 Okay, so alright so what we are going to do now is look at a proof of, try to look at a proof of Furwitz theorem and try to look at probably an application of Furwitz theorem. So let me again recall what Furwitz theorem says, I mean in short it says that if you have a sequence of analytic functions that converge to a function normally in a domain then a 0 of the limit function is obtained as an accumulation point of the zeros of the functions in the sequence beyond certain stage, okay. But if you want it more precisely this is a statement, if fk is a sequence of analytic functions that converge to the function f normally in D, mind you normally means that the convergence is uniform not on the whole of D but it is uniform on compact subsets of D and suppose, of course suppose each fk is analytic and if z0 is a 0 of the limit function f then there is a small disc surrounding z0 such that beyond a certain stage n all the f sub k for k greater than or equal to n they all have the same number of zeros counted with multiplicity in this disc as the multiplicity of z0 of the 0 z0 namely the order of the 0 z0 of f and as you may cross smaller all these zeros of the various fk's they converge to z0, okay. So this is for which so I mean for example you may ask what is the use of such a theorem, okay so I will tell you one use one application of this theorem just in words this using this theorem you can prove that if each fk is 1 1 1 to 1 as a map then the limit function f if it is non-constant it is also going to be 1 to 1, okay. So you see normal convergence is already something very strong because it is actually uniform convergence on compact subsets and you know uniform convergence is a very powerful thing, okay. You know we just proved in the last lecture that since each fk is analytic f is also analytic that was because of the uniform convergence on compact subsets due to Morera's theorem and application of Morera's theorem but what you get also is that if each fk is 1 to 1 then you get f is also 1 to 1, okay that is an application for example. Of course you assume f the the limit function is non-constant so if you have a sequence of 1 to 1 analytic functions injective analytic functions analytic functions are injective as maps, okay functions which do not take two different points to the same value then the same property of injectivity is passed down to the limit function provided the limit function is not the constant not a constant function, okay. So that is an application of this, okay and you know why this is important it is important in study of classes of analytic functions for example people often study the class of functions analytic on the open unit disk and they also normalize it by putting conditions like the function as the value 0 at the origin and its derivative at the origin is 1, okay and then they study special classes of functions which are for example 1 to 1 on the unit disk and this leads to lot of geometry and all those for examples for example in all these condition under such conditions you can show that if you have a class of functions with a certain property then any sequence of such functions which converges normally to a limit function then the limit function also has the same property. So it again belongs to that class and that is true for example for the class of all univalent functions that is 1 to 1 functions on defined on the unit disk which are normalized to taking the value 0 at the origin and whose derivative at 0 is 1 that is for example an application of this, okay so it is very important for function analytic purposes. So what I am going to do now is you know try to give the proof of this theorem so you know so you see the proof of the theorem is basically mind you it is of course you know the general theme is we are looking at 0s of analytic functions this is the general theme which we are now discussing about and you already know how to count the number of 0s of analytic functions inside a closed curve, okay that is precisely given by the well residue theorem in fact more precisely by the argument principle, okay. So you know so let me so if you recall if gamma is a simple closed contour in D then 1 by then 1 then the so called the argument principle gives 1 by 2 pi i integral over gamma d log g z t z or d log g number of 0s minus number of poles, okay where n of 0 is number of 0s of g n infinity inside gamma and I say inside gamma it is in the region enclosed by gamma and gamma is given and what do I mean by region enclosed by gamma it is a region that lies to the left that lies to your left if you walk along gamma in the prescribed direction mind you whenever we write integral over gamma whenever you write a path of integral already there is an orientation, okay usually we take the anticlockwise orientation so that the region inside gamma is actually the region that is enclosed by gamma, okay and n infinity is number of poles of g inside gamma and where and of course g is meromorphic in the interior of gamma of gamma and is non-zero and analytic on gamma on gamma, okay this is what you want this is the argument principle, okay. So the logarithmic derivative if you take the integral and divide by 2 pi i you will get the number of 0s minus number of poles and of course mind you you can also recall that 1 by 2 pi i integral over so further 1 by 2 pi i integral over gamma d log g z d log g is actually 1 by 2 pi into the change in the argument of g z change in the argument of g over gamma, okay so this is another interpretation which is