 So, I am just writing down the theorem statement of last time f as an essential singularity at z naught if and only for every w. So, says that there is a sequence of numbers ever getting closer to z naught such that as you take f on each one of those numbers you get closer and closer to w. Now, what I want to clarify because we in the last two theorems also we saw a characterization of other types of singularities to understand exactly what this characterization means. So, in the first case which was removable singularity the characterization was that there is a small enough neighborhood of z naught in which f is bounded. So, which means that you look at z naught here let me pick a different color for this and then pick a tiny neighborhood which is maybe distance delta such that in this neighborhood the value f z is bounded by some constant c in this entire neighborhood even on every point in this neighborhood is less than equal to some constant c this is removable. For poles the condition is that if you look at z naught and again in small enough neighborhood of delta then in this entire neighborhood was two you cannot see maybe you can go there. Here in this neighborhood what we are saying is that if you take any here we have to show that the characterization is that it is unbounded which means that no matter what constant you choose c there is a small enough delta for corresponding to that constant such that inside this disk f of z is bigger than c everywhere right. And the third one which is the topic of discussion today in this we are saying that any w then for all epsilon there exists delta and there exists z z minus z naught less than equal to delta which is z is within this disk and f z minus w is less than equal to epsilon. So, that is the meaning of this characterization that you can if you want to be epsilon close to w there is a small enough neighborhood of z naught where you will it be epsilon close to epsilon close to w or pick any epsilon pick any delta and then there is a z inside the delta disk where f z is epsilon close to now these three are of exclusive conditions one is not equivalent there is no overlap between these two because this is saying that inside a small disk you can find very large value for f z as well as very small value for f z. So, this distinction are clear particularly this characterization. So, let us prove this now the proof was very similar to the proof we had for Poles characterization. So, let us assume that there is a w. So, we will prove the contra positive there is a w such that w is never reached inside the delta neighborhood of z naught. So, that is w is not of this kind. So, what does that mean in this picture what it means is that there is an epsilon such that no that is not good enough because suppose f z takes value somewhere on the complex plane then this condition will be satisfied because for every epsilon there is a delta you can choose a large enough delta. So, that that particular value is within the delta neighborhood what we want to say that very in very very small neighborhood of z naught you can find the value w or something close to w that is very important. So, if this does not happen for some w what does it mean it means that in a very if you look at a very small neighborhood of z naught every point inside in that disc will be at least epsilon away from w that is there is a small delta such that and there is a small delta there is a small epsilon such that in that delta disc every f z is epsilon away from w. So, that is let us just write it down such that there exists epsilon there exists delta and f z minus w is greater than equal to epsilon and now we will show that of course f is analytic in the neighborhood that is given as a condition. So, now we are going to show that at z naught f as a point. So, f is not an essential singularity and as I said the proof is very similar to the last time. So, the way the proof goes is you define let g z to be 1 over f z minus w and consider what can we say about g z inside this disc what is the absolute value of g z inside this disc less than equal to epsilon 1 over epsilon which means g is bounded inside this disc there is a constant 1 over epsilon and at every point the absolute value of g z most 1 by d epsilon and notice that g has a singularity at z naught because f z is undefined f z naught is undefined. So, g z is also undefined. So, there is a singularity at z naught, but because g z is bounded this is a removable singularity. Now, if g there is removable singularity which means that g z naught is can be uniquely defined and g is analytic inside this disc completely. Now, if g is analytic on this disc completely g z is of course, g z naught is also bounded now g z naught may be 0 or may not be 0. So, let us say g let g z be z minus z naught to the n h z with h z naught not equal to 0 if g z naught is 0 it is a 0 of some order finite order and that is what I am writing h z naught is certainly non-zero all right. Now, what is f just recall f is w plus 1 over g z and writing this we get this for g z for f and this equality holds inside that disc. Now, what can we say about f h z naught is not 0. So, 1 over h z naught is non-zero is bounded. So, 1 over h z is therefore, analytic inside the disc. So, f what is f z naught is a polar order and well it actually may not even be a polar polar order and because that will depend on no it will be polar order sorry it will be a polar order and because this is this now the you can write 1 of h h z naught as a power series and then this is a gives you a Lorentz series with the lowest or the minimum degree being minus n. So, which means that if f has an essential singularity at z naught then it necessarily has this property that you can reach as you tend towards z naught you can reach any value that you wish which is completely bizarre and wild behavior kind of hard to measure although we did see an example last time that e to the 1 by z as at least in these two directions you get two very different values and in fact, you think little more about it you can you will realize that you can achieve any value at least for e to the 1 by z in following a particular path and these are the only three kinds of singularity that can exist because the definition was exhaustive removable singularity was at all negative coefficients are 0 pole was at some finitely many coefficients negative coefficients are non-zero and essential was infinitely many negative coefficients are non-zero. So, that is the only there are the only three possibilities that can exist. So, we have covered all