 OK. So we continue, and we use the notation we assumed so far. We take this power series summation ak z minus a. We assume that the radius of conversion is r. We also denoted by f1 of z, the series obtained in this way, ak z minus a, k minus 1, k greater or equal to 1. We already observed that the two series have the same radius of convergence, so also the second one has the radius of convergence r, and then in some sense we decided to split fz as snz plus rn of z, where sn of z is the expansion up to the order n, so it is a polynomial, and the remaining part, rn, is the tail in the power series expansion. So sn is the following, so it is a0 plus a1 z minus a plus a2 z minus a squared plus an z minus an. So we observe as well that if we define s prime of z to be a1 plus 2 a2 z minus a plus n an z minus an minus 1, we can also, sorry, this is s prime of n, right? You can also define f prime f1 of z to be the limit of s prime n of z, okay, by definition. So as I said, we wanted to prove that starting from a power series function which is defined as a power series expansion with the radius of convergence r, this function is complex differentiable. So that we consider the incremental ratio, and this is given z0 such that when first of all, as I said, we can take a equal to 0, and this is without loss of generality, then we can take z and z0 to be smaller than rho which is smaller than capital R, the radius of convergence, okay, so everything, if z and z0 are given with these assumptions, this function here is well defined for z different from z0. For any z, this is defined, this is a number, this is a complex, so this is the difference of two complex numbers, and this is the quotient of two numbers, okay. On the other hand side, we want to prove that this becomes very close, in the topology we introduced, to f1 of z0, and so we can consider the difference, and we have to show that the modules of this difference tends to 0 as z tends to z0. And to do so, we recall that f of z0 was split into two parts, the final part up to the degree n and the infinite part, the series from n plus 1 to infinity. So we rewrite everything like this, and then we divide everything again, so this fraction in this way, then we subtract and add s prime and z0. So here what is left is the following, rn of z minus rn of z0 over z minus, okay. Because of the triangle inequality, the modules of this is the modules of this, but is smaller or equal to the sum of the module i, this plus the modules of this plus the module d. And if we show that these three modules i are smaller than epsilon, epsilon over 3, of course, if you want to be very precise, so you show for z tends to, for z close to z0, then we show that this incremental ratio has a limit as it tends to z0, f1 of z0, okay. So we are left to conclude the following. So this is smaller or equal to, with the notation so far introduced, what I said first is, okay. This is, in real calculus, formally the definition of the derivative, and this can be made smaller than epsilon if z is close to z0. This comes from, this is also infinitesimal as n tends to infinity, since we are considering what that f1 of z0 is the limit as n tends to infinity of s prime n. The difficult part is this, which is an infinite sum, so it is the series. We have to see if this series converges to something which can be considered small. But remember from the definition, rn of z is the sum of nc to the power n, sorry, k, with k greater or equal to n plus y. So it's, as I said last time, it's the t, so the part, the remaining part, not the polynomial, so that if we consider this difference, this is in, evaluated in z0, so if we consider the difference, and this is what we are doing, we obtain the following, so rn of z minus rn of z0 is in fact summation k greater or equal to n plus 1 ak z minus, sorry, z to the power k, correct? And so remember that z to the power k minus z0 to the power k can be factorized in the following way, z minus z0 times z to the power k minus 1 plus z, power k minus 2 z0 plus z naught k minus 2 plus z naught k minus 2, k minus 1, sorry, all right? Therefore, when considering this quotient here, we divide the factor z minus z naught, because z is different from z naught, it's taken different from z naught. And this is summation of ak this, correct? So now I take the modules of this, this is the modules of this, and become smaller or equal to the modules, or to the sum of the module i, right? And so this is summation sum of ak, and then I have here, how many, do you know how many summons are in this parenthesis, k elements, right? So I have that module of z and module of z naught, remember, are both smaller than raw, and raw is smaller than r. So when I consider the modules, modules of z and then the modules of z naught, I can substitute because I'm considering an inequality here, instead of z, modules of z and modules of z naught, modules of, sorry, raw, okay? And notice here that the power of the module i is precisely k minus 1. So it becomes smaller or equal to raw, k minus 1 times k, correct? Okay, okay. And this is precisely the kth element of f1 at raw. And this is convergent, f1 is convergent. So this number here can be made smaller than epsilon, cannot divert, it is actually convergent to something which is zero, because raw becomes very small. So this is more than epsilon. So summing up these considerations, we can say that actually it's true that for any power city expansion, for any power city expansion f of z, and when considering the, what we say, the derived power expansion, we have the following that this is smaller than epsilon as z minus z naught is smaller than delta and n is sufficiently