 So, where were we? So, that is the theorem we stated last time and I had a problem with this which is right this should be continuous the partial derivatives need to be continuous for this equal as to hold. So, the proof of this theorem I am going to leave up to you to work out and see and convince yourself. As I said that we are going to be interested in functions which are at least differentiable on the complex plane and therefore, this is a very critical or important theorem for us. Now, let me put define specifically the given name to the functions of the kind we are going to be interested in before that I need to define domain. So, this is the kind the type of sets we will be interested in D is a domain on complex plane if it D is an open set and such that any two points in D can be connected by a curve with descending formal way of saying something very simple that D is a region on the complex plane which is not which does not consist of disjoint parts. It may have some holes in it which is fine. So, this is the most general kind of sets we are going to sets of points we are going to be working on. In fact, more specifically we can rule out some funny boundaries also one can have boundaries which are not continuous or happen behave in a bizarre way. So, let me write that down as well. So, we are going to be interested only in the domains of the kind whose boundary now a boundary of a domain is this outer curve as well as things inside which throw out the piece the points inside. So, the boundary of a domain consist of a finitely many piecewise differentiable curve. So, this is one curve this is another curve this is another curve and they are differentiable. So, they do not behave in any funny way one could have things like this which is fine only finitely many. So, whenever I say domain I would mean a set of points of this kind. Now, let us come to our main definition we call function f to be an analytic function on the domain D if it is differentiable everywhere on D and this derivative is continuous. Now, this condition is actually superfluous as long as if f is differentiable on the everywhere on the domain D it actually follows that f prime is continuous, but we will not worry too much about it let us assume that this is the case I will not even going to prove why this is superfluous because it is not important. So, it is the analytic functions that will be our objective interest and study. Now, the reason I have defined an analytic function on a domain instead of on the whole complex plane is that the many of the functions that we will study are not analytic over the entire domain over the all the over the entire complex plane, but they are sort of piecewise analytic on different regions of the complex plane. A simple example is let us just take f z to be one word and let us see if it is analytic. One thing is clear in any domain which contains the point 0 this is not analytic because diverges at point 0 what about other points at least it does not diverge. So, it may be analytic how do we show that this is going to be analytic we have to verify that f is differentiable and verify that f prime is continuous and to verify is differentiable we have to use this theorem. So, let us say so we can write f z in this form this is my the reason for writing it in this form is that this is going I can write take this to be the function u and minus of this to be the function v and I need to verify the Cauchy Riemann equations for this. So, what is del u by del x and del v by del y. So, are these two equal they are equal and similarly let us look at del u by del y that is going to be and del v by del x that is going to be and this is negative of this. And so, what is f prime it is differentiable. So, that is good to know that then what is the derivative of f it is a usual derivative yes, but one has to actually justify that over reals yes we know how derivative how to calculate derivatives whether the same laws apply over complex numbers also is something that needs to be actually verified and that process I will lead to you it actually turns out that all the laws of differentiation which hold over reals just carry over to the complex numbers. And at some point later we will see a general principle why that happens, but for now let us just continue using the usual laws of derivative differentiation. So, therefore f prime z is minus 1 by z square and we need to just verify one more thing that this is continuous which is of course, the case for all points except z equals 0 this is continuous. And the second order the partial derivative these are also continuous as long as x and y both are not 0 which is something we are ruling out anyway. So, this together implies. So, take the complex plane and take out the point 0 this is a domain it is an open set with one boundary inside and this function is analytic let us look at another example is this analytic and if yes over what domain. So, we do the same thing this is u this is v and with this we can verify actually almost directly del u by del x is 2 x del v by del y is 2 x. So, they are equal del u by del y is minus 2 y del v by del x is 2 y. So, that is negative of this both are continuous functions over the entire complex plane. Therefore, f is differentiable over the entire complex plane is derivative again following the same principle is 2 z which is continuous function over the entire complex plane. And therefore, in fact, there is a name for this property functions which are analytic over the entire complex plane are called entire. So, you would naturally expect entire functions to be a simpler kind of function which do not behave strangely at any point. And that is indeed the case and there is a very nice theory of entire functions which tells us a lot of properties about of these. We will not know, but when we go into zeta functions that at some point we will have to touch upon entire functions also and at that point we will see some of the interesting properties of this. Well, so some should we look at more examples that is going to be somewhat time wasting. One should expect there are some general theorems one can prove about what kind of functions