 Okay, today we will be talking about complex integration as announced in the previous lesson. So we start from a very natural generalization of line integral for complex valued functions. This is chapter complex integration. So we have a function f define an open set u of c and to c consider the variable to be z okay and the components of f to be u and v as we have already established several times. So now we take the interval a, b on the real and we want to define this which means that f is restricted on the interval on the real axis. We assume that u of course contains a, b otherwise it is not meaningful okay. So the natural way to define this integral is the following. You consider the integral of u of t, t plus i of the integral of v of t correct. Which means that the output of this operation to start from an interval from a complex value function is a complex number if it exists of course. For instance if we assume that f has continuous components then these numbers exist right. Some properties, so basic facts, properties of this integration well two basic properties are somehow condensing one sentence this integration is linear which means that when you take the integral of the sum of two functions to complex value functions and you take the integral of this sum and this sum means that for any t you sum the complex number corresponding to the values of the function when calculated in t of course so it is meaningful. This is a, b, f, t, d, t defined as before plus and this is the sum in the complex the integral of a, b, g, t, d, t. I invite you to verify that this is true and the second property is that when you take the function f of t times c complex number then this is this has an integral which corresponds to c times the integral of a, b, f, t, d, t. So this is left as a very simple exercise followed by just the definitions nothing and this will be done here because then we apply again. So take c to have alpha as real part and beta as imaginary part I cannot use a and b because a and b are in the interval this time. So when I consider c f of t I have alpha plus i beta v of t plus i v of t right which is also alpha sorry alpha u of t minus beta v of t plus i beta u of t plus i alpha v of t. Therefore the integral of c f t d t is the definition of the integral of the real part of this number u of t minus beta u of t and t plus i the integral of beta u of t plus alpha v of t and t is it. So here I also recognize that this is c times u of t you see this alpha and beta this is this is a real number and I already know that for the real for the real constants you can take out the constant from the symbol of integral because it is r linear. So this is true for this pair of summands and here I have what plus i then I have alpha v t and then beta is with a minus in front which represents also replaced by i square so it is i take out if you want and i. So I have i c integral of v t which is this is a real number this is a real number so it is c integral according to the definition we gave so the second property proved and as I said it will be useful for this important inequality we have to prove want to prove. So the inequality want to prove is the following so the integral so the modules of the integral of the line integral is smaller or equal so this is a real number right this non-negative real number on the right hand side I put this which is the integral of a real value function so it is another real number so we can compare the two numbers of course there is nothing to prove if a b the integral of f along the interval a b is zero so the order modules is zero we will intally because this is true but how can we prove this in general do you have an idea the Riemann sum okay this can be a strategy but this is very long I never well I never used Riemann summation Riemann sums to define the integral I somehow applied what we know from real integration which means of course Riemann sums okay I think that well this is possible it's very long so the good idea is to consider to apply the properties we have found so far so we can okay this this okay in some sense this is what we will do this the same inequality same notation same inequality is true for real value function all right so we'll restrict our consideration to the real part in particular or something with a constant in front and then we obtain the desired result but the idea is of course to to use the inequality of course of the real and imaginary part you cannot use anything else so okay as before take as a constant this number which I write this way this is a complex number correct and theta is arbitrary an arbitrary real number right and then we look for an estimate of the real part of c integral a b f t dt okay this is a real number this is well defined correct and this is a real number correct now we know that this is the real part of the integral of a b c f t dt because of the property we have shown before and this is nothing but the real part of the integral e minus i theta f t dt now from and this is well this is just replacing what we know well you use the prop one of the properties linearity of the integral of the line integral over the complex numbers for complex value functions and this is just replacing c with now remember that by definition when you take when you consider the real part of the of this line integral you are considering the integral of the real part of the function right for definition remember that the real part is the integral of ut dt while imaginary part is the integral of v t dt so this means that we can also somehow put the real part inside okay so this is what students normally describe during oral exams okay so we take the real part the real component of the function we have written here right okay now but the real part this is okay this is a real value function and one of the first estimate we have considered for complex number is that the real part of a complex number is smaller or equal of the modules of the complex number right and now we can continue by saying that well