 Okay so let me recall what we did in the last lecture, so if you remember the theme that we are discussing is about zeroes of analytic functions okay and as you know the residue theorem allows you to compute the number of zeroes okay and that is through the so called argument principle and then we saw Roush's theorem which tells us that if you take an analytic function and change it by small amount that you may that is you add a smaller function to it, a function that is smaller on the boundary curve in of course in magnitude or modulus then the there is no change in the number of zeroes okay and then what we discussed in the last lecture was Hurwitz's theorem, Hurwitz's theorem which says that a zero of limit of analytic functions is coming from zeroes of the functions in the limit okay. So let me just recall that, so here is Hurwitz's theorem, so you assume that fk is a sequence of analytic functions which converges to the function f normally in normally on a domain D, so let me recall that domain is an open connected set, it may not be bounded and of course subset of the complex plane and all these fk's are analytic functions defined on D and this f that this statement that fk converges to f normally means that the convergence is uniform on compact subsets okay and then what Hurwitz's theorem says is that if you take a zero of f let z0 belonging to D be a zero of f of order m0, so as I explained in the last lectures it will follow that f is analytic okay because of normal convergence and normal limit of analytic functions is again analytic okay that is essentially because of the uniform convergence on compact subsets and since analytic function has zeroes which are isolated okay you can always find given any zero you can find a disc surrounding that zero where there are no other zeroes okay, so suppose I pick zero of the limit function and suppose zero is of order m0 okay then there exists a row greater than zero such that for k sufficiently large fk has exactly m0 zeroes in mod z-z0 strictly less than row and z0 is an accumulation point of such zeroes as if you want row tends to zero okay. So this is Hurwitz's theorem and I explained proof of this theorem and basically the proof of course use the argument principle okay. In fact I mean the basic idea of the proof was the idea of the proof I gave last time was you just calculate 1 by 2 pi i integral over mod z-z0 equal to row of d log fk and show that this tends to 1 by 2 pi i integral over mod z-z0 equal to row of d log f where of course d log fk stands for fk prime the first derivative of fk divided by fk dz and similarly d log f stands for f prime by f the first derivative of f divided by f the logarithmic derivative dz okay and of course the more serious point of the proof was that you will have to show that this is defined this is defined and then this converges to that and of course argument the argument principle will tell you that this is actually m0 and this if you call this as mk then the argument will tell you that mk then of course this m0 is of course the value of this integral is m0 which is the number of zeroes of f in this inside the inside the region bounded by this circle and of course you know you do not I mean you choose this disc as I told you in such a way that there are no other zeroes of f and similarly this quantity if you call this as m sub k that will be the number of zeroes of fk inside this disc and this argument tells you that this mk converges to m0 and but then mk being a sequence of integers when you say sequence of integers converges to an integer it means that beyond a certain stage the sequence of integers is just that constant integer which is the limit okay so that means that mk is equal to m0 for k sufficiently large and that is the conclusion of the theorem and of course argument if it works for certain row it will start it will work for smaller rows okay or if you take smaller radii it will work so this is m0 implies that mk is equal to m0 for large k okay and of course diagrammatically what this means is that you see if z0 is the point here where you have a zero of f this is the disc centered at z0 radius row then you can find all these zeroes of these are zeroes of fk and all these zeroes they converge to z0 as you make row smaller okay. Now what I wanted to discuss is the two things I wanted to discuss one is that there is another proof that you can give which actually uses Roush's theorem okay and then I also wanted to discuss about the application of the application which says that if you take normal limit of univalent functions and the limit is non constant then the limit is the limiting function is again univalent where of course univalent means one to one so let me first do that so let me look at this application again if fk converges to f normally and each fk is univalent on D then f is either constant or univalent on D but of course univalent means one to one which is also called a re-injected okay. So this is an application of Horowitz's theorem and I just wanted to look at this proof see so I know that each fk is one one I want to show that f is one one and of course the result says that f is one one provided f is not constant so you assume f is not constant okay assume f is not constant so you know I want you to remember that the moment I say that f has a 0 of order m finite order I am assuming that f is not constant okay because if f is constant then f has to be identically 0 and if f is identically 0 then basically you do not you will not get a you will not be able to find disc surrounding a 0 where there are no other zeros because every point is a 0 okay so you must understand that this Horowitz's theorem applies only to a non-constant it applies only to the case when f is you know non-constant analytic function okay. So even the theorem on the set of zeros of an analytic function being isolated assumes that you are working with an analytic function which is not constant for non-constant analytic function the zeros are isolated okay so non-constant is always there at the back of all this okay so assume f is not constant but it is important that is the reason why I am insisting sometimes we might be careless enough not to write it or insist but it is very important it is there in the background. So assume f is not constant suppose f of z1 is equal to f of z2 is equal to omega naught I will have to show that for z1 z2 in D I will have to show that z1 equal to z2 and what do I do I apply Horowitz's theorem so here is my z1 and you know there is a row and there is a disc surrounding row so that I can find zeros zeta I can find a 0 zeta i of zeta i is 0 of f of z-omega naught okay so what you must understand is fk converges to f so fk of z so I should be fk so fk of z converges to f of z so fk of z-omega naught converges to f-omega naught okay and it is again normal convergence. So I am applying Horowitz's theorem to not to f not to this but I am applying it to the sequence with minus omega naught added on both sides okay. So I can find a 0 zeta i of fk of z-omega naught and of course you know I will let me write it as Ki for Ki sufficiently large and you know the same way so there is also the point z2 I again take a similar disc of radius row of course you know this row here is chosen so that z0 z1 is the only 0 of f of z-w0 in this disc and here also I am trying to choose row of the same type that is 1 in which in this disc z2 is only 0 of f of z-omega naught but what I want to tell you is that to begin with this row 1 these rows may be different this may be row 1 that may be some row 2 but then I am saying take the minimum if you want take the minimum row and do it for the minimum value of row okay and so that row is the minimum value okay and what you do is here again Horowitz's theorem will tell you that I will get an eta i so I will get an eta i0 of f of Ki of z-omega naught for Ki sufficiently large and you see the little point to note is that I am choosing the