 Cauchy's theorem right and we split this proof into three parts part one to divide the domain suppose this is the domain into sort of generalized triangles not necessarily straight lines but smooth curves that is important the three sides must be smooth curves. So, one can for example do it something like this and so on you can say that yes it should extend f should extend smoothly to the boundary yes it may I will prove it but Cauchy's theorem equation smooth and boundary but in boundary it can be analyticity at a point that automatically in the domain it will be going to boundary. So, boundary may not at boundary it may not be analytic. So, let us prove it carefully then fine let us prove it carefully. So, first step you agree that we can split it in this fashion three shapes and integrate over each one of them and add them up. So, whatever is the summation of integral is the final integral. So, the problem boils down to integrating over a three this shape triangle shape and step two is that is step one step two is to look at a specific triangle the real triangle and look at the. So, let us call this triangle t look at the integral over delta t of epsilon and this we can split as a integral over delta t u plus i v dx plus i dy which is integral over delta t u dx minus v dy plus i integral over delta t of similar quantity and let us focus on this we can write integral over t over the boundary of t u dx plus minus v dy as integral over a double integral actually of 0 to 1 and 0 to 1 minus x let us say del u by del y plus del v by del x this should be plus and now we claim that because of Cauchy Riemann this is this is equal to negative of this. So, this is 0 however as you point out this integral is going from not only through the triangle, but also hitting the boundaries of the triangle fine. So, let us so we need to be careful about the value of this double integral. So, let us do the following let us integrate for x going from epsilon to 1 minus epsilon and y going from again epsilon to 1 minus x 1 minus x right little careful here. If x is epsilon this is 1 minus epsilon x is 1 minus epsilon this is epsilon yeah this is fine this is 0 because this is entirely inside the triangle where the Cauchy Riemann holds you think the boundary. Yes, I agree. So, then there will be problem, but this is not a problem this is fine inside is. So, now what this holds for every epsilon greater than 0. So, now take the limit of epsilon going to 0 these functions are continuous as you because of the smoothness of f on the boundary the extension of f. So, all we need is continuity to show that this integral when you take set epsilon to 0 remains 0. See as long as you think inside is continuous and we take a limit going to a certain point. So, as yes yes yes one of the side is complete sure that is see inside is as you extend. So, now let us take the extreme case let us take this triangle and let us assume that f is analytic inside, but not at the boundary, but only thing less we can assume is f extends smoothly to the boundary. So, the same objection that you raise are here for all three corners of this yet because of continuity of these two functions at the boundary this limit holds. So, that complete step 2 because this other integral you can do the same thing you will get something similar, but the other Cauchy Riemann equation will come into play. So, that complete step 2 now less the final step is that we have this triangle 0 0 1 0 and 0 1 and let us take an arbitrary this general triangle which has three corner points connected by smooth curves. And what I want to do is to translate this result to this boundary. So, one thing we can immediately observe is that there exist. So, there is clearly a map from points on this triangle to points on this shape and this map can be made continuous as well as differential because everything is if the boundaries are smooth. So, that is important and there are only three boundaries. So, map this point to this point this point to this point this point to this point these are the only three non-differentiable points which you made map to three non-differentiable points. And all the other this H is map to this H and there is a smooth way of mapping this and all the insides is map to inside this H is map to this H this H is map to this H. So, there exist therefore, and let us call this smooth map this is actually a pair of functions tau sigma from this is triangle t and let us call this t hat t 2 how do you prove this here you can formally prove this, but right now I am or for I will only appeal to your intuition. You can just look at real analysis the basic chapter the real analysis you can prove this just by using differentiability and continuity of between this and this of these regions. So, intuitively what we can do is you take this point map it to this and just move along mapping these points along this H and you can stretch see this length may not be same as this length you can stretch the map a bit and also keep turning around according to this curve the thing is that neighboring points get map to neighboring points right and the tangents are also move very smoothly because this is smooth. So, if you look at neighboring tangents they are also their values are very close to each other. So, that shows that it is at least mapping from this H to this is continuous and differentiable. Now, once you have fixed this then look at points here very close to this just map them points very close to this similarly look at this and map it to this. So, you can now look at vertical lines here and map them to lines like this or curves like this. So, that is basically the intuition here the whole domain is map to the whole domain very smoothly this is the x coordinate this gives you the y coordinate this is a 2 dimensional map. So, that is all I am going to say for this map I am sure all of you are convinced that there exist some map like this if you really want to get a formal proof go and read some book on real analysis. So, now with this in place let us look at this integral I want to show that this is 0 assuming of course, that f is analytic inside