kinds of isolated singularities the first one does not quite is silly kind of a singularity the last one is a bizarre kind of singularity. So, the middle one that is the pole is what we really understand at least classically as a singularity because this essential singularity is something which is not found in real analysis this only found in the complex analysis because of this strange behavior. So, from now on we will not talk about essential singularities because they are too hairy to analyze and more importantly in our subsequent analysis of zeta function these do not arise which is very fortunate thing. So, we can stick with singularities which are more manageable and in fact, there is a whole class of such functions which have only poles and removal singularities as their singularities and this is these functions are called meromorphic. So, f is a meromorphic on a domain d if on every point in the f is either analytic on d on that point or as a pole at that point. So, clearly this is a generalization of analytic functions we are also allowing poles to occur inside the domain and these are the class of functions will be interesting. In fact, we will even more or most of the time we will be looking at functions which are meromorphic over the entire complex plane. So, now that we have gotten rid of essential singularities and say that you only focus on poles you already seen the examples of poles earlier you also seen the advantage of expressing now I can call it the meromorphic functions as a Lorentz series around the poles. So, there is actually one very important usage of Lorentz series which and singularities also which we will make use of extensively in our subsequent analysis. In fact, I am going to give you examples of this usage which is really remarkable and that is for integration of functions. So, let us say let us just write integration keep switching between these. So, suppose f has a pole at z naught. So, let us write f z as a Lorentz series around z naught. Now, this Lorentz series may not be valid over the entire domain, but it is valid only on some punctured disk because it is a pole it is an isolated singularity. So, that can puncture that analysis as there is an analysis on which this is valid and this inside circle is just a point and now let us consider this integral. So, this let us say this is valid for what is the value of this integral it be 0 no there is a pole inside. Let us use the fact that there is a Lorentz series expansion of f around this point. Now, we use a uniform convergence to swap the integrals with summation a k of course, can be taken out and then inside we have z minus z naught to the k. Now, let us look at the integral inside what is its value for k greater than equal to 0 what would it be a 0 because this is a polynomial which is a analytic in the entire disk and therefore, by Cauchy the integral will be 0. What about the negative powers what about say k equals minus 1 you will get d z over z minus z naught. So, there is this infinite sum of integrals look at the first one what is its value 2 pi i 1 over 2 pi i what is Cauchy's integral formula Cauchy integral formula is that value of the function inside a disk is 1 over 2 pi i times integral around the disk the function divide by z d z and this is. So, what is the function a corresponding the constant function. So, its value is always 1. So, this integral is actually 2 pi i let us move on second derivative yes what is it 0 why that is right the value of the derivative of the first derivative f prime z equals integral f w divide by w minus z whole square d w. So, that is precisely that form this is the first derivative of function 1 well derivative that is 0. So, that is 0 third one they are all derivatives that is it now this is remarkable fact that integral of a Lorentz series around its disk of convergence around a pole of course, around its disk of convergence is just as 2 pi i times a minus 1 just that one coefficient one term which defines the integral value no other term and now we can use this to great effect. But, before I use this let us generalize this event there are 2 poles inside of a function inside a domain this is the case when the function has one pole in a domain what if there are 2 poles. So, here is the domain and let us say there are pole number 1 is here let us have colors available. So, I can pole number 1. So, what about this some function f exactly. So, suppose I want to integrate this function around some closed region say this region this integral around this region is same as integral around these 2 regions because here is completely analytic. So, the integral is 0 and you can shrink these 2 like this. So, which means that you take f express your Lorentz series around first pole look at the a minus 1 express it as Lorentz series around the second pole get that a minus 1 corresponding a minus 1 add them multiply by 2 pi i that is your integral value. So, as a result or as a conclusion of this what we get is the following theorem. So, before I define theorem let define notation which is called the residue of an meromorphic function at some pole z naught that is equal to a minus 1 which is the comes out of the Lorentz expansion around z naught. And note that this also covers the case when there is no pole at z naught then you will have a power series expansion around this which means a minus 1 would be 0. So, then the theorem says let f have poles at z 1 z 2 to z t inside the integral of f around the boundary of the domain is there is the sum of residues at all the poles of f multiplied with. So, it is this so the power of this theorem lies in the fact that it allows me to compute the integral of a meromorphic function around a domain very easily just look at the poles identify the poles complete compute the residue at each one of the poles and that is the value of the integral do not have to worry about what is the contour and whether how does a function behave over the entire region of the domain just the behavior at the poles is what determines the integral. But there is something even more interesting see if I know the poles of f explicitly then this allows me to a while I am already repeating myself if I know the poles explicitly already said that this allows me to compute the integral. There are some very interesting examples of this let us take the simple one who knows the value of this integral this is a totally real integral x going from minus infinity to plus infinity 1 over 1 plus x square this is a classical integral the inverse yeah it is pi or pi by 2 one of this do you