 I think we can start. So we continue considering the singularities of holomorphic functions. And in particular, we'll study in details a little bit more the case of poles. Last time, we classified the singularities. There are only three possibly. One is the not interesting case. If you want, it is the case of removable singularities. Then there were poles and essential singularities. We characterized them. Let me just remind you that last time we considered half holomorphic, say, and this, which from now on will be the punctured disk centered at A. I repeat just for the sake of completeness that I mean this. So we are dealing with a domain, which is disk, and we remove the center, and the center is A. So locally, all singularities are like this, right? And assume that A is a pole for F. So a pole for F means, well, we have given characterization of a pole. But essentially, FOZ has a principal part. Remember, we describe the principal part, and then it is summed to a holomorphic function. So here's something like C minus M over Z minus A to the power M plus C minus M plus 1 over Z minus M. Sorry, C minus A over M minus 1, then plus C minus 1, C minus A plus something which is not surprising what you expect to have. That is to say, this part here is, in fact, the standard power expansion at A. So if all these CAIs are, in fact, zero, the function is holomorphic, or it has, as we said, a removable singularity more correct. So there is a removable singularity if, in fact, can be extended holomorphically at A. But if 1, at least, is not zero, these CAIs are necessary. There's a singularity, and the function is not defined at A. But it has a pole because the number of CAIs is finite. Otherwise, we would have such a singularity. This is the characterization. So let me just consider the case that A is a pole for F. What if we consider the integral of F of Z over the close curve? And of course, I always omit that it is sufficient regular to make this calculation blah, blah, blah. Close curve in D dot A R. Remember, this was called the principal part of the thank you F. And this was, well, the not surprising parts, in some sense. It was the standard holomorphic power expansion. So notice that, as we already done several times, we can split the integrals we are considering here in this way. So we have the integral of a gamma of F Z D Z as the integral of a gamma of the principal part of F D Z. If you allow me to use this notation, plus the integral of a gamma of summation pk C minus A to the power k is Z. And this gives you 0. Why? Because this is the integral over a closed curve. This is holomorphic and the disk. So we are exactly in the same situation as we considered before taking into consideration singularity. So this is 0. But here we have what? And finite number of summands. So in particular, so let me write this explicitly like this. C minus m Z minus A to the power m denominator plus. I can split because the integral is linear minus m plus 1 over C minus m A m minus 1 plus D Z plus. So this is 0 because it's a constant, so it's holomorphic. This all, but not the first one. So the one with the index 1. And in fact, also giving zero contribution to the integral. Because as we already noticed, when considering the first examples of the complex integration, then we observe that when you have this, we calculate it on a closed curve. Each of this, when k is different from minus 1, each of these functions has a primitive, remember? So also this function is primitive and therefore the integral is 0. So the only contribution is this, is given by this factor. So that we can say the important thing I'm summarizing here. That if f is holomorphic dot a r and a is a pole for f, then for any closed curve and for any closed curve gamma in d a r, we have the integral, you can't see anything. I'm sorry, maybe this is, it's much better, right? So you have to wait a little bit because I think that my hand, the color of my hand changes the focus and probably also the diaphragm. Okay, so be patient, please. Well, this is not zero, otherwise the function f will be holomorphic, of course. But then we can be more precise and say, well, from this calculation, we can say this is this integral, all the others have in fact zero contribution to the integral. And so this is C minus 1 of what? 2 pi i, the index, sorry, n gamma a, right? So this coefficient is very important and is called, for some historical reason, residue of the pole a. This is the residue and this number is in general, it is not zero. On the other hand, the above given description of a pole, local description of a pole, we can also say that if f has a pole at a, then we can write fz to be z minus a to the power minus m, okay? Times g of z, where g of z is holomorphic. Remember, we have a principal part, then holomorphic part, then we can rescale the coefficient in such a way that we have this factor, okay? Which is the leading somehow term, okay? This is, largest m is like