t hat. Now, again I will use that to write it as del u dx minus v dy plus i delta t hat and something similar expression. So, I will only show that this is 0 and correspondingly one can show that this is also 0. So, let us look at this integral and since you know going through the corner the boundary of t hat is same as going through the boundary of t and as you move along map everything using tau sigma. So, I can rewrite this integral as running through the boundary of t, but before that let us just write here u is tau and let us have 2 more coordinates here. So, here I am assuming that there is one coordinate system given by alpha and beta on that coordinate system the triangle t exist and I am mapping using tau and sigma to the x y coordinate system on which the t hat exist. So, this can be rewritten as u sigma dx is this is standard differential analysis minus v tau sigma del sigma by del alpha no, no, no, no these are 2 dimensional maps. So, u x both x and y vary for a point a point is specified by both x and y coordinate and for a point tau gives the x coordinate of the corresponding point sigma gives the y coordinate of corresponding. So, they both in general will depend on both the coordinates fine. So, we have this expression. So, let us just rewrite this a bit delta t let us collect everything for d alpha you get u tau sigma and I will shorthand this by tau alpha minus v tau sigma and this is sigma alpha plus minus and now let us recall this integral v u dx minus v dy integrated over the boundary of t is 0 as long as u and v satisfy Cauchy Riemann. So, the coordinate system has changed instead of x y now we have alpha beta. So, you have this as your nu u this as your nu v except that the sign is. So, the negative of this is nu v and if these 2 satisfy Cauchy Riemann with respect to alpha and beta then the integral will be 0. So, and the for Cauchy Riemann the kind of requirement is del u by del y should be negative of del v by del x. So, this is my u. So, del u by y is beta now del by del beta and I need to calculate u tau alpha and I need to show that this is equal to this del by del alpha u tau beta and if I can show this then the integral is 0 and everything works out. So, what is this left hand side del by del beta this is equal to del u by del beta what is del u by del beta u is over sigma and tau and sigma. So, I can rewrite del u by del beta as del u by del tau del tau by del beta which is tau beta plus del u by del sigma sigma beta yes. So, I need to show that this is equal to this. So, here I am just expanding the left hand side negative no this is already. So, this is v is minus of this and then what I need to show is that this is minus of this. So, two minuses they just cancel out that is plus tau alpha plus u tau alpha beta tau alpha there will be differentiated with beta is tau alpha beta. Similarly, this would be del v by del tau. So, let us short end this further I can write the u tau. So, this is u tau tau alpha tau beta ok long expression and what about the RHS using the same expansion we get u tau tau alpha u sigma sigma alpha tau beta alpha beta minus v tau tau alpha. So, let us look at the difference between RHS and RHS and see what terms cancel out this cancels this. The second term does it cancel no third this cancels with first term no fifth and sixth they both cancel. So, the difference is u sigma tau alpha sigma beta minus v tau alpha tau beta minus u sigma sigma alpha tau beta plus v tau all right now apply Cauchy remain here. So, this is u tau sigma. So, just a change of coordinates v tau sigma this differentiated with respect to sigma this differentiated with respect to tau. So, that is negation of this this is 0 and that completes the proof ok. Now Cauchy's theorem is the most fundamental or most important theorem in complex analysis everything that we are going to prove almost everything is going to flow out of this seems very simple theorem. Of course, you can see that it is quite interesting that integral around any closed contour is 0 for an analytic function, but its applications are really remarkable and we will see a sequence of such examples. Let me show you one immediately is a Cauchy's integral formula. So, if s is analytic on domain D and extends smoothly the boundary. So, this is a very standard assumption that we are always going to make then for every point inside the domain D f at z equals this integral. This is integral over the boundary of f of w divided by w minus z. So, it is very interesting that integral over the boundary is determining the value inside the boundary in a very simple fashion really all you need to do is divide f by w minus z ok. And the proof of this as I already remarked almost immediately flows out of Cauchy's theorem let us see that this function is not analytic inside the domain because of this z is inside the domain. So, when w takes a value z this diverges. So, it is no longer analytic we need to handle this except for point z this function is analytic at all other points ok. So, what we do is say this is the domain D and this is the point z. So, let us define a new domain out of it which is formed by striking out a very tiny circle around this point z. So, just cut this out from the domain you get a new domain. So, let us call the new domain and this is circle of let us say of radius epsilon and let us give it a name C epsilon. So, take out the circle C epsilon from the domain and define a new domain D epsilon. Now, in the domain D epsilon this function is analytic therefore, by Cauchy if you integrate along the boundary of this domain you get 0. Now, let us see this integral which is along the boundary of D epsilon. In what way the does do the boundaries of D epsilon and D differ the boundaries are identical except for this additional circle which is present in D epsilon and since it is inside. So, I think I remarked this already that the positive. So, whenever we talk about integral over a boundary we have to fix a orientation to traverse that boundary. So, for us the positive orientation would be clockwise on the outside. So, the form the intuition is that as you traverse the