actually the reason why it is called argument principle if you compute this integral what you will get is 1 by 2 pi into the change in the argument and you know since gamma is a closed curve the change in argument will always be a multiple of 2 pi so dividing by 2 pi will give you an integer and what is that integer that integer is exactly number of 0s minus number of poles that is exactly what the argument principle says, okay. So this is the reason why it is called the argument principle, okay now what is and of course you know how you get this is basically from the residue theorem you apply the residue theorem to the logarithmic derivative of g which is the function g dash of z by g z so this d log g is actually g dash by g this is what it is, okay so maybe I should also add a d z to be very precise d log g is g dash by g d z, okay or in other words d by d z log g is g dash by g, okay. So fine so now why this is so important is because suppose your function is not meromorphic but it is actually analytic then there are no poles so what you get is you get the number of 0s, okay so it helps you to count number of 0s inside a closed curve, okay that is why this is important and we are worried about number of 0s, okay you want to count numbers of 0s. So this is the starting point and you see what this will tell you is that you can now apply this to calculate what is the number of 0s of the limit function f and what is the number of 0s of each fk so you know so you know you will get that integral over mod z-z0 is equal to rho so you know I have not told you what this rho is, I have not told you what this rho is but take a disc centred at z0, okay mind you z0 is a point of d where f has a 0 the limit function f has a 0 and the 0 is of order m0, okay that is our assumption. So take this point z0 and take a circle centred at z0 radius rho and orient this circle in the anticlockwise sense so that the region inside the circle is actually the region inside the circle as we would normally think of it and what you do is over this if you integrate d log fk what you will get is number of 0s of fk inside the circle mod z-z0 less than rho. Of course the assumption is that fk should not vanish on the boundary, okay provided fk not equal to 0 on the boundary mod z-z0, okay because you know to apply the argument principle the function should not vanish on the boundary, right and I can write a similar statement for f mod if I integrate mod z-z0 over equal to rho of course when I write this it is with this positive orientation anticlockwise orientation if I integrate d log oops I think I forgot to put 1 by 2pi, okay so let me do that, yeah it is very common to forget 1 by 2pi or 1 by i or an i or something like that so you have to keep track of this. So d log f of course I am saying it is common but I am not saying it is correct so it should not be done on purpose so this is the number of 0s of f inside mod z-z0 strictly less than rho provided f is not 0 on boundary mod of z-z0 equal to rho, okay this is what you have. Now let us go back and so I will have to tell you what this rho is I have to find a rho such that mod z-z0 equal to rho on that circle fk should not vanish f should not vanish I need a rho like that. Now the point is see here is where a technicality comes in because fk converges to f normally and since each fk is analytic f is analytic that is what we have already seen. Now since f is analytic you know the 0s of an analytic function are isolated, okay so this z0 is an isolated 0, okay that means you can find a disc surrounding z0 where there is no other 0 of f, okay this is where technically I am using the fact that f is analytic and the 0s of an analytic function are isolated, okay and you know if you go back to your first course in complex analysis that is actually due to what is called the identity theorem. The identity theorem says that if an analytic function has 0 at a set at every point of a set which has an accumulation point, okay and the accumulation point is in the domain of analyticity then the analytic function has to be throughout 0 it has to be identically 0, okay. In other words if you have a sequence of if you have a set of 0s of an analytic function which converges to a point where the analytic function is defined and analytic then obviously that point will also be a 0 because of continuity and therefore for that point every neighborhood of that point will contain a 0 that 0 will not be isolated the limit point which is a 0 will be such that every neighborhood of that limit point however small a disc you take there is some other 0 and that cannot happen for an analytic function unless it is completely 0 identically 0, okay. So of course you know in all these things I am not assuming that the function f is a constant function because of course you know if you assume f is a constant function then that means and if you assume f as a 0 of order m0 at z0 then you know it really does not make sense because if it is it will tell you that f is 0 at z0 and then it will also tell you that since it is constant it is 0 everywhere so I am certainly not looking at the case where f is constant, okay. So f is certainly not constant so the 0s are isolated, okay and I am using this fact now because the 0s are isolated I can find a row such that f does not vanish not only the interior but also on the boundary circle so that is how I choose my row, okay. So choose row such that f does not vanish for all z satisfying 0 strictly less than mod z-z0 less than or equal to row so this is