this, okay? Remember that m is known as the order of the pole. I think that I already told you last time this, okay? So m is, of course, m. So it's not one of the m's, it's the largest, okay? Then we can factor out m and then consider the other coefficients and factors. But then, of course, this g of z which comes out from this factorization is holomorphic. It has power expansion, it's complexity, okay? So keeping in mind this, and remembering what we have done for the zeros, what if we calculate this ratio, okay? So we have to differentiate the f of z which is, in this expression, very easy to calculate. So m minus m, z minus m, minus m, then what? Minus one, times g of z plus z minus a to the power minus m, g prime of z. And then I have z minus a to the power minus m, g of z. Which means what we have here is minus m, correct? Okay, plus g prime of z over g of z. Which is similar, except for this minus in front. To the case of the zero, remember that when a was a zero, we factor out z minus a to the power m. m was the multiplicity of the zero. And we show that in this ratio, the part which was not so trivial for the calculation of the integral was like this, with a plus from the front. So if now I take the integral over gamma of f prime over f, d z, and gamma is a closed curve, as usual I'm assuming it's sufficient to regular to make this calculation. And reasonable, it means that it is defined where the functions are defined. Then I have minus m, integral of one over z minus a, d z, plus the integral of a gamma, g prime of z over g of z. What we know is that g is holomorphic and g of a is not zero. We also have this. So this integral here on the right side is the last term, and the final expression is zero. Because it counts no zero, okay? So this is zero. And here we have a minus in m times what? The index of a with respect to gamma. In other words, we have this fundamental fact. So if f is holomorphic in d of c, okay? Assume z1, zk's are zeros of d, and w1, wk are the poles of f, sorry, of f in d of f in d. So I consider just a finite number of zeros because if there were infinitely many, there can be infinitely many. In any case, then I take the integral of a closed curve so that I will, as you remember, consider just the subsets of zeros and of poles which are contained in the bounded complement of the curve. Remember? So I take curve gamma, homologous to zero, closed curve, and I don't care about poles and zero which are outside. So outside means in the unbounded component of the complement of gamma. Why? I'm not considering them because I know that the index with respect to gamma of poles of zero or any point is zero. So that what we are interested in is what is inside, okay? So then I can restrict to the case of a finite number of zero and a finite number of poles. Otherwise, those which are, because I have to say, if I consider the closed curve and the subset which is as a contour, this closed subset, this closed curve, if there were infinitely many zeros inside, these zeros would converge to a point either inside or to the boundary. But we are assuming that the closed curve is not passing any zero or pole, okay? Which is not passing through zeros and poles, okay? So in the previous, in our previous consideration, we considered curve gamma in the dot, which means they cannot pass through point A. So they send the pole, right? So what I have is that when I consider the integral over gamma of f prime of z, f of z, and then I multiply 1 over 2 pi i in this integral. This gives you what? This gives us what? We already know if the function were holomorphic, this gives us the multiplicity of each zero times the index. In this case, we have also poles at it. So it's more generic situation. And we have seen locally for one pole, but then we repeat it to each pole and repeat it k times. So this is the difference of what? Of, let's say, write summation of mjm gamma zj. This is multiplicity of zero zj. And I know that this sum is finite sum because of what I said before. Minus summation, and then I put here maybe m tilde l and gamma wl, where this m tilde is the order of the pole l. Sorry, wl, wl. So this number is the difference of the indexes of zeros times multiplicities minus the difference and poles times order, sum of poles and all. If you prefer to avoid this notation because this can be somehow hard to remember, then you sum the index so many times as the number of the inverse image of zeros or as the multiplicity. And you sum the index so that you can in some sense consider a multiple zero as a zero counted several times as a simple zero. So in general, this result is known as the argument principle. And then we'll see some application of this argument principle. If this integral is zero, it doesn't mean that the function has no zero and no poles. It can be as many zero as many poles with multiplicities, for instance. And that's what we'll take the application we'll see in detail. So now just to give you some