this is smaller or equal to the integral of and the product of complex number has a more as modules which is the product of the module i and the modules of this number here is one so that this is the integral a b f t dt right now whoever proved the inequality we were looking for so far however from this we can conclude quite easily because we have the freedom of choosing theta as we like right so in particular so remember this is the inequality the real part of c times the integral is smaller or equal of the integral of the modules of f good now this number here can be also written at this is a complex number right when it exists of course everything is the modules of this complex number times what e i the argument say arg of the integral if we use if we use the polar notation this is the module this is raw time cos say mu plus i where mu is the arg of the number correct now i have the the the the possibility to take as t as sorry as theta to be actually arg of this i put arg maybe the same arg right this is a number right this is a real number so theta was arbitrarily chosen in the previous consideration it it cannot in fact it doesn't change the inequality here it doesn't affect any any change in our consideration now we are considering this result and therefore for theta equals to arg when you take c then remember that here we have this right so this is e minus i theta integral sorry modules times e i theta if theta is this right so this cancel this because they are one the inverse of the other and then we have the inequality the inequality we have the inequality right and this will be very important for many estimates now the next step is to consider instead of the interval a piecewise continuous at least path in the plane which is the natural extension and generalization of the interval so consider gamma to be piecewise assume that it is differentiable okay in season so we have that gamma is defined in an open set of the reals and takes its values in c and it is well sufficiently regular except for some odd points okay so here you have tangent vector in the real sense okay but here of course this is a corner point okay now assume that well gamma of a b is contained in u and f is a complex valued function taking its values so define in u so then we can consider f of gamma of t of course huh so the composition is well defined and it turns out this composition is a function from interval into the complex plane so we define the integral over gamma of f z and of course z is the variable now so I have to put d z here because we are integrating and and the complex plane this turns out to be by definition the integral of this function here call it if you want g of t with respect to t correct so the interest in between a and b because gamma was defined so it is it is reduced to a line integral again but of course we have to take into consideration that we have gamma here as new variables so we have a change of variable so it is properly considered and the new chain in a new variable so this is a line integral and this is an extension for nothing new okay of course I believe you are all familiar with this stuff we can also make another change of variables you can take another variable tau for instance and the change of variable between t and tau the interval has to be of course a change of variable so it has to be a more or more increasing function right you can substitute one into the other and take into consideration this fact without great efforts I guess because it is essentially real integration what we observe as a consequence from the definition is that well you have like for the interval since the interval is a subset of r similarly the curve the curve gamma as so on the curve you have the same orientation of the interval so you start moving from the point gamma of a and you reach the point gamma of b since t is increasing between a and b you imagine that the point is moving from gamma a into gamma b so there is a preferred sense of on the curve gamma and what is natural to do is well if you reverse this sense this order well you put a minus one in front so so what is natural to use as a notation is the following so you take minus gamma with the convention that this means the same set of points but the curve is okay starting from bay and moving so if you want it is reparameterized in such a way that gamma of t so sorry so gamma tilde of t is gamma of how can I write it it could be minus t but this is not them b b b plus when it is zero right so b t a sorry one minus t right what's that so t zero this is no anyhow you know what I mean you know we have say assume just one piece and this is gamma of a and this is gamma of b right so we move from a into b on the parameter set t here and this way then we start from here and move back to here so that the point is that this interval is not zero one okay so t is not moving okay it's like you're okay probably it's like this a minus t sorry it is sorry anyhow with minus gamma I mean the same the same function but you take the opposite orientation on the interval okay so with this in mind and remembering what happens for the real integration so if you reverse the end points of the interval which is what we have done right you obtain the opposite of the interval okay and that's that's what we have also because we also have the possibility to join different path okay have gamma then take eta and so on or gamma one gamma two gamma one gamma three and so on and so forth and if you imagine that the end point of gamma j is the green is the same as the starting point of gamma j plus one okay you in fact have a collection of pieces or if you work more in general you take gamma to be the union of gamma j's and the integral of a gamma of f sorry fz here is fz sorry fz right fz dz is summation of the integral over gamma j of fz dz right about notation which is also important to to recall we use dz sorry this is seven so remember the z was normally written as x plus i y right so what is dz well dz is if you want by definition is this