the fact that means I have put 0 strictly less than because I do not want to include z equal to z0 because I add z equal to z0 because it does vanish it is a 0 z0 this is 0 of f but I do not want apart from the centre of the disk I do not want any 0s for f I do not want any 0s even on the boundary and such a disk I can get that is because the limit function f is analytic and the 0s of analytic function are isolated, okay fine. So this is one point of the story then the other point of the story is of course you know I need to also worry see the moment I assume this integral is well defined because for this integral to be well defined mind you d log f is just f dash by f dz and I am dividing by f and I do not and I am integrating on the circle, okay so what I am dividing by should not be 0 on the circle this is true, okay. So this integral is defined but I have to worry about why this integral is defined, okay. Now the answer to that is the following, okay so f does not vanish on this boundary, okay that means that mind you if you look at mod f mod f is a continuous function, okay and this boundary is a compact set you know a continuous real valued function on a compact set is uniformly continuous and attains its maximum and minimum value. So if you look at mod f mod f will attain a minimum value and a maximum value on this compact set which is the circle and since it is never 0 the minimum value cannot be 0, okay so that is exactly what I want I just want to say that the f is bounded away from 0 on the boundary circle. So let me write that since mod f is continuous on the compact set mod z-z0 is equal to rho mod f attains its maximum its minimum is set and that minimum is a positive, okay. See a continuous so this is a you know if you want it is a theorem from real analysis first course in real analysis it is the if you want in the simplest form you take a real valued function a continuous real valued function of one real variable if you take the image of a closed interval the image will again be a closed interval topologically what you are saying is that since the first of all since the you are taking the image of an interval and which is connected and since the function is continuous the image will also be a connected subset so that means the image will also be an interval and since you are started with a compact set because a closed interval is both closed and bounded so it is compact and the continuous image of a compact set is compact. So what you will get is the image of a closed interval is again an interval which is compact which means it is an interval which is closed and bounded which means it has to be a closed interval. So what it means is that if you take mod f mod f if you restrict it to this compact set mind you the circle is both closed and bounded as a subset of R2, okay. If you want you can re-parameterize it as a interval on the real line after all you have to just take the parameterization z equal to z0 plus rho e power i theta where theta lies from 0 to 2 pi, okay. So then you will see that mod f will the image of this circle under mod f will be a closed interval, okay and that closed interval will have a minimum and a maximum value a left end point and right end point the left end point is this minimum and that will be positive because it is a value taken by f on the boundary and f does not vanish on the boundary that is our assumption it vanishes only at the centre, okay. So call this minimum delta call this value as delta, okay. So what you get is so what you get is the following you get that so what you get is so for mod z-z0 stick to less than rho less than I mean sorry equal to rho mod f of z is greater than or equal to delta, okay yes. So you see so you have this so I was thinking for a moment because I will have to justify why this this integral makes sense. So you see the fact is that this integral makes sense for k sufficiently large, okay. So let me come back to this so you have this, right. Now you see since fk is converges to f, okay fk converges to f and it is uniformly on mod z-z0 equal to rho in fact less than or equal to rho because it is the convergence I have told you is normal that means the convergence is uniform on compact subsets. So in fact it converges uniformly on the closure of this disc that is the open disc along with the boundary circle. So since it is a uniform convergence, okay what will happen is there exists an n there exists a n1 such that k greater than or equal to n1 implies that the distance between fk of z f of z can be made I should say lesser than epsilon for a given epsilon greater than for any given epsilon greater than 0 independent of z of z on independent of z, okay this is what uniform convergence means. Uniform convergence means that the value of if you take any point z which lies in the domain in the set you are considering in this case it is the closed disc namely a point z satisfying mod z-z0 less than or equal to rho then mod fk of z-fz the distance which is the distance between f of z and fk of z that can be made lesser than epsilon, okay you are just saying that fk comes to within a distance of epsilon from f if k is sufficiently large and this does not depend on what z you choose and this non-dependence of z is the uniformness of the convergence, okay. If it is non-uniform then you know this n would change along with z the fact that this n you are able to given an epsilon you are able to get an n1 which does not depend on z is the uniformness of the convergence, okay. Now you see you know what this