large, right? Good. So first, a first important step in comparison is made. So assume that we consider as a class of function, we are interested in the class of function which are expressed in power series expansions. These functions turn out to be differentiable in this, in this sense of the limit of the incremental ratio of this, and this is finite. Good. Furthermore, we have this. From the derived power series, we can continue the same trick and consider the derived of the derived power expansion. There is no reason to consider this as the last. And applying the same argument, we conclude that actually it is, again, differentiable. And then we apply the same trick and argument to the, the right and so on and so forth. So the only power is z infinity. And there is an expression, if we use this to be the kth derivative, this is the kth derived series. So this, evaluated at the center of the power expansion, at a, this is now to be what? Well, thinking, we have an example. If we evaluate the function at the center, we have only a naught. If we evaluate the derived function in a, in a, we obtain a one. If we evaluate the function derived, derived in a, we obtain two a two. In general, we obtain k factorial ak, right? Okay. So this is k factorial ak, which is also important to know. And in general, what we have, in fact, is the following. That fk, so the kth derived series, derived series, has these expansions, something like factorial ak plus k plus 1 factorial ak plus 1, c minus a plus. This is also important because when we apply this to our consideration, for instance, to the function exponential, we obtain the following, important example. Remember that a of z was the function. One of the example of power expansion of, of complex, say, complex value functions defined in terms of power series expansion with radius covered in infinite. And the coefficients were given as follows, 1 plus c plus d square over 2, factorial plus zm over m factorial. The coefficients were real, but well, it's just by chance for some sense, okay? Now we can say, well, this function is differentiable in the complex sense. And we know that, that is, it's the derived function is the function which has the following expression. 1 plus z plus d square over 2 factorial plus zm over m factorial, okay? So we actually have that function exponential solves this equation, differential equation with initial condition, like in the real case. Well, do we have other examples of function with the real coefficient which can be, can be extended to the complex case? Well, for instance, why not? Let's try with trigonometric phi, see? Cosine sign, we remember that cosine of x was 1 minus x square over 2 plus x4. By the way, x is real, right? Sorry, two factorial, sorry. Yes, the coefficients were real, so again, just by similarity, we can say, well, we extend to the complex case and put this way. This is a definition, right? 1 minus z square over 2 factorial plus z factorial minus blah, blah, blah. And similarly, we find sine of z to be z factorial plus, okay? This is by definition another function. So I invite you to verify the following. First, verify that the radius of convergence of, so that r, radius convergence sinus and cos sinus is infinite. Again, and furthermore, verify that the derived series of cos is sin, sorry, is minus sin and of sin is cos. Like in the real case. So this is an exercise. Now we have a relationship between cos, sinus, sine and exponential in the complex case. Nothing similar is in the real case, okay? Another similarity with the real case, the form would take cos square of z plus sin square of z, and this is equal to 1. What we don't have is the following. You consider e i z plus e minus i z. Remember, the i and minus i is the property that i square. So for all positive powers, odd or even, they are negative positive with i in front or minus 1, right? So I take this and divide by 2 and this is, prove it, cos of z. Prove, simply substitute in the definitions of the two power series, okay? And similarly, 2i e i z minus e i should be this, right? This is, and from the two, I quote here, we conclude the following. We have first that e i z is 2 cos z minus e. We have e i z is 2i z plus e i. Then I sum the two relations and I attended the e i z as cos z is very important to know. In particular, when we restrict z to the reals, you obtain the famous Euler relations. E i t is cos t plus i sin t, okay? Very important. So what is difficult to prove also in the real case? In general, what is difficult to prove when you start dealing with power expansions is the fact that when you multiply, you have a mess. But what is surprisingly true is that, in fact, for some power series expansions, like the exponential, you have good properties. So that the coefficients are mixed, but they are mixed well. This, as a consequence, implies, for instance, that if you take e to the power a plus b, this is equal to e power a times power b, following the traditional power rules, okay? Rules of, you learn in primary school, right? But it's difficult to prove it in general, all right? You have to make some calculation, to make some consideration about the convergence, so on and so forth. So what is not surprising is that in the complex case, you actually have the same property. But in order to prove this, I want to make some more general consideration about derivation in the complex world. You will see that this property has a consequence of the properties of derivative and derivation in the complex setting. So let me, first of all, use this notation, which will be the same for the entire course, which is quite standard in many books. So we have a complex value