if you want by definition but this can be also obtained because imagine at this point here depends on t this is x of t plus i y of t right so if you take the derivative with respect to t then you take the differential okay you have the dz of t right over dt dt is dz right that's how dz of t over dt is dx of t over dt plus i dy of t over dt which means that this is dx plus i dy and similar dz bar dz bar is the this is called form right this is the form with the minus in front of y okay which can be considered as formally as the conjugate of the previous or the previous form now what do we mean by considering this the integral of a gamma not with respect to z but the dz bar this has to be defined or to be calculated right well if you replace everything in this in this notation this is the conjugate of the integral of the conjugate just verify if you want right and now we have also the possibility to express the integral over gamma of fz dx well dx is related to dz in this way right plus i dy remember dz bar is dx minus i dy and this is one half the integral over gamma fz dz plus the integral over gamma over fz pardon me on the here fz of dx okay now that's what i'm saying right so this can be expressed in terms of the integral the integrals we have a difference so far so i invite you also to to write okay this is a matter notation we have we are done i guess so now remember that f of z is normally considered as function of u and v depending on z fine therefore i can also use this expression to calculate the and dz is dx plus i dy so that i can reduce again to some stuff which is more real than than the other okay there in the other in the other definitions that it is intrinsically it is complex but then you use the line it it becomes real well it is two I force a counterpart here let's see this is after some modification from obvious calculations is uz v minus vz x so uz dx okay dx and dy correct yes plus i integral over gamma vz dx plus uc which means that we can also consider the integral of a gamma of fz dz to be expressed in this way the integral of a gamma of function p of z x plus c dy of course z is x plus i and y and if you want to be more precise and use this equivalent so this is a compass value function depending on x and y then integrate with dx plus i something else similar so it's up to you in some sense you can define this integral either by considering this which is also something meaningful in the real in the in the sense of the reintegration or use the other opposite opposite but equivalent statement now what is well known I guess and in the in the real integration theory is the following so assume that we want to prove I can also this is I right yes but I can put inside I call it and put it in a semi integral it's I because well it's not so important but I can put I in front of q right and I want to say something about the possibility to do to to express this integral without taking into consideration the path but only the end points of the path okay this is quite say a standard result in playing in a real analysis with more than one variable and you ask well is it true that if you take the integral along a curve gamma this integral at the same value if you take the curve the integral over another curve with the same starting and ending point is it always the case no it is not and it's it can be characterized okay there's also a physical interpretation okay okay yes conservative fields are related to this okay do you have example of conservative fields gravitation good electric field yes good so in physics this is applied several times and it has equivalent it can be expressed in equivalent ways but the integral does not depend on gamma but only on the end points or the integral over closed curve which means that the end point is the same as the starting point the integral over the over a closed curve is zero but this all is characterized in the following way so let me change and say the following and so I have that depends only let's say on the end points gamma a and gamma b but not on gamma that is to say I have gamma a here gamma b here this is gamma if I replace gamma with another curve sigma with the same starting point so such that sigma of a is gamma a sigma of b is gamma b and then the integral over gamma of f sorry of p dx plus q y is the integral of a sigma of p dx plus q okay if and only if there exists a function say capital U such that U is differentiable and U as derivative with respect to x equal to p and derivative of U with respect to y is q this is a theorem in an equivalent way in many books especially in standard mechanical again sorry standard classical books from mechanics from so standard calculus this is also summarized by saying that this form here p dx plus q dy is an exact differential okay so every time you have a function like this in physics it assumes the role of a potential okay so if you want to rephrase the statement once you this theorem is proved that the in terms of physical interpretations the field is conservative if and only if if it has a potential right now let's see how this proof goes the sketch of the proof right are you familiar with this stuff I said something you already know yes I hope so now let's see okay never heard this word well this is just a word in in physics it means well you have a function with this property we are mathematicians so we prefer to define it in terms of the partial derivatives and so on but it has a meaning okay so if you want you have a field and this this field so to say for instance to move an object from here to here you have to make some work right you have to spend some energy but well this rubber now has increased potential gravitational potential so I have my my muscle here work against gravitational okay so I'm doing some work and this object here achieves the potential in fact it falls down okay so this is in potential the energy and this is the energy spent and when it arrives here all the energy the potential energy the gravitational energy is spent well when it