extra terminology, a function which is holomorphic and has singularities. Only poles will be called a meromorphic. These two words, holomorphic, these two adjectives, holomorphic and meromorphic are Greek words. The meromorphic means function with singularities which are either removable. If they are all removable functions simply holomorphic and the singularities can be only poles, not essential singularities. So a function which is meromorphic can be extended to the singularities by putting that, well, I assume that the pole, if f is meromorphic and w is a pole for f, then we can extend f to a function f tilde, which agrees elsewhere. But at w, assume the value infinite. If you want to extend it to three months, you just have a more general notation for meromorphic functions. So let me show you one of the first applications of the consideration given for poles and for meromorphic functions in general. Of course, any holomorphic function can be considered as a special case of meromorphic function. So no poles, if you wish. So the great difference between poles and essential singularities. So let me see if I can state this, which is known in some book as Ruchet's theorem. Assume that f and g are meromorphic functions in d, right? Let gamma be a closed curve d, not passing through zeros and poles of f and g. So I have this situation. I have domain. I have some poles of one function, some zeros. And then I take a curve gamma. Since these are poles and zeros are isolated, I can always find enough room to move this curve. Then I assume that this inequality holds along gamma for any t in i. So gamma is a function like this. i is the interval where I consider it's an interval in r. We have that f of gamma of t minus g of gamma of t. So the values taken over the functions f and g, meromorphic along the curve, this is a finite number because these numbers are all complex numbers, right? This is strictly less than f of gamma t, right? So this is an assumption. This is meaningful because the curve gamma is not passing through any poles. So then, this is just the assumptions of then. The number zeros minus the number of poles of, sorry, of indexes of zero, the number of indexes of poles counted with multiplicity respectively order of the two functions f and g are the same. But I have to be more precise of the zeros. And pole can't be multiplicity and the bounded region c minus gamma. I can also meet this. But of course, I have to repeat then that what is outside the curve gamma is not counted for our considerations because the index is zero. So probably the notation I gave you so far is not very convenient. And let me just use the standard notation. When I have a meromorphic function, the number of zeros is normally denoted by z of f. Zeroes with multiplicity. Sometimes I use this same symbol for the set. But well, this should be somehow the cardinality of the set. Of each zero, time is the multiplicity minus. And then normally you write this way. p of f, to remember that p is the pole. So this sums up this difference. So this is more precisely summation of mj and gamma zj. And this number here is summation of m tilde j and gamma wj with the previous notations. So in short, if I adopt this notation, the theorem stays the following. Take the curve, consider this, which of course depends on gamma. I admit gamma, but I have to remember this is depending on gamma. This difference of zeros minus poles counted with multiplicity for the curve f, for the function f is the same for g. This is Rouscher theorem statement. Now, probably it's longer to describe the theorem than to prove it. First of all, let me remark that this inequality we have along the curve gamma, strict inequality, implies this is for any t in i, that f has no zeros along gamma. Because otherwise, it would be 0 greater than something which is greater or equal to 0. So that I can divide. And I take this, therefore, this is true. What do we have here? Well, we have the ratio of two meromorphic functions. Is it meromorphic? So we know that the ratio of two homomorphic functions, think about the exercises I gave you. When it is defined, it is homomorphic. But in general, it is not defined in the same domain. So what are the problems? The problems are where the denominator vanishes. So in this case, there are the zeros of f to be taken into consideration. And in fact, this adds some poles to this ratio. Because we have the poles over G and then other poles. But what we have is a finite number of poles in any case, and not extra singularities. So this function here is meromorphic. What we also have from this inequality is something interesting. When we restrict to t in i in the interval, then we see the, well, let's call this function if you want h. h of gamma of t is the image of the curve gamma with respect to this meromorphic function. What do we know about this meromorphic function h? Well, we know that when t varies in i, gamma of t is not passing through any of