function of complex variable. So we assume that f is defined in an open set u of c and its values are in c, okay? So we take as variable z, which is, since c is like r2, also written as x plus iy. This is the real part and why the imaginary part of the complex number z, correct? So similarly, the function f has two components, the real component and the imaginary component, right? Therefore, we write f of z to be the sums of two real valued functions of complex variable. For them, typically u and v. So I call here u is a function from real valued function and v as well, real valued function. Sometimes you can also find this, just to use the same symbols for the variable and for the components of the function. Now, after using these notations, we can consider the function from u into c as a function from an open set of r2 into r2 with two components, instead of calling them as you normally do in the real case, f1, f2, we call them u and v, right? Now, let us assume that f has complex differentiable z0. This means that the limit as h tends to 0 of f z0 plus h z0 over h exists and is finite. And normally denoted by f prime of z0. If you want, there's a definition. We say that f is complex differential of z0 if the limit of the incremental ratio exists and this is finite, nothing different from the real case if you want. Pardon me? You? Yeah. Yeah, sure. If you assume that f is defined in u, z0 has to be in u, otherwise there is no meaning. So you start from a function defined in an open set. You take a point inside this open set and you consider the limit. Okay, since h tends to 0, we can assume that h is in a smaller neighborhood of z0 in such a way that this smaller neighborhood is containing u and this is not a restriction because we are assuming that h is so that this is defined, right? Do we have examples of functions which are differentiable in the complex sense? Yes, we do. All functions which are defined as power series. So we have examples. But what does this definition imply? This definition implies many things. First of all, there is no assumption about the choice of h. H can be any number tending to 0. So remember that on the real line, when you assume that x, a real number, tends to a value x0, you can approach x0 from the left or from the right, correct? So in fact, we normally define the left limit and the right limit and we said that if the two-linear limits coincide, then this limit can be different. In the complex plane, z0 is a point in the plane and we have infinitely many directions along which you can converge to this point. So we are not assuming that a privileged direction is assumed, nothing is assumed. So for instance, we can restrict our consideration to an incremental ratio along the reals or along the pure imaginary numbers. So let us take as h tends to 0, h real. In the definition of conflict differential ability, nothing is assumed about h. The only hypothesis you have to take is that h is becoming closer to 0. So in this particular case, we have that considering f of z0 plus h and using the notation real and the imaginary part of z0 to be x0 plus iy0 and z in general x plus iy, we have the following. This is f of x0 plus h plus iy0, but remember that f was also as u, when we consider the two functions u and v, the real and the imaginary part of f, be dependent on z or if you want to look at the real structure behind the complex numbers depending on x and y. So that if I take the incremental ratio with the assumption that h is real, we have the following. We have that this becomes u of x0 plus h y0 plus h y0 minus u of x0 y0 or if we divide it properly, we have u of x0 plus h minus u x0 y0 over h plus i. So the study is reduced to the real case because this is a real value function and these are the two components of the real and imaginary part of the complex variable. Here we have a ratio of real numbers and this is represented in the real calculus by what? The derivative with respect to x at x0 over the function u. So if you want to use this, this is notation. And here I have i times well the same derivative with respect to x of the component, second component and evaluate it if it's not. For the sake of simplicity, since we will have many derivatives with respect to x to y and so on, I prefer to use this as a shortcut in our considerations. I put an index x and with this I mean derivative with respect to x if you don't mind. This is somehow quite common especially in differential geometry in order to avoid all these notations. So the next step is of course to consider what if instead of taking as I said h to be real h a purely imaginary. So in this case I have something different if you think we have the same pattern in some sense but this time the increment is made along y instead of along x. So I take h tilde to be something like this, i times h and h real. In this case the increment ratio starts from is the following which becomes using the notation so far introduced that as follows x0 y0 plus h remember that we have h tilde in the denominator plus i v x0 y0 plus h minus v x0 y0 2 over h tilde and h tilde is i h in other words this becomes okay i and i cancels. So the second part becomes what you expect to have so the function so the increment ratio of the function v with respect to y what about the first term minus i very good minus i because we have well 1 over i is in fact minus i so I have minus i and this is the same h okay here and and I call this appears and it's cancelled by the i here and h in the denominator here gives you this so when I take the limit as h tends to 0 so also h tilde tends to 0 this is the only way to make h tilde tends to 0 right correct. So I have that first terms tends to okay minus i you will expect to y x0 y0 plus v y sorry x0 y0. But the limit exists so the limit as h whatever h you take as soon as it goes to 0 the limit exists and is the same right independently on the choice of the way you are approaching 0 so the two expression have to be the same so we have this important relations and the calculation with h real we obtain it f prime of z0 when h tilde was i h and h real we have obtained what f prime of z0 is v y of z0 minus i u y of z0 since the two the limit is the same these two expression are the same and we obtain this important u x at the z0 is z0 sorry for z0 and u y z0 is minus these two conditions are called Cauchy-Riemann equations we have to so this is the real part and this is the imaginary part so u y is minus v x so we can summarize what we have done so far in the following proposition if f is complex differentiable that's enough then f satisfies Cauchy-Riemann equations that's enough. So the real and imaginary parts the two real valued functions associated to f have partially with respect to x and to y such that these two condition are satisfied okay well important consequences of this first of all here you see we have made a direct calculation and obtained this property from this we also say we go and can also say that basic rules of their derivations in the complex settings are inherited from the real case so that if you sum two differentiable function at z0 then the derivative is the sum of the derivatives right because of the the factor here we are derivating with respect to x one real value function okay then we have also liveness rule chain rule everything works fine this is very important so that we don't have to somehow we don't have we don't have to to to to get bothered with you know the the standard of stuff okay so we can use the same techniques we know from the real case probably most of you already know everything also in the complex case but well this is a next simple explanation now I want to characterize complex differentiable functions we have examples so the class is not empty which is also important to now you know on the empty set there are several theorems very beautiful theorems but normally we have a theory with something inside you're too young to appreciate what I'm saying but there are some cases in which you have very long statements and so the only condition was simply implied that there is nothing to to talk about okay okay so the the next task is the to prove the following to prove that any function complex value functions with real and imaginary part which are differentiable so in the in the real in the real sense which are the C1 C and which satisfy Cauchy Riemann equation they are in fact complex differentiable so that we can characterize complex differentials are those functions which satisfy Cauchy Riemann equations okay so for the time being we have the following power city expansion so the general functions as considered by ancestors our ancestors okay they were proved to be complex differentiable and we also know that they are still power city expansion and we can apply several times that we have important relation but okay this guarantees that we have something concrete to work with now starting from a complex differentiable function from this very simple calculation we have shown that in fact they satisfy some equations so the real and imaginary part satisfy differential equation which are called Cauchy Riemann equation now we start from the opposite side take two complex take one complex value complex valued function and consider the real and imaginary part to be sufficiently differentiable at least see something okay C1 so that they have the derivative and the derivative are continuous and assume that they also satisfy Cauchy Riemann equation can we prove that they are complex differentiable yes we can by using all this stuff we are we are characterized in several ways conflict differentiability okay so let me let me continue by saying the following now assumed f from an open set u of c into c components as I the real and imaginary part to be u and v to be such that u and v are c1 so continuous partial derivatives and satisfy Cauchy Riemann equations now I want to prove that this in fact implies that f well this is done at one point okay we assume that the functions are c1 and enable to c0 okay and the Cauchy Riemann equation are satisfied as c0 we want to prove that function f is completely differentiable at c0 okay now first of all assume assume take this incremental ratio at the point x plus i y so x y okay this is purely real this is a real valued function of two real variables x and y how can you express this difference in terms of linear part and the rest can you do this yeah okay how ux so h so you use the linear part right and so if you want this is different application of the notion of differential of gradient okay of a real valued function so we have that this is in fact ux x y times x by time h sorry plus ui x y sorry k plus call them call the function okay and this function here as a property that this tends to zero as h and k tends to zero good but we can say something more we can say that as h plus i k tends to zero so it is it tends to zero faster than h plus i k so it means okay the the speed is okay similarly vx plus h y plus k minus v x y as vx x y h plus v y x y k plus psi h k with a similar property as h k tends to zero remember this is well a node how do you say small o of h and k right or the modulus of h and k now we have these two relationships for the real and imaginary part of the function f we have to consider the following minus f of c over this is the complex increment we are considering i don't know how can we write it could be well w w has real part h and imaginary part k okay good so this is as before x plus h y plus k plus i vx plus h y plus k minus u x y minus i v x y over h plus i k correct so this minus this is something which can be expressed in terms of a linear part and the rest as we did before this this two u x plus h y plus k minus u of x y is in fact u x h u y x y k plus v h k right and similarly with an i in front we have the difference we evaluated before v x plus h y plus k minus v x y so this is plus i v x x y h plus i v y x y k plus sorry say h k over h plus i k and so far we have only used the regularity of the function not the fact that the functions u and v satisfy consumer equations but now we split the two well we don't consider five sign because they're tending to zero as h and k tend to zero so we're not interested in this we're interested in this part here and we have u x x y h and then I remember that v y u x and v y are the same right right so I can factor out u x and take i k h and i k and then I have also that u y is minus v x correct so u y is minus v x so here I have minus v x plus i v x h plus five sign sign h plus i k but this is not important because this will tend to zero right now minus v x is also plus i squared so I factor out i and I obtained finally that this is u x x y I simplify this and this then I have plus i and I collecting i I have v x h plus i k v x x y h plus i k over and then what is left is here in other words when consider this incremental ratio which is the general setting with the assumption that u and v real and the minor part of the function have satisfied Cauchy-Riemann equation of course they have to be at least with partial derivative otherwise you can even consider to solve the the to be to be satisfying the Cauchy-Riemann equations we obtain here that as h and k tends to zero or as h plus i k tends to zero this tends to zero and what is left what remains here is precisely the derivative of f as in the previous expression so the derivative of the real part with respect to x plus i the derivative of the imaginary part with respect to x so we I consider concluded that this proof and then any complex value function with sufficient to regular real imaginary part satisfying Cauchy-Riemann equation is in fact complex differential and this is one important consequence of the definition another important consequence is the following oh yeah assume that the function f as real imaginary part turn out to be of class c2 u okay so f is if i sorry u i use capital u for the open set and normal letter u to be okay i'm sorry but well this means a set and this means a function in any case u and v both of class c2 so they have also second partial derivatives with respect to x and with respect to i there were two y now consider this a second derivative of u with respect to x is also this according to my notation okay in short and also this for instance okay will be the notation i use so far correct x and y x and y okay and of course all the others y x and y y all right what is this let me see if you are familiar with this this is the laplacian of what sorry ah sorry yes true the laplacian of you sure sure sorry the laplacian of you is normally denoted by okay this symbol this is the laplacian this second order differential operator is called laplacian this is just a name how are called the solution of a laplacian equal to zero harmonic functions very good so this is completely real stuff right okay what is one of the important consequences of the fact that u and and being the real imaginary part of a complex value factor they satisfy Cauchy-Riemann equation those are consequence we obtain that these two functions are in fact harmonic functions how can we see this but remember that we have this relation u of x is v of y okay all right the Cauchy-Riemann equation once again and u of y is minus v of x if i'm not mistaken right therefore if i take u x x this is as v would be y x okay and if i take u y y this is like minus v x y if i sum up the two on the left hand side i'll paint this u x x y y on the right hand side i obtain v y x minus v x y however if you are sufficient to regular that is to say c2 is enough they have to be with partial derivatives which are continuous that's not then schwarz schwarz schwarz-lehm the theorem implies that the order of derivation is not important for the problem okay so the partial derivatives do not depend on the choice of the order of x and y so this is zero this is the same how do you call the theorem i call it schwarz theorem it should be short or something oh there are several schwarz in mathematical history there will be also schwarz-lehm we'll see in in complex and otherwise it's not it's not i don't think it's the same schwarz by the way the same man i mean in in germany it means black so it's a very common family name anyhow we conclude this for you and similarly i invite you to prove that this is also true this implies okay this is Laplacian Laplacian equal to zero is a consequence of fishery manipulation so in particular we can say that f complex differentiable implies its components harmonic functions as a matter of terminology complex differentiable is quite a modern way to say this to indicate this in fact for several reasons for historical reasons they are more often called holomorphic function okay so holomorph a holomorphic function means composite differential so that's the the standard terminology used in complex analysis harmonic function harmonic and harmonic functions are instead say a real calculus real analysis terminology okay harmonic and the also the the origin of this name this specific name is has some reasons okay related to property of a true use the the harmonic analysis you can you can find some well something related to wave equations and so okay question so far so you might ask why why all this long very long privilege to to very direct proof of something related to exponential and complex cosine and sine well because we also hope to use the tools we have so far to do so and in analysis without derivation without this specific tools you can do very little in fact this tells us a lot of things you are the entire theory related to the derivation all right well for instance so important consequence in real analysis is that from the study of this of the study of the in the increasing or decreasing of the function depends on the positivity or negativity of the derivative right because it is a real number it can be positive negative or zero unfortunately we cannot hope to have anything like this in the complex case because well complex numbers are not positive negative or but they can be zero can be zero so what if the function is such that it's it it is complex differentiable so it's harmonic and it has its derivative equal to zero what's your opinion about this yeah you know in an open it's constant like in the real case can you can we prove it directly without making many calculations no it's connected not convex yeah sure sure okay normally and this is quite natural to think of the sets as open and connected normally functions which are holomorphic are supposed to be defined in a domain domain means open and connected subset of c why is it quite natural well so far we have very few example but if you think that power series functions define functions which are holomorphic the radius of convergence can be big large or very small but it doesn't matter the the set of definition is a circle is a disk therefore it is an open and connected set yes so since we are using this property to be assumed for for the domain opening open set is natural because we have to to deal with the derivative so we have to have a neighborhood of a point right why connecting this is important as well well because probably the argument relies upon something which is topological okay so let us see it in detail if you don't mind or do you have do you have another another way to think about it any opinion well we should also try to find a shortcut remember the derivative the compass derivative is expressed in terms of the u x plus i v x right so if it is identically zero there is no increment of the real part the imaginary part of the function along the real axis but it is also minus u y sorry sorry minus v y so v y minus i u y so it is also constants along the imaginary axis and any two points can be joined by a polygon by a path with vertical or horizontal lines as segments this is also proof you are right but this is a step too too long for the moment yes all right yes but you have to prove the the cushy integral formula first right yes but in fact in in in a few hours say we have this special to the cushy integral formula because as a consequence of the cushy integral which is very peculiar in complex analysis will resemble something which is say the opposite way what we have done we will consider power set expansion and prove to be the complex differential or homomorphic now we start from the output and equivalently to have to have a really imaginary part satisfying cushy Riemann equation now we start from function which is complex differential or satisfied cushy Riemann equation is it true that it is complex analytic so it has a power expansion and it's true in fact but it spends some cushy integral formula so we introduce this important tool so I don't know maybe tomorrow or the day after tomorrow so but if you if you you Sure. OK. Thank you. OK. Your colleague is trying to use another argument. That is the very detailed argument to prove that the function is counter-differentiable and its derivative equal to 0 is in fact constant. He said, why don't we use the following argument? We consider the level set. We take a point inside this open and connected set. Consider the value of the function. So if I'm not mistaken, please tell me. So we consider u to be open and connected. Take a in u, f, of course, holomorphic, and assume that f prime, so the derivative is 0 for any z in u. So your colleague is saying, why don't we consider this value, f of a, a is a point in u, f of a is well-defined, and consider a counter-image of a, inverse image, a to be the set of z in u such that f of z is equal to f of a, correct? So I think that, well, a is, of course, not empty because there is at least one point. And since it is a level set, it's the inverse image of a function which is, I didn't say it, but of course it is continuous. We have partial derivatives as a component, so that it is continuous. Continuity is a consequence of differentiability also in the complex case. Because, well, I didn't say, I didn't write it explicitly. If you consider the incremental ratio and you know that the limit existence is finite, then, of course, f of z tends to f of z naught, as z tends to z naught, right? It cannot be anything else. Otherwise, the limit is not finite. Anyhow, since it is continuous, this is also, in short, f minus 1 of a. So it is the inverse image of a point, and it is a closed set. So a is closed. What is left to prove? If we prove that a is also open, we have a non-empty opening closed set, so it is the entire u, which means that for any z in u, f of z is equal to f a. So it discounts, and we're finished. So to prove this, we have, well, to prove that a is open, thank you for your observation, we do the following. So we have u, we have a, and we take an open ball of radius r centered at a, and with a small radius in such a way that it is contained in u, OK? I want to take z in b a of r and prove that z is actually in a. If this is done, I'm finished, because I've proved that any point in a has a neighborhood in a. And so a is also open. So how we conclude? So this is the last part of the argument. OK, I take z, and I take a. And I take the line, the segment, connect z to a. So remember that z and a can be considered as vectors. So I can use, well, standard stuff like this. Well, what is it? T z plus 1 minus t a, right? And t real, actually t real in between 0 and 1. So when t is 0, this number here is a. When t is 1, this number here is z. So this is the segment connecting a to z, right? Then I consider f of t z plus 1 minus t a. And call it g of t. So a and z are given. I take z in the ball. a is the point we'll consider from the beginning. So what varies here is the parameter t, which is real parameter. So g is a complex value function defined on 0, 1. Good. Now what can I say about, well, one point I forgot to say. Why can I take t z plus 1 minus t a to be a good variable for f? Because I'm considering z to be in the ball centered at a, right? It is complex, correct. So the entire segment is in the ball. And so f is defined for any t. So g is well defined. Now I can see that g of t minus g of s over t minus s. So this is incremental ratio. Is it for g? So t z plus 1 minus t a minus f of s z plus 1 minus s a over t minus z, t minus s. Then I notice that if I write t z plus 1 minus t a, which is the first variable for f in here, then I subtract what is in here minus s z plus 1 minus s a. I obtain, notice that I have a and a and a minus in front. So n, a and a cancels where I have t z minus s z. And then I have minus t a plus s a. So I have t minus s z minus a. Are you with me? Just doesn't matter. It's very simple calculation. So that I can rewrite with this observation, rewrite this as f of t z plus 1 minus t a minus f s z plus 1 minus s a over, and I write t s z minus a, which is exactly the difference of what we have here and here. And then I multiply for the same amount, and I do I have a t minus s. Right? So this cancels this, and this is t z plus 1 minus t a minus s c plus 1 minus s a. And this is an incremental ratio for f. Now when t tends to z, this tends to 0. And this tends to what? To f prime at s z plus 1 minus s a, which is 0. Correct? Good. So in the last few minutes, I conclude by saying that as t tends to s, we have the g t minus g of s t minus s tends to f prime of s z plus 1 minus s a times z minus a. But this is 0. Because this point is inside. It is in u in particular, but it is 0. Which means that this function here is constant. And g of 0 is g of 1. So g is constant. And g of 0 is what? f of a, and g of 1 is f of z. That is what I wanted to do. And this concludes very good. Because we are assuming that the set is open and connected. So we found an open and closed set inside a connected set and no name. So it is entire. Now how can we apply this to have the relationship we were looking for? So consider this function as a function of z. z is a complex number. And e, the power z, is the complex exponential. So this function turns out to be a complex differential. And this turns out to be a complex differential as well. Correct? Because, well, you just shift the center of the power expansion of u of z. So we have all the hypotheses. So what about the derivative? Is it 0? Are you all? It's e to the power c. Correct? But not because we are using the property. Because we made a calculation. I'm not cheating you. I'm saying, well, if you want to be very formal, use what we know, know what we want to know. You're right. But it's just a test just to verify that we can do this. Now we apply Leibniz's rule. Because we have a multiplication of complex differential function. And we apply Leibniz's rule. So it means that we have to differentiate first this times this function. And then this function times the derivative of the other. But the derivative of e of z, for definition, is e of z. So it's e of z times c, c minus z, plus, then I have e of z, the function, times the derivative of this. Remember that, as I said, we can use standard rules. So we have a composition of functions. Here we have minus 1 in front. Minus 1 is a coefficient. And then e c minus z. So it is 0. So the function is, in fact, constant. And in particular, as I said, 0 is e c minus 0. And from 0, it is the value that 0 is c of the power c. So we have this relationship. And this concludes today's lesson. Then we have that e z times e c minus z is e c. I call z equal to a, c minus z equal to b. So c is a plus b, correct? So we have a e a times c d, right? So in particular, just if z is x plus i y, e z is e x plus i y, and it is x times e i y. But e i y is, this y is, remember real number, as x, times cos y plus i y. And once again, we obtained something interesting. We obtained that the modules of v of z is e x to the power x, which is a non-negative real number. And y represents the argument. Pardon me? Pardon me? e to the power x? Sorry, it's strictly positive, right? Correct, correct. So e of z, correct. Sorry, sorry. Right, sure. That's what I want to say. e of x is strictly positive. And therefore, we can say that e of z is never 0, never vanishes. And as you will see, this is somehow the extremal properties. The only function, well, we'll see what happens in the future. But it's the only value which is missed by the function e to the power z, complex play, right? So I think we can stop here and see you tomorrow morning. I prepare the copies for you tomorrow, if you don't mind.