know how to integrate it you do a substitution for x equal to tan theta then d x is something and then you do the integral right that is what we are doing let us do it through the complex integral. So, the first step is to think of this as not being done over a well this is certainly being done over integral over reals but I am going to change this and change the perspective and think at least the x being taking value over the complex plane. But I have to then see what is the region over which I integrate and here is the region or the domain that I pick up this is the complex plane. So, let us identify two points large at large distance plus r to minus r and the region that I define is this and then a semi circle centered at 0 of radius r. So, that is a region that is my domain for this function. So, let us call this domain d then I am going to instead of this integral I am going to consider this integral what is this integral hold this domain let us apply this residue theorem we should be what are the poles of 1 over 1 plus z square plus i and minus i. So, does any of this pole lie inside this half disk plus i yeah r is large let us assume r is large because we have to evaluate the integral from minus infinity plus infinity we are going to send r to infinity at some point. So, i lies inside this there is just one pole around this pole what is the Lorentz series expansion of 1 over 1 plus z square 1 over 1 plus z square of course, can be written as 1 over z minus i times z plus i. One of the simple way of calculating this a minus 1 if you think about it think in terms of Lorentz series you multiply because it is a pole you multiply the whole thing by z minus z naught. Let us postpone this I will come back to this later let us first discuss computing the residues. So, let us say f z is k going from minus n to infinity a k z minus z naught if n is 1 that is a simplest case of pole it is a simple pole at that point of order 1. Then the residue is a minus 1 and we get z minus z naught plus a 0 plus 1. So, the way to calculate residue would be let us take limit z going to z naught f z times z minus z naught sorry this is minus 1 multiply f with z minus z naught and take the limit as z goes to z naught you will get a minus 1 all other terms will vanish. The same trick you can apply for higher values of n, but the problem is that let us say for example, if n is 2. Then if you consider limit z minus goes to z naught f z times z minus z naught whole squared what do you get which is not the thing that you want you do not want a minus 2 you want a minus 1. So, what do you do take a differentiation once you have this this is a minus 2 plus a minus 1 z minus z naught plus higher powers differentiate this whole function and then take the limit and you get a minus 1 and now you can generalize this for higher values of n you just have to take a higher order differentials. So, that is the way to get it good. So, now come back to this example this is what is the order of the pole at z equals i 1 clearly. So, what is the residue then you multiply this function 1 over z minus i times z plus i with z minus i you get 1 over z plus i and take the limit z going to i. So, you get 1 over 2 i. So, this is 1 over 2 i times 2 pi i that gets you. So, this integral is pi, but this this is integral is not quite this. Now, let us split this that is a say consider the same thing in a different fashion this is an integral which is of the kind you want its whole minus r 2 r with this part being completely real because why is the imaginary part here is 0 in this on this line plus this is imaginary part of z equal to 0 look at this strictly greater than 0 well this should be strictly and look at the second integral what I will do is this is the kind this is the integral that I want once I send r to infinity this is the sort of the annoying integral which I do not want. So, any integral of the kind that we do not want you like to send it to 0 as r go to infinity. So, let us see does it go to 0. So, to check that let us just take the absolute value of this this is less than equal to what is absolute value of 1 plus z square when you integrate move around this circle of radius r well this minimum or the largest value it will take is 1 over r square minus 1 the largest absolute value that will happen when you have hit the here the absolute value starts with 1 plus r square and as you traverse it keeps shrink reducing and when you come here it hits 1 minus r square or r or then in the absolute term it is r square minus 1. So, it varies between r square plus 1 and r square minus 1 and since we are interested in an upper bound we will take this smallest of these values which is 1 over r square minus 1 d z and now you can do the integral is that it is a half circumference that is pi r. Now, when you send r to infinity why is it r square minus 1 oh yes you are right sorry I made a mistake yes this here is also r square plus 1 here it is r square minus 1 here because r e to the i pi by 2 and when you take square you get r square e to the i pi which is minus 1, but you cannot go below r square minus 1 and so the limit as r goes to infinity this integral is 0 which is very convenient because then when r goes to infinity this is the integral we want and we know this is pi. So, in this very simple way you get the integral you can try out more things like this, but I want to give you show you one more example which is something we will make use of later which is the delta function. So, let me define this or not the delta function it is like a step function you want to function like this it is equal to 1 between 0 and 1 and after 1 it just drops to 0 and stays at 0. So, this is a non-differentiable curve and the question is can we besides this simple definition can we get a more let us say more nice definition of this in terms of some complex integral and again it turns out that yes we can and using the same residue calculus. So, let me give you the definition first and then we will see why it is so. So, this curve is precisely given by this integral of x to the s by s ds and the integral is on the complex plane at any c greater than 0 and along this vertical line passing through c from going from minus infinity to plus infinity any questions on this c can be anything greater than 0 it does not matter. So, it does not matter where you place this line as long as on the positive x side by x to the s x is a real number x is a complex what is meant by this complex powering well this I should have x to the s is simply e to the s. So, we will prove it in the next class.