the zeros of G of f and the pole, so it is a curve. This is a curve called sigma of t. So we don't know very much about this curve, but the distance from 1 is smaller than 1. This is the only fact. So 1 is here, and 1 is the distance from the origin of the point 1 on the real axis, point 1. So if I take a circle centered at a, well, what I can say is that the image of this curve gamma with respect to h, as a distance from 1, smaller than 1, so it is inside here. This is sigma. Maybe it's not like this, but it is far away from 0. Enough far, or far away. Enough far because of this inequality. Are you with me? Good. Now, with this geometrical description of the set of the curve of sigma, we can also say that we're in the complement of sigma, or say along, more correctly, in the complement, sorry. In a small neighborhood of the set sigma i, you can define the logarithm because we are far away from 0 and we are in a ball, correct? Remember that the logarithm can be defined, for instance, if you want the stunt one. So you remove a slit like this from the plane. We are here with something which is inside here, but not passing through 0 and not reaching this half plane. So that we can define the logarithm. So logarithm, the complex logarithm, is well defined in a neighborhood of sigma i. Sigma i is this, right? It's this curve. So I can see the small neighborhood and define the logarithm here without any problem. So one branch of the logarithm, the principal branch, is defined. The problem of defining something which depends on the choice of the previous considerations for the square root of z is when you go to consider, when you are considering, say, a domain which contains 0 and which goes around 0. So this is not our case. So logarithm is well defined. Good. Now I consider then as h of z is g of z over f of z and then I restrict it to the curve gamma, right? So I take h prime of z, which is g prime of z times f of z minus f prime of z g of z over f square of z. In other words, it is g prime of z over f of z minus f prime of z over, sorry, times g of z over f square of z. So that I have that when I consider h prime of z over h of z, this is, so I have g prime of z over f of z minus f prime of z over f square of z times g of z. And I have to divide everything times g of z over f of z. So this is, first part reduces to g prime of z over g of z because I cancel f of z. I'm considering everything in general when it is meaningful. Then I restrict to the curve gamma when z of t is gamma of t, right? z is gamma of t. And I can simplify everything because gamma is not passing through zeros and poles, right? And then here I simplify f and I simplify g. So I have f prime of z g of z. You see this? This is a simple calculation. Sorry, f of z, right? Sure. f cancel 1 f from here, g and g cancel. So when I consider therefore this number here, over the curve gamma, which is meaningful as I said, right? This is the integral over gamma 1 2 pi i over g prime of z g of z minus d z minus 1 over 2 pi i integral over gamma of f prime of z over f of z d z. And here, here and also here, you can recognize what we have in the argument principle theorem stuff, right? So this would be the number of zeros of g referred to gamma minus the numbers of poles of g, right? And here, the numbers of zeros minus the numbers of poles, correct? On the right hand side, I can calculate. So I have the difference, right? What do we have on the left hand side? You might say, well, the number of zero of h minus the number of poles, but I have something more. I can say that, well, this function here is that, what I mean, it has a primitive. It has a primitive because it is that, well, I can write this as logarithm of h of z, the derivative of the lower y. Because as I said, I can define in the neighborhood the complex logarithm. So this guarantees since this integrand function has a primitive, the integral is 0 on the left hand side. So this is 0 because of this. And therefore, we have the equality. So number of zeros of f minus number of poles of f is the same as number of zero of g and the number of poles. Some application. Well, one basic application is not so trivial. It is another proof of the fundamental theorem of algebra. So let us, 11, right? 