is moving part is still gravitational and part is cinematic right sorry it's kinetic kinetic okay energy because it moves only hits it hurts okay it has energy when finally it arrives here so so consider this as a level zero for our consideration okay you can rescale everything so the energy potential energy is zero but also it it does this object is not moving so the entire energy is absorbed in terms of internal energy so in essentially heat energy but it's just a name right so we now consider more abstract aspect so we have to consider once you have a function with this property the integral improve that this integral does not depend on the curve vice versa if you assume that the the the integral does not depend on the curve but only on the end points we will so construct a function you with this property and that's what it will be very important for our consider future consideration okay let me continue so this is so assume that we have you of course to be very formal I have to specify where it exists of course we have to assume that the function you has to have derivatives so partial derivatives with x and y okay now the integral of a gamma of p dx plus q dy as then replace an integral of du over dx dx plus du dy dy and this is by definition the integral over a b becomes a line integral right of du dx x prime of t dt plus du dy y prime of t dt which is also the integral to a b du dx x prime of t plus du dy y prime of t dt which is also by definition the derivative of u with respect to t when u is restricted along the curve x t weight so from very basic calculus this when you have here an expression which is the derivative of function well this is nothing but u of x b y b minus u of x a y b y a sorry which means that what is important is to consider the starting and the end point evaluate this potential there consider the difference but not the curve so this number here is independent of the choice of the curve it depends on your own the end points of the curve vice versa so assume that the integral does not depend on gamma but only on gamma a and gamma b with some views of notation I said that well x and y are in fact the components of gamma if you want okay when you restrict on the interval a b gamma of t is x of t y of t which varies okay in the previous slide I use this okay I assume that this is a natural way to indicate if you want you can say well gamma of t is in fact x t plus i y t but then as usual this is identified with the pair x t y t in the plane R2 we are considering which is probably more familiar to you to work with so now now let me make a small so we have here our set omega where functions are defined everything works fine and we have this end point here right gamma of b and here gamma of a so we can choose any curve connecting gamma a to gamma b and if the curve is piecewise differentiable and so on the integral is defined independent of the curve because this is our assumption okay so the curve has to have some regularity but for the rest it doesn't really give any effect on the value of the integral so in particular we can assume that well as a curve we take a polygonal path connecting these two points with segments parallel to the axis okay this is in the Gauss plane assume that this is x this is y so in particular we take polygonal path like this okay for instance then another one is like this or more sophisticated stairs okay so assume that the well and and the choice is completely arbitrary now assume we consider with the last segment parallel x axis which means that y is fixed in the last part right so and we can keep y constant the last segment if it is parallel to x axis now we define u of x y to be the integral of any say say this point is uh more rotation used here okay x p y b gamma of p x y dx plus q x y u y or gamma is polygonal path okay the choice of the polygonal path it just because we want to prove the second property of u but u is well defined take you take sorry state start from one point and you define the function capital u and the following way take any curve not necessary polygonal but for how it is useful okay which connects the starting any starting point in gamma to this point and consider the integral since the integral is independent on the curve but only on the end points okay this function is well defined okay just to come back to the previous example the starting point is equivalent to the side in this stupid example to decide where the potential is zero just to rescale okay I decided that this is the starting point so for instance at that point the function u of is zero the potential is zero change the starting point you have another potential okay correct so for instance this is zero for my consideration it's not zero if I if I take this point here right this level here is level good so in the last in the last segment what we do have is the following y is constant and c is moving so that u of x y and the last segment is something like depending on x only p x y the x plus a constant because y is constant to some plus something where from this if you just move x they have an integral over x or something while the derivative is just the integral function and similarly since the choice of the curve is completely arbitrary you can also come to join the same point starting and ending point with a polygonal path with a vertical segment at the end and you obtain the analogous yes please no no no no this is a constant to x not probably but when then you take the derivative is not depending on x no okay let me let me say it again probably to have it's worth saying a few words so we are in gamma this is a starting point call it z not gamma of a okay and then we connect this point z in gamma of b using this segment in the last part okay of the polygon up to here when you have one other say when something like this but you can take whatever you want okay the value of this integral is given and depends only on this point correct because we are assuming so so when you actually consider the last part of the polygonal path