are analytic which is true. And one of the real basic ones which allows us to construct more and more analytic function is this theorem. So, if two functions f and g are analytic over some domain d then all their arithmetic combinations are also analytic except there one has to be careful when looking at f by g that g should not become 0 so that it diverges. So, as long as that does not happen f by g is also analytic. The proof of this I will leave as assignment problem it is very simple you just have to blindly follow through this the same technique you use f to be u 1 plus v i v 1 g to be u 2 plus i v 2 just do the operation and verify whether Cauchy Riemann equations over. So, I keep giving this assignment problem and at some point I will ask for you to submit those assignment solutions to those assignment problems. So, do not postpone solving these problems because if you do that you will probably even forget what were the assignment problems and if you come and ask me I will also would have forgotten. So, do not take that risk because I will remember while looking at at least one of you who would remember the problem. So, that so I will grade according to that, but those who have forgotten about some problem will then lose. So, now with these we are in good shape what immediately follows from this is that all polynomials are analytic. In fact, all rational functions are analytic over the domain where the denominator does not vanish. So, pick a domain in which q does not vanish at any point and then p by q is analytic over that domain well this is good to know, but if we keep our goal in mind the zeta function that is not a rational function it is an infinite power series actually. So, we will need ability to talk about powers infinite power series in order to analyze the zeta function. So, we will very soon come to analyzing whether a function given as an infinite power series is analytic or not. Well it is not you are right actually the way it was given is not even a power series it is an exponential it is an exponential series which we can write in terms of power series. Yes, so there is still lot of work to be done before we start discussing properties of zeta function. Now, very brief detour here look at an analytic function it is differentiable everywhere and not only that its derivative is continuous which means that if you think of the take the analytic function and look at its derivatives at every point on the complex plane. And so let us say this is some domain and you look at the look at the derivative take any point and look at the derivative of the point at that derivative of the function at that point and think of that derivative as giving you a vector which is what derivative does is a tangent vector right. So, at and look at the derivatives at the neighboring points again as a tangent vector. So, what the analyticity of f implies that as you move from any neighboring points towards this particular point and look at the way the derivatives of these tangent vectors evolve they change very smoothly and converge to the derivative value at this point. So, this has a very direct connection with the flows on a two-dimensional plane which I mentioned somewhere else earlier also that when a fluid flows on a plane then the with every point you can associate a derivative which is the velocity of the fluid at that point it has a direction as well. Now, if it is a flow without any turbulence or source or sink then this flow derivatives will we have this form where you can smoothly going from one point to another. So, that is the correspondence between flows in two dimension with analytic functions. In fact, analytic functions are heavily used to study these flows. Now, we come to perhaps the most important theorem on analytic functions almost everything subsequently follows from this theorem. If D is a domain and function f analytic over D then if you look at the integral of f over along the boundary of. So, delta D represents the boundary of the domain. So, here should say bounded otherwise this does not make sense it is a bounded domain not going to infinity anywhere then it has a boundary well defined you integrate f over this boundary you get 0 on function f to be defined on boundary yes that is root point that is a good point. So, I should write and suppose not only we want f to be defined on the boundary you want its extension to be smooth because D is an open set. So, you can with staying by staying within D you can reach closer and closer to the boundary and in the limit you will hit the boundary and there the function of f the value of f should be what the limit tells you. So, then this integral over the boundary is 0 this is on one hand very remarkable theorem that it does not as long as it is saying that f is analytic does not matter what other properties f has its integral is going to be 0. On the other hand it is if you think change your perspective it is not so surprising because what this is saying essentially again if let us take a simple example suppose D is in domain like this then what this theorem this so this is the boundary of D what this theorem is saying is that if you integrate f along this let us pick two points on this and you say integrate f along this path or integrate f along this path the resulting values will be the same because this is equivalent to the statement because it is if the two integrals are same then the integral of f along this boundary is this plus negative of this and that cancels out. So, essentially this theorem is saying is that along the boundary it does not matter which path I for take I take can take any two points and I can follow any path along the boundary to reach from one point to another the integral value is the same. In fact, since f is analytic over this entire domain we can say take fix these two points and take any path from this point to this within the domain D this integral value is going to be the same in respective of this path and this is not surprising because we have when we define the fact that f is differentiable we said that it does not matter which direction you approach f is going you are still going to end up with the