12. So application. Another proof of the fundamental theorem of algebra. What is the trick point when applying this theorem? Well, we have to find the correct inequality we have to use, right? Because no, just because you take f and g or maromorphic, all of them are, it's not difficult to find pairs of all of them. But with the inequality needed. So start from a polynomial P of z. And of course, we assume that the polynomial is not constant, so of degree at least 1. And it's not restrictive to assume that the polynomial is monic because otherwise we divide it. So the first, the leading coefficient is 1. This means monic in case. So assume that the degree of P is greater or equal to 1. And that the an is 1. But it's not without losing any generality in our considerations. So the polynomial is like this. a m minus 1 z n minus 1 plus a 1 z plus a naught. This is our polynomial. The polynomial is holomorphic function. Very good. But it applies, the same consideration also applies for the case of holomorphic function because as I said, holomorphic function represent a subset of maromorphic functions. So in this case, we don't only have 0s, no poles. Call this P n minus 1 o z. That is to say P of z is z n plus P n minus 1 o z. n minus 1 reminds you that this polynomial here has degree. Good. Now take this P z minus z n. That is to say P n minus 1. And consider its modules. So this is smaller than z n modules. At least for z is far away from the origin. Well, how you can see this? Well, you see, you have that, sorry. This has degree n minus 1 P n. So if you divide by z n P n minus 1 z over z n, as z tends to infinity, takes the modules distance to 0. Because the degree of the denominator is larger than the degree of the numerator. So for sure, this can be made smaller than something. And so the numerator is smaller than the denominator. So in a sufficiently large, what is this, by the way? For z greater than r, what is this set? It is the complement of a disk, right? Of a closed disk, to be more precise. Is it an open set in C? It is. If I want to use the point of view of the Riemann sphere, can I consider this to be a neighborhood of infinity? Yes, it is. If you are not, say, somehow familiar with this vision, say, well, what is this? Well, this is neighborhood of infinity. Transform everything from outside to inside, using 1 over z. So the stereographic projection, and we have a disk, open disk at the origin. So the point of infinity is mapping to origin. And then we have neighborhood. Anyway, if we take, as a curve, circle of radius r, and r is sufficiently large, then we have this inequality. This inequality holds for any z. So in particular, p of gamma t minus gamma t to the power n, smaller than gamma t to the power n, if gamma t is, say, r e i t, and r is greater than capital R, right? And of course, t varies n to pi. So this is our i in the previous notation, so the interval. But then, count the 0's of this polynomial. How many 0's does that? Does 1, 0 count in multiplicity? n, right? So there are n 0's. Or if I say, I use the previous notation, that the z f, when f is z to the power n, is n. And p, the number of faults, is 0, all right? So apply Ruchet theorem to this pair of holomorphics, in particular, meromorphic function, pz and zn. This is our, say, g, and this is our f. For f, we know everything. So we calculate explicitly the 0's. There is 1, 0 with multiplicity n. About p, we don't know anything in principle. But we can conclude, since this inequality holds, that there are n 0's also for p. No pulse, p is a polynomial inside one of this large disk. So all the 0's are in the plane. So in fact, p o z is n 0's, q e d, right? All right. So another, maybe, trivial application is the following. Assume that we have function holomorphic. And assume n d. And d is bounded domain of c. Then I take a and d, and I take g of z to be f of z, f of a, minus f of z. It's the same, right? Assume that we have this, that f of a has a modules, which is greater than f of z, for n is z. Remember that this leads to the fact that the f is constant because of the maximum model of theorem, right? Remember this. The holomorphic function, any holomorphic function, cannot have a point inside the domain of definition, such that its modules is strictly greater than the others, right? So the maximum of the modules of the holomorphic function can be taken only on the boundary, all right? So then I take this, g of z, minus f of a, is the modules of f of z, right? Which is smaller than f of a for our assumption, right? So what I conclude is that, well, assume that f is holomorphic, g is also holomorphic. So there are no poles in this case. So the number of zeros of f of a is the same as the number of zero g of z. But there is only one zero for g of z. It's a, right? And here there is none. Contradiction, unless f is constant. So if we assume this, we have a contradiction from here. Let me also point out that last time we used the, which is important to see the same proof with different approaches. So last time we saw the proof of the maximum model of film as a simple consequence of the open mapping theorem, which is a very deep result, which characterizes the topologically holomorphic functions. But we can have the same result by applying the Cauchy integral formula, which is also a very important result, and which is good for holomorphic function to be used several times, as we have done several times. So assume that we have exactly, as in the previous case, a point in the domain d, but it's suffice it to consider a small disk, because we have local integral Cauchy formula that this is guaranteed. So for instance, in dar, ethylomorphic, and this is the