it depends on your next not on y but the constant here depends on this value here and it is independent max it's a constant number so which represent what if you want using again the example which represents the the work you have done up to a certain level okay then you add something new yeah it's the one it's only one parameter you can choose okay but we have two parameters in our example so the constant I used I used I indicated here is what you have done up to this point where the the last segment the horizontal segment starts then on here you take the derivative with respect to x but since you have a constant added to an integral which depends only on x the constant disappears so it's zero right and what is left is the integral function that is to say p of x if you repeat the same argument used by using but substituting the last segment and you are allowed to do this because the integral does not depend on the choice of the path you take as last segment in the polygonal path for instance a vertical segment again you take whatever you want up to here as a path you integrate in this number here gives you a constant number then the last the last contribution is just the contribution along the vertical line so it depends only on one parameter y so you have a constant plus something which depends on y so the first one depends on x it's a constant second part depends on y and the derivative with respect to y is just the integral of this the second the second function because it depends it's integrated with respect to y this is what I wanted to say so now it is natural to ask well this is something okay your probably already seen in in calculus courses in physics interpretation so this implies in particular this implies that the integral over a closed gamma sometimes it's indicating physics like this sequitation right of p x y d x plus q x y d y is zero gamma's closed curve that is gamma way is gamma which also means that the work have to do to rise this rubber from here okay so there the energy I have to spend okay is the same I or if I so closed curve means well the starting point is then why is it well of course because well you're this is gamma way a equal to gamma of b take any point different from this to this point take gamma of c huh so imagine that you are running along this curve using this orientation so and you have the integral from here to here and then the integral from here to here correct but then this is the opposite of the integral from here to here you see this so split the curve closed curve the closed contour into two curves they call gamma one and gamma two right so the integral over gamma one between gamma a and gamma c okay it's like the integral from gamma a to gamma c along this gamma two but you have to reverse the sign huh that is to say if you sum both the integral is zero because one is the opposite of the other same value but different different sign and this is very important okay now the question is the natural question which can arise now if you are considering this is a course in complex analysis well how is this related to complex value function so this is this is nothing but well real analysis and in fact the question natural question is when the integral of this does not depend on gamma but only the end points in fact we write what we have here we have actually p and q written explicitly because remember that d of z is dx plus i dy so that p x y is f of z and q x y is i f of z do you see this in fact this is the integral of f of z of d z or the integral of f of z dx plus i dy or the integral of f of z dx plus i f of z dy over gamma and this is actually p and this is q as usual with abuser notation I say depending on x and y set over depending on z but there is this correspondence okay so I have this so the question is when is it well I have the answer the answer is when when we have another function u the potential such that the answer is when there exists a function u such that u of x is f of z in our case p right and u y is i f of z which is our q in this particular for this particular choice of the form right but then u is u is a complex value function assume that u is a plus i b for instance then you have that u of x is a x plus i b x and u y is but we are saying that you see these two conditions imply that u of x is i u is minus i right or yes minus i u y correct no f of z times i so i u of x is u of y this is correct that is i a of x minus b of x is i y plus i b of y or this and these are the same and this and these are the same that is to say a of x is b of y and i of y is minus b of x which is easily seen to be in fact the condition the Cauchy Riemann equations for capital u so the answer is yes it is possible that the integral of a gamma of a complex value therefore z d z in d z is independent of the curve gamma but depends only on the endpoints or if you want if you take a closed gamma a closed curve gamma the integral is zero but this is when there exists the potential which is holomorphic so these are then u satisfies as this equivalent say the u satisfies Cauchy Riemann equations and this means that u is this holomorphic rephrasing again we can say well so the integral over a closed gamma a closed curve gamma of f of z d z is zero if f is the derivative of a holomorphic function do we have examples yes we do have examples of functions which are derivative we have the notion of derivation in complex sense for instance if i consider this function as a function of z a is given n is given is it true that considering the integral over gamma a closed curve this is zero this is an exercise well it is a basic example but yes the answer is yes of course until n is different from minus one why well because you see this function here the integrand is in fact a derivative with respect to z what we are just using standard rules of so if you want of derivation and integration okay and it is well one over n plus one that's why n has to be different from zero of z minus a to the power n plus one correct so the integral is zero so