actually tells you see so in fact see in fact you see you also you in fact since mod fk also converges to mod f, okay see if fk converges to f then mod fk will converges to mod f will converges to mod f so what will tell you is that the values of the distance between mod fk and mod f can also be made as small as you want independent of what z is, okay. We can make given epsilon greater than 0 we can make we can we can get mod fk of z minus mod f z less than epsilon for k greater than or equal to n2 depending only on epsilon and not on z, okay, right. So what this you know it is a lot to write down but for the moment just do not look at all of this but just think of it like this fk converges to f so mod fk converges to mod f and this is this convergence can be it is uniform but you see mod f is greater than or equal to delta on the boundary. So what it will what this tells you is that since mod fk the values of mod fk approach mod f and since mod f is greater than or equal to delta you can make mod fk greater than delta by 2 if you want, okay. If you choose k sufficiently large that is what I want so there exist n3 so I am writing n1 n2 n3 so that you know you do not get confused so there exist n3 such that k greater than or equal to n3 implies mod of fk of z it can be made greater than or equal to delta by 2 for all z such that mod z minus z0 is equal to rho this is what I want and why I want this is to this is just to tell you that you see this integral is also well defined see what is this integral this is 1 by 2 pi integral over this circle of fk prime over fk tz the logarithmic derivative and this logarithmic derivative always has the function in the denominator, okay it is a derivative of function divided by the function and therefore for it to make sense it function cannot vanish so I need to make sense the idea I need to make sure to make sense of this integral I need to make sure fk does not vanish and how do I ensure that fk does not vanish by ensuring that fk is greater than or equal to some positive quantity on the boundary and for that I am saying I can choose a case sufficiently large and that comes out of this that is all I want. So the moral of the story is that therefore you see for the rho I started with not only is this integral well defined also this integral is well defined, okay and therefore both integrals make sense, okay and mind you this integral what is the value of this integral the value of this integral will now be m not because the value of this integral is equal to the number of zeros of f in this circle inside this circle but I have chosen the disc in such a way that there are no other zeros because of the isolation of zeros. So the only zero is at the center at z not and that zero is of order m so you have to come mind you whenever you count zeros or poles you should always count them with multiplicities. So this integral will be m not, okay and you know what I have to actually prove is that as k tends to infinity this also tends to m not that is essentially what I have to prove, okay. So that should more or less tell you how the proof is going to end note that both integrals 1 by 2 pi i integral over mod z minus z not equal to rho t log fk or k greater than or equal to I think that I use n3 and 1 and you know so let me call this as let me call this as mk let me call this as mk and 1 by 2 pi i integral over mod z minus z not equal to rho d log f is equal to m not are well defined, okay. You see now what I want to tell you now you know what I have to prove I am claiming that I just have to prove that you see I get this sequence mk, okay I get this sequence mk mk is what mk is the number of zeros of fk inside that circle inside that disc, okay and you know you can expect that since fk converges to f this integral will converge to this so you can expect that mk converges to m not, okay that is natural to expect and that is exactly what happens mk converges to m not. But then mk converges to m not means that beyond a certain stage mk is exactly m not see if you have a sequence of integers which converges to an integer then the sequence of integers must beyond a certain stage be equal to the constant sequence, okay. See when you say a sequence converges to a value what it means is that beyond a certain stage the values of the sequence come very close to the given value. If you now if it is a sequence of integers which converges to an integer what you are saying is beyond a certain stage all the integers are very close to this integer but one an integer very close to an integer has to be the same as that integer, okay. So if you just prove that mk tends to m not then you are done what it will tell you is that maybe you will have to choose a bigger value of n such that for k greater than or equal to n mk converges to I mean if you prove mk converges to m not then you can choose a index n such that for k greater than or equal to n mk is actually equal to m not, okay. What will that tell you that will tell you that fk has as many exactly m not zeros in that disk which is the assertion of Hurwitz theorem, okay that is part of the assertion of Hurwitz theorem. Then what you must understand is whatever I have done so far will not only work for row but I can replace row by half row then I can replace it by one third row then I can replace it by one fourth row. So you know that means I can shrink this disk smaller and smaller and smaller and everything works the same argument works. So what it will tell you is that for that value of n beyond which mk are