boundary the points inside the domain should be on your left. So, that is the positive ways of traversing. So, whenever we say that integrate over a boundary of a domain we say integrate over all the curves which define the boundary in this fashion that you traverse these curves to ensure that the points inside the domain are on your left. So, for example, this traversal will be in this direction the traversal here would be in the opposite direction clockwise similarly traversal here will be clockwise. So, when you traverse the boundary of D epsilon that is the direction of traversal for each one of these defining curves. So, I can write the traversal of D epsilon as traversal of D minus traversal of C epsilon because when I say traverse C epsilon that is in counter clockwise whereas in D epsilon we traverse C epsilon clockwise. So, that is the negation of the integral and this is 0. Notice that I can split it like this because on all these boundaries this function is well defined it does not diverge it only diverges at z. So, this implies that integral of f w over w minus z d w on the boundary of D equals integral of f w over w minus z d w on the boundary of this circle C epsilon. Now, what is the circle C epsilon? C epsilon is a circle which whose center is point z radius is epsilon and it traverses this in counter clockwise. So, and the distance between so and w is follows the circle right. So, w is therefore, exactly a distance epsilon from z always. So, unless whenever you traverse a circle it is always good to change the coordinate system and move into polar coordinates. So, we let w be z plus epsilon e to the i theta that precisely captures traverse the points along this circle right w plus epsilon is a radius from w and e to the i theta as theta goes from 0 to 2 pi traverses the whole circle. So, then this equals 0 to 2 pi f z plus epsilon e to the i theta what is w minus z that is epsilon e to the i theta but I will fix it that formula and what is d w is epsilon i e to the i theta d theta this all cancels out you get an i outside. Now, I think you can see the proof now we send epsilon to 0 as epsilon tends towards 0 where does this quantity tends towards f z when epsilon is very very close to 0 this is more or less f z. So, actually let me be a little more specific let us write it as 0 to 2 pi f z plus f z plus epsilon when epsilon is very close to 0 this is very close to 0 try you can put an upper bound upper bound which and that upper bound tends towards 0 as epsilon tends towards 0. So, this whole integral will therefore, tends towards 0 as epsilon tends towards 0 because the quantity inside this tends towards 0 and the integral is finite this d theta or sorry this is d theta this is d theta at most is integrates around values 2 pi. So, this whole integral will tend towards 0. So, and we can send epsilon to 0 without affecting the integral on the left. So, what left is left out therefore, is f z is 1 over 2 pi i integral around this needs to be tends to 1 over 2 pi. Yes no along as you integrate along the boundary this is not w is never equal to z. However, to apply Cauchy's theorem if you are integrating over a boundary of a domain this function the integrand has to be analytic inside the on all points of the domain which it is not. Yes how do you formally put a bound. So, we use basically just use continuity this is continuous. So, as epsilon tends towards 0 there is some delta which bounds this value that is simple continuity and this delta will also tends towards 0. So, this whole integral in absolute value will be bounded by 2 pi delta and when you send epsilon to 0 delta tends towards 0 and therefore, this goes to 0. Yeah math 101 those of you who did this who have done this course one of two I thought it is no it is this continuity epsilon delta that is in 101 right. It is and it is quite a it used to be a very famous course because the first year students first class they are hit by this continuity in epsilon delta and half the class does not understand what is happening. I do not know how it happens now probably the same, but it is actually useful. You can see the continuity is very useful in analyzing function f is analytic around f is analytic on z and around z eventually I am going to show that any analytic function on a domain can be written as a Taylor series. That is going to be a very key theorem on analytic function, but to prove that you need all of this putting limit inside integral also requires. Yeah there are many things one has to be careful about when moving things inside or outside integrals or infinite sums and I am sort of not worrying about them too much here because otherwise they will just get stuck in them. So, whatever appeals intuitively we will just go along with it. There are some points we need to be careful about yes and then I will highlight those points and one point that you raised that was very useful that even though they are on the boundary the function is not analytic one can still extend the integral all the way up to the good. So, what is next? So, you might say what is interesting part of this theorem says that you can get a value of a f on any point inside the domain by integrating along the boundary, but that in itself does not sound like such an exciting result, but what follows is really exciting. If f is analytic then x is infinitely differentiable the mth differential is given by this integral formula again this constants are a little suspect, but we as we derive this we will figure out the constants. This is remarkable because this is completely different from real analysis a differentiable function which is what an analytic function is does not have to be second order third order fourth order differential here it is if it is first order differentiable is differentiable all the way and there is a nice formula for calculating that as well. This is a nice correspondence between the differentiation and integration here as well that you can calculate the differential using integral or vice versa actually the calculate this integral using this differential as well. So, we will prove it in the next class.