any potential well this is of course also for a equal to zero we are done what if n is equal to minus one okay this is maybe the example we have to deal with yeah but the logarithm you know is somehow not function okay first of all i want to show you that the integral is not zero over a closed curve which means that you don't have a potential okay so take n equal to minus one so that is to say we consider the function z minus a minus one okay this is allomorphic yes except for z equal to a right it's well defined it cannot be even continuous so in a so it's easily seen that cannot it cannot be holomorphic in a right because we know that if it is completely differentiable and it is also continuous there i can have your case now take as gamma the closed gamma to be a plus r e i t which is a circle centered at a of radius r t of course varies between zero and two pi this is a closed curve this is the closed curve consider in our consider in our example so we have what we have a function defined in any point but not at a and we take as a curve closed curve a circle centered at a of radius r there are very many but take one okay and i want to calculate the integral over gamma of one over z minus a d z correct which is the integral over the two two pi substitute okay z is remember a plus r e i t that is d z is what r r i e i t d t what is z minus a well is r e i t from here right correct so i have that we have to substitute d z then i have r i e i t d t and then consider r e i t r cancel r r is of course positive otherwise we are not considering a curve right d i t is not zero so so what is left here is the integral of the t between zero to the pi with i in front so it is two pi i which is very far from being zero so it's not always the case that you have a potential but in several cases you have a potential any time any time you have a primitive function in the sense given before that these to say if you can find a holomorphic function whose derivative is the function and the integration now the next task is to generalize well this previous example the previous example and this example here is not just by accident one example if you remember the very generic class of function we were considering or the class of function which are expressed in terms of power city expansions right and in the power city expansion exactly this was one of the elements which appeared with a constant in front but this is zero the constant is not a problem so you can also prove that a n z minus a n is zero right put an a n in front the only difficult fact well if you have a polynomial we are done if you have a finite power expansion which describes and defines the function the analytic function well it's fine but when you have an infinite sum well you have contributions of infinite summands which singular zero but as you know this is not sufficient to conclude that it is zero and it will be actually true that if you start from a holomorphic function okay any holomorphic function is such that the integral of a closed curve is zero the previous counter example if you want is somehow the easiest counter example which is in fact a function not holomorphic in an entire plane but the entire plane except from one point and the curve is surrounding this point all right so before considering a very generic case let me just catch you one okay one result which is known under the name of a Gursa theorem and according to alphors uh no uh easy proof can be found in the entire complex analysis theory so the the theorem i want to prove is the far take f holomorphic in a rectangle rectangle r the rectangle is a subset of c and includes also the signs the edges okay of the so it is an open set and can be the domain of definition of holomorphic function but since we are considering also its edges and we want to consider in fact the integral over r of f of c d z we have at least to assume that f is continue up to the boundary but so with some lack of general we assume that f is continuous on the closed rectangle which means that in fact it is holomorphic in a small neighborhood of the rectangle itself because remember the definition of holomorphicity requires an open set okay you have to have the possibility to calculate the derivative so you to move a little bit around the point okay correct now i want to show that this is the case this standard case of the function of which is holomorphic of a rectangle guarantees that the integral of this rectangle is zero how do we prove this so this is Gursa's well first of all let us indicate this all right we have to prove this right call this number eta of r it's a complex number we have to prove that this complex number is zero and the method to prove it is by induction some sense and using the bisection procedure which is quite common and also in real analysis so the observation is a following when i calculate the integral over the the rectangle of course i'm assuming that there is a verse you are running around this rectangle okay this there is a known orientation so for instance you start from here we'll be here here because this is a closed curve and this is a generic holomorphic function so we are not assuming that f is specific it's a generic now we are requiring some extra hypothesis for the curve the curve is a rectangle it's a closed curve all right so there is the following the rectangle into four sub rectangles by subdividing the edges half edge half edge okay and then we have four rectangles call it call them r1 r2 r3 and r4 and according to the orientation of the the previous rectangle we also orient the other one as you can see if i calculate the integral over the four rectangles there's a contribution here which coincides with the contribution over the bigger rectangle then a contribution here which is cancelled by the contribution along r3 each internal edges has a contribution of this positive and the other is exactly the negative the same here here is positive for r2 negative for r1 you see this correct so that i can write that the integral of r of fz dz