all equal to m not all the zeros of fk, okay as I shrink row they are all going to come closer and closer and closer to z not. So they are all going to converge to z not, z not will be an accumulation point and that is exactly what Hurwitz theorem says all the and a zero of order m of the limit function comes from zeros of order m there comes from m zeros of the functions in the limit beyond a certain stage and these m zeros they actually coalesce together to give you the zero of order m in when you take the limit that is exactly what Hurwitz theorem says, okay. So I will do that so it remains to show that mk tends to m, okay this is what I will have to show and the answer to that is again as you will as you can expect it is again uniform convergence, okay. So the answer to that is again uniform convergence and how does one prove, prove it it is pretty easy you see you know basically I am trying to show that this converges to this so which means I will have to first of all look at the you know the integrand I have to show that this minus this is eventually zero, okay which means I will have to show that the integrand minus this integrand is essentially zero, okay. Now so you see so now I am going to make use of the following thing if you look at d log fk minus d log f this is just fk dash by fk minus f dash by f tz this is what it is, okay. And if you look at the modulus of this quantity modulus of fk dash by fk minus f dash by f I look at the modulus of this quantity, okay. Mind you when I am writing such an expression I am assuming that whatever values of z I plug in fk and f cannot be zero, okay. So for example this makes sense on this boundary circle mod z minus z not equal to rho because I have chosen it like that, okay. And of course I have chosen k, k has to be chosen sufficiently large namely k has to be greater than or equal to n3, okay. So if you look at this what you will get is well you will get fk prime is a simple calculation the kind of calculation that you should have been already used to in a course in uhhh first course in any first course in analysis. And you know well you see uhhh see this is less than or equal to you see mod mod fk uhhh mod f is greater than or equal to delta. So 1 by mod f is less than or equal to 1 by delta mod fk is greater than or equal to delta b2 so 1 by mod fk is greater than equal to 2 by delta. So this lesser equal to 2 by delta squared okay into this quantity modulus of fk prime f minus f prime fk and you know how to handle this okay it is a very simple trick that you always use to for example prove you know very simply how you prove a product rule for differentiation that kind of trick very simple trick so let me write this here mod of fk prime minus by fk minus f prime by f is lesser equal to 2 by delta squared so of course I should write here on mod z minus z not equal to rho for k greater than or equal to n3. So if I continue that here so let me rewrite what I have written there I get fk prime f minus f prime k and of course you know the trick is this is 2 by delta squared you add and subtract an obvious term fk prime f minus fk prime fk plus fk prime fk minus f prime fk so you see I am adding and subtracting this term okay and then I group these two and these two so this becomes 2 by delta squared modulus of fk prime into f minus fk plus modulus of fk sorry fk into fk prime minus f prime like this and now this is lesser equal to 2 by delta squared mod fk prime into mod of f minus fk this is by triangle inequality mod fk mod of fk prime minus f prime is what I get okay and now what I want to say is that this can be made less than epsilon for k sufficiently large this can be made less than epsilon for k sufficiently large and why is that so that is because you see mod fk prime see fk fk converges to f okay and the convergence is uniform therefore fk prime will converge to f prime because under uniform convergence the derivative if a sequence of functions converges uniformly to a given function okay and if of course you can actually check again by if you want to use of Morera's theorem that fk primes will also converge to f prime you see fk's are all already analytic okay and since fk converges to f normally you prove that f is analytic but then you see f is analytic means f is infinity differentiable and each fk is analytic so it is infinity differentiable what is the relationship between the derivatives the derivatives of the fk's namely the fk primes they will converge to f prime okay this will this can again be checked by a Morera theorem kind of argument if you want okay basically it is uniform convergence so fk prime converges to f prime so this is bounded you know convergent sequence is bounded and since the convergence is uniform this is uniformly bounded so I can simply make this less than some constant the uniformness I am using because I do not care what the z is so long as z lives on that boundary circle okay similarly I can make this less than or equal to some constant because of uniform boundedness okay and this also can be made as small as I want this I can make as small as I want because this is also uniform convergence this is also uniform convergence see the fact is if fk converges normally to f then the derivative of fk converges normally to derivative of f the second derivatives of the fk converges normally to the second derivatives of f and so on and so forth this will happen this will go on and on and on so because of the uniform convergence I can make this as small into some constant and this again