same value. So, there is a direction invariance inbuilt in the definition and so it this theorem just extends that invariance to longer paths instead of localized invariance which is what the differentiability implies this is saying you know on a long longer scales there is an invariance. So, this is such an important theorem that we will prove it this also gives a sense of the type of proofs which are used in complex analysis. So, we start with the domain D and triangulate it. So, which means that and like this basically you make lots of triangles covering the entire domain of course, you cannot cover the entire domain with this because the boundaries particularly they will be curved they are not straight lines. So, in the boundaries you might have a situation where what you get is not a triangle, but something which has at least one of the sides being a curve that is something we will accept. What basically I want is division of this entire domain into small pieces still finitely many which have only three sides and as far as possible these sides are straight line, but if not then they can be curves. So, that is the meaning of triangulate for us here it that is yes we need to prove, but that is easy because this is a connected domain by definition domain is connected and it is bounded that is given by the in fact you do not even need connectedness for this because as long as if it is not connected it since it is bounded it will only have finitely many disjoint regions. So, for each region you can do this triangulation. So, this triangulation basically implies that breaking. So, what I am going to do is to look at each one of this triangle you know triangle in the sense of this and integrate f over the boundary of it and what I will show is that f on each one of this boundary integration is 0 and that is going to be sufficient to prove this because once you add up all these integrals of course, I have to fix an orientation of traversing these regions triangles as well and by convention we always traverse in the counter clockwise. So, that is considered positive traversal of a region of a boundary clockwise traversal is also used at times that is considered as a negative traversal. So, we will traverse all of them in counter clockwise and so when you have a neighboring triangle here this will also be traverse counter clockwise and if you look at this particular curve or edge it is once traverse in one direction second time is traverse in the other direction. Therefore, the integral some of the integrals along this will be 0. So, if each integral along each of this boundaries is 0 when you add them all up the resulting integral is of course, 0 and the boundary that you get of the resulting integral is precisely the boundary of this everything else cancels out. So, therefore, all that I need to show is for take any of such region this triangle in this general sense the integral along the boundary of this triangle of f is 0. So, for that we will start with something a very simple actual triangle and we will show that so let us say this is a point 0 0 this is a 0 and this is and we will show that that integral is 0. Notice that I have also given the coordinates of the corner points that is also something we will take care of later. So, let us call this t and what is delta t of f z. So, delta t is a boundary of t traverse counter clockwise there are three edges defining this boundary. So, we can break this integral down into three pieces integral going from. So, now I have to first transfer let us say yeah let us just say delta t this is our complex sense. So, let us translate this to the two dimensional real structure. So, f is u plus iv. So, let us write it u plus iv and dz is dx plus i dy. So, just collect the real and imaginary parts separately this becomes u dx minus v dy plus i u dy plus and since this is all real and this is all real we need to show that this is 0 as well as this is 0. So, let us consider the first one now we can talk about the three this as a two dimensional plane and break delta t into three parts in the first part only x varies it goes from 0 to a y does not vary in the first part. So, dy is 0. So, for the first part we only get u of x 0 dx in the first part y is 0 what do you get in the second part where make it even simpler let us make it 0 1 in the second part we go from 1 0 to 0 1. So, that is the line x plus y equals 1 or y equals 1 minus x that we need to follow. So, x goes from x goes from 1 to 0 and you get u x and what is y? y is 1 minus x dx minus what about y? y goes from 0 to 1 and you get v 1 minus y y dy and what about the third region in third region x is 0. So, dx vanishes minus y goes from 0 to 1 sorry 1 to 0 and you get v 0 y just rewriting it now look at this expression inside this this is u x 0 minus u x 1 minus x I can write this as there is just a trivial rewrite partial derivative with respect to y dy integrate from 0 y this is of course, u x y going from 0 to 1 minus x. So, that is just the difference I can see where we are going this problem with sign somewhere there is a problem with sign del u by del y this should be del v by del x. So, just interchange the order of integrals here. So, 0 to 1 x 0 to 1 minus y y same as 0 to 1 y and 0 to 1 minus x. So, this integral is same as this integral the limits there is just swap this and add this you get this this is 0 because f is analytic therefore, it satisfies Cauchy Riemann equations and according to one of those this is negative of this. So, this is 0 and therefore, the whole exactly the same proof works here except that here you will be using the other Cauchy Riemann equation. So, that shows that at least along the boundaries of this specific triangle is one unique triangle 0 0 1 0 0 1 the integral is 0. So, we will continue this analysis next time no it is a integral see this was a this was a area integral double integral x going from 0 to 1 and y going from 0 to 1 minus x. So, you basically are integrating inside the region, but you do need that the limit in the limit they should not some funny things should not happen, but that is already guaranteed by the fact that f extends smoothly over the boundary.