulus of convergence is analytic essentially the same proof that we used for power series carries over do you remember the proof how did we show that any power series inside is this of convergence is analytic. So, what we did was we looked at truncation of power series just limit the sum to some finitely many. So, that is a polynomial polynomials are analytic we know and then when you take the limit because of uniform convergence the analyticity is preserved because that is by Morera's theorem then you use because every polynomial is analytic. So, the integral around any rectangular is 0 and then you take the limit of the these polynomials which is the power series and due to the uniform convergence we can swap the summation and integral and therefore we get. So, that is the proof for analyticity of power series and the same proof really goes through here. So, I will not prove that I will leave that due to the workout instead what I will prove is just like we showed for power series that around if you are given any analytic function and around any point inside the domain we can represent the analytic function as a power series on a disk. Similarly, given any analytic function and given an analyst then inside any point sorry given any analytic function on a domain given any point inside that domain we can represent the function as in Lorentz series inside an appropriate analyst and the proof is actually pretty much mimics the power series proof, but there is one nice twist to it that is why I want to show it to you. For example, look here this is a Lorentz series convergent on this domain this is an analytic function also on the domain inside that punctured complex plane this is a Lorentz series for the same function and it is around the point z equals 0 that point is not inside the domain is actually singularity. So, this Lorentz series expression this one is around a point which is not in the domain yet can be expressed as Lorentz series around that inside the domain it is already a disk. So, that is not that interesting anyway yeah then it is anyway Lorentz because it is a power series or it is a Lorentz series so it is not anything interesting. So, given any z naught let us say let us try to write a statement and we will try to prove it f z can be expressed. So, f is analytic on d take any point z naught f z can be written express a Lorentz series around z naught, but its analysis of convergence will be inside the domain that we cannot escape from. So, how do we prove this? So, suppose this is the domain and this is the point z naught which is actually outside. Of course, if I take z naught here then there is no analysis of convergence and this point less to no analysis of convergence which fully lies inside the domain. So, it forces me to pick a point which is covered by a boundary of the domain and then I can try to define analysis of convergence. So, how do we handle this? Well, let us pick an analysis which consists of two circles with centers at z naught lying completely inside the domain and let us say this is alpha and this is beta.