something small into some constant and if I choose the index large enough I can make this whole quantity very small I can make the whole and since this is already a constant I can make the whole thing less than epsilon k is sufficiently large so what this tells you is that fk prime over fk converges to f prime by f uniformly so that is fk prime by fk converges to f prime by f uniformly on the boundary okay this convergence is uniform alright but then you know if you have uniform convergence if a sequence of functions converges uniformly to a function then the sequence of integrals will also converge because uniform convergence allows you to interchange limit and integral so this will tell you that 1 by 2 pi i integral over mod z-z0 equal to rho of fk prime over fk dz will converge to 1 by 2 pi i integral over mod z-z0 equal to rho f dash by f dz that is because if a sequence of functions converges uniformly to a limit function then the sequence of integrals over a path will also converge to the integral of the limit the integral and limit can be interchanged so that is why you get this but this is exactly the statement that mk converges to m0 okay so that tells you that mk converges to m0 that means mk is actually equal to m0 for k sufficiently large okay that is mk converges to m0 which implies mk is equal to m0 for k sufficiently large so that finishes the proof of theorem so let me add all this works for any rho prime with 0 lesser than greater than rho prime less than rho and that is the end of the proof of theorem okay that is the end of the proof of theorem so let me explain quickly an application of this theorem so here is an application so that you will appreciate the power of this theorem application let fk converge to f normally on D and assume that each fk is 1 to 1 so there is a very special name for such functions they are called univalent functions univalent okay. Then if f is not constant then f is also univalent that is an application a normal limit of univalent functions is univalent so long as the limit is non-constant that is a beautiful application of Hurwitz's theorem and how does one prove this I mean you have to show f is not constant you have to show f is univalent you have to show f is injective so you have to show that if f takes two values to the same value then those two values are the same that is what you will have to show and how does one prove it is just two lines suppose f of z1 is equal to f of z2 is equal to omega not for z1 and z2 in D. So I should think in the following way saying that f of z1 is equal to w not should be thought of you have to think in terms of zeros so you should think of z1 as a zero of f of z minus w not and you must also think of z2 as another zero of f of z minus w not. So then z1 and z2 are both zeros of f of z minus w not right and but then what is going to tell you Hurwitz's theorem tells you that there is a sequence of zeros of fk of z minus w not which converges to z1 and there is another sequence of zeros of fk of z minus w not which converges to z2 okay but that should tell you that z1 has to be equal to z2. See by Hurwitz's theorem there exists a sequence so let me give it some name let me call this as zeta i of zeros zeta i tending to z1 with zeta i a zero of fk of z minus w not and a sequence are eta i tending to z2 with eta let me call this eta j with eta j is zero of fk of z minus w not this is what Hurwitz's theorem says. Hurwitz's theorem says because you know fk minus w not will tend to will converge normally to fk f minus w not fk minus w not tend converges to f minus w not normally that is the reason okay and you are applying the Hurwitz's theorem to this sequence fk minus w not okay. So you see I have to say that limit of the zeta i is the same as limit of the eta j's okay and why is that obvious that is because the fk's are univalent see that is where I have to use the fact that the fk's are univalent see fk so maybe I will use the same index I will use the same index okay I will use the same index because this zeta i and zeta i are zeros of fk's okay so well what I will get is basically fk of zeta i is zero is fk of zeta i is a zero of fk of z minus w not means fk of zeta i is w not okay and that will also equal to fk of eta i because eta i is also a zero of fk minus w not. But then this is true and as fk is univalent okay mind you I am taking k sufficiently large since fk is univalent okay what this will tell you this implies that zeta i is equal to eta i now if you take the limit as i tends to infinity on limit of zeta i is z1 the limit of the eta i is z2 so you get z1 equal to z2 okay so this implies limit zeta i is equal to limit eta i that is z1 equal to z2 so that tells you that if f of z1 equal to f of z2 and z1 equal to z that is the proof that f is univalent and that is the proof okay. So if you have sequence of univalent functions that converge normally to a limit function the limit function is not constant then the limit function is also univalent okay so that is and of course you know in all these things where I have used the non-constancies it is because a converge serum does not apply when the limit function the limit function is constant okay. So converge serum does not apply when the limit function is constant and here the limit function is f minus w0 if f minus w0 is constant that means it is the same as saying f is constant that is not allowed because I have already assumed f is not constant okay so I can apply converge serum because f is not constant alright so we will stop here.