is the same as the integral of of the same function over the four sub rectangles are you with me because the only contribution which do not not cancel each other are the contribution on these sides here so that the contribution of the edges of the previous say it's a of the starting rectangle i'm say previous because then the procedure of course explains you why we are now restricting and focusing on one rectangle and then going on because we have now this in a quad so we take this and this is smaller or equal of four times one the the integral of one of the rectangles okay i don't know why i don't know which one but one of the rectangle is greater than the other so these are one of the rectangle one of the integrals is greater than the other i take this and i have this inequality okay this is an equality take the modules this is smaller or equal of the sum of the module i of the rectangles but there is one which is bigger than the others and i say four times the modules of the larger rectangle if there are two okay choose one okay if they are all equal okay choose one and then repeat the same procedure on this rectangle subdivide this rectangle into four small rectangles and repeat the argument all right so i say that this is r1 the first but not because it is the first in this sense in the subdivision but it's the first in the procedure right and i can repeat it how many times i want right so i have a chain of rectangles i r1 r2 rn each one is included in the previous one and i have this in a quote the integral of the starting rectangle is smaller or equal or four power n of the rectangle at the level n of the dc so each time this number is divided by four the preview inequality i had four here then i repeat four four four four right so imagine if you have a length and you can okay zoom and you see the same image every time okay at any scale you repeat the same argument you notice that well this is the length of the diagonal of the first rectangle and the second step the diagonal is one half the the edge is one half the previous edge and so on and so forth what do we have we have a net if you want huh of of rectangles and well by the the component this sequence will eventually converge to a point inside the clothing inside the the rectangle okay okay so z naught has the limit of the sequence on j's is it correct so we are moving but somehow reducing the the the the the dimension of the rectangle we are considering up to a point well at this point our function is holomorphic right because it is inside well z naught is in r it's a complex set so okay it's a close set in particular good what does it mean it means that well we can we can take f of z minus f of z naught over c minus z naught for z and in r okay and we know that this is very close to f prime of z naught if c minus z naught is smaller than delta huh because the function is holomorphic you can also rewrite this in a quote in this way the square modules of f of z minus modules of z naught minus f prime z naught times z minus z naught is smaller epsilon times i observe that the integral over r of f of z d z is modules right was it modules or i don't remember if i put modules by v modules right or precisely huh and then i have also to be the integral of r j now from this from a previous inequality i also have that the integral of this stuff is smaller or equal of the epsilon integral of z minus z naught yes minus yeah i mean yes correct this is done with respect to z d z correct so i calculate this over more precisely i would have chosen this notation for the drn all right not the inside but well r is the entire domain and the the edges are drn right means along along the contour okay this does not give you any contribution to the integral because we know this is zero this is a constant don't don't be worried about f prime this is a number this is a constant and this is well just because of the previous example just z minus z naught to the power one if you want so it has a primitive function and the integral is zero and similarly also the integral of a constant is zero because you have always a primitive of a constant so and i want to finish but i'm running my time i'm over my time 21 okay so i have that on the one on the one side i have the integral and this after some consideration is greater of the modules of this right this number here well if you think of z naught is here i have this is rn so i have z running around here right so this number here is at most the diagonal the largest distance between two points on the boundary is dn so this is smaller or equal to and then i have well if i replace this by the end then i have just the integral of the boundary of rn so this is ln well n is the sum of the length of the edges right but look every time when i subdivide and i cut the starting the the rectangle into four small rectangle the edges the length of each edge is the one half of the previous one and similarly also diagonal so at stage n you have if d is the diagonal of r of the starting rectangle dn is d over two to the power n and ln the sum of the edges is similarly l over two power n so i have and this concludes the proof i guess this is 21 and this is epsilon ln pn greater or equal to epsilon integral over rn c minus z naught dz is greater or equal to the integral of dn over fc dz modulus and this is remember is the integral at r over four to the power n remember the nth integral then this is d over two n and this is l over two n right in other words i have epsilon dl over four to the power n is greater or equal to eta r four to the power n and since epsilon is taken as small as we want we conclude that this number here is zero okay because these are two finite numbers and epsilon can be taken as small as we want this is the only possibility so we conclude that in fact four special curves closed curve the contour of our rectangle the fact that f is holomorphic in the rectangle guarantees that the integral of this contour is zero all right and i stop here for today thank you and sorry for the delay