assumption, a is a point where the modules of f is greater than f of c, say, for any z in dar. This is always the case, because, well, this should be called a local maximum for the function modules of f, right? Since a is inside a domain of definition of f, and f is holomorphic at a, then we always have the opportunity to take a small neighborhood of a in the side of the domain of definition. Now, in particular, restrict our consideration to a curve gamma. So the curve gamma is gamma of t is like this a plus epsilon eit, say 2 pi it, just to be on the safe side. And epsilon is strictly smaller than r. So I take a circle inside dar, and I can define f of a, this time won't forget this, I hope, the integral over gamma of, remember, f of c, c minus a, dc, right? 1 over 2 pi i, I don't have to put the index 1 here, because, well, it's 1, right? Because the index with respect to a, with respect to gamma, the point a is 1, right? Then this is just, as I said, this is nothing new for you, but just a way to see the same proof, to say the proof of the same result using another tool, which can be interesting. So this is an equality, and this holds locally. So in particular, if I have this assumption, generally we have it also locally, so we can apply it. Then, taking the modules of f of a, this is, well, the modules of the integral on the right hand side, and this is smaller or equal of 1 over 2 pi, 2 pi, sorry. Now this is, sorry, this is the same as, and then I replace gamma of t into this. So this is gamma of t, so that I have f of a plus epsilon e i, 2 pi i, 2 pi i t over, the integral is between 0 and 1, and then I have epsilon e 2 pi i t, because a and minus a cancel, and then I have here, instead of dixie, I have what? Epsilon 2 pi i e 2 pi i t dt, right, is dixie, one restricted to gamma, epsilon 2 pi i, right? So I have epsilon 2 pi i, sorry, e 2 pi i dt. So epsilon and epsilon, yeah, cancel, right? Am I right? What is missing? Hope nothing, is it correct? So to conclude, I have f of a as integral over 1, 2 pi i, sorry, 0, 1, then I have f a plus epsilon 2 pi i t, and then I have e 2 pi i, of course I forgot t, that was probably what was, yes, I know, so this cancel this, and this cancel this, right, sorry, but the 2 pi i t. So this cancel, and then I have this, and then I apply the inequality, so this is already gone. Inequality integral, the modulus of the integral is smaller or equal to the integral of the modulus, and this will lead to a contradiction, because this number here, this is a real number, because of this inequality is certainly smaller than f of a, right? So I have that, this is f of a, it's an equality. This is strictly smaller than this, sorry, this is the same as smaller than this, contradiction. This is an interesting application of Rouscher theorem, and well, finally, let me just catch the next topic. So as I was saying, everything in complex analysis somehow important locally, we define a function to be allomorphic in a neighborhood of a point, because we need to define something like a complex differential, so some disks are required to be the natural domain of definition, in a set of point, so it's not point, it's local, but when we have a singular, we cannot use points. So neighborhood of points, we have to use neighborhood, puncture this, or neighborhood of singularity is punctured. But unfortunately, the singularity has a point as the same problem, so it's better to enlarge what happens in the neighborhood of a singularity. We've seen that if you take a neighborhood of a singularity and the image is dense and the complex plane, then necessarily it is an essential singularity. So what is natural to consider is not only this, but we prefer to take a small neighborhood of the singularity and then enlarge if you want the point to a disk and consider the annulus. So now the next task is to try to define power expansion in an annulus, which would be the natural extension of what we have found for holomorphic to the case of meromorphic functions or functions which have a singularity at A, more in general, right? So either A is a point where the holomorphic function is defined, or if it is not defined and then A is a singularity for the function, it is natural to consider neighborhood of this singularity and consider what is the puncture disk without this neighborhood, that is to say the annulus. So consider the annulus, central at A and the radios R1 and R2 is the set of point Z in C such that R1 is smaller than Z minus A as smaller than R2. So all the Z's such that this inequality holds R in A and vice versa. So this defines annulus. Now we consider two circles in A, R1, R2 centered. Therefore gamma 1 of t can be parametrized in the following way, A plus R2 EIT, and t varies in 0 to pi. And I assume that R1 is smaller than R2. So we have two circles inside the annulus. This is R1 and R2. Since we are in the annulus, we have this in a quote. R1 is smaller than R2. And capital R1 is smaller than R1. And capital R2 is greater than R2 in between. Because of this choice of the parametrization, these two curves are oriented like this, moving from 0 to 2 pi. And the index of A with respect to gamma 1 is 1, with respect to gamma 2 is 1 over 2 pi i in the index. I always forget 1 over 2 pi i. But what I meant was the following. Whereas all the other points outside the annulus, but not in this bounded component, of course, has more index 0 for both curves. So I also noticed that geometrically, I can deform gamma 1 into gamma 2 and vice versa. They are homotopic equivalents. So remark gamma 1 is homotopic to gamma 2 in A R1 R2. So that I notice also that if I take gamma 2 minus gamma 1, and when I consider in gamma 2 minus gamma 1, I mean the same gamma 2 and then gamma 1 in the inverse sense of orientation of gamma. So if you want to put minus t in front, as a parameter not in front, minus t. This is also true. This depends on this. In fact, I have this and then this. But then this set this, sorry, this is formal notation. It means one curve considered with the positive orientation, one with the negative. But this is not a closed curve. This is not a closed curve. So I add and subtract. Well, look, this is somehow formal. But the segment, one of the segment connecting, segment of one curve connecting gamma 2, 2, gamma 1 in this way. So I enlarge the previous picture a little bit. And I try to show you what we are doing. As you can see, since everything is a homotopic equivalent, it's not important. They are circles. So you are correct. But you can say that two functions are homotopic in general. Nothing to say. This is not the path. These two functions are homotopic if you can. Well, there is no starting and ending point. But now, I'm also answering you. And I choose one point and one other point on the other curve, on the other circle, to be honest. And I take this to be lambda. So I put lambda, say this is gamma 2, and this is now gamma 1. So what I'm doing here is I move from here. Start from this point. Go around, then I'll say maybe this lambda is this. Then I go this way, then I move this way, and then I come back. So now I choose a point. Thank you for your question. But in general, you can define homotopy for continuous functions in topological spaces without starting and ending point. If you have a path, you say, homotopies of path, then you normally require that the starting point and the ending point are preserved in general. So what I'm saying is that then finally, this new curve gamma here is a closed curve in the annulus. And because of this position, and it is homologous to 0. Homologous to 0 means precisely what I've written here. And the sense that if you consider any point, say for instance, a, take the index of a, and then any other point inside of this disk. I have an index 0, so because take the integral, this and this. So over this segment, over this curve, one contribution is cancelled by the other equivalent. And then you have 1 and minus 1, 0. Outside 0, so it is homologous to 0. So that I can finally say, why don't we apply Cauchy integral formula? Because now we finally have, we are in the setting of Cauchy formula, Cauchy integral formula. And we take a z here and the two curves. The two circles, sorry, the curve is 1. So I can say, well, take 1 over 2 pi i. And this is what? Is f of z times n gamma z. And this number here is what? Is the index of the curve gamma defined here for the point z? But z is here. So it is in the unbounded component of a complement of gamma 1 and in the bounded component of gamma 2, correct? So this integral, this index is 1. Therefore, we have this, OK, this can be omitted. We have this important formula. And remember that as we have done for the holomorphic case, starting from this and using some uniform convergence of elements, we have found that any function with this integral representation can be described in terms of power series. And then we'll see what happens, OK, next time. Four functions defined in with this integral formula inside an n o's. So the idea is that we have to recover all the situation with singularities. So we have to expect to have power series expansion, but with index 6 going from minus infinity to plus infinity, so that we include all the cases, OK? So if you start from only from a certain point, so from a certain a, it's a certain value minus m, then we have power, OK? If we start from m equal to 0 in the summation of a power expansion, then we have a holomorphic line, so the singularity will be removable. But then if we have an infinite principal part, so that the coefficients are not all 0 from a certain, so definitely 0 from n minus m to minus infinity, then we have an essential singularity. So we'll see that with this tool, the important tool of Cauchy integral formula, we can describe all the singularities and the expansion of a holomorphic in a neighborhood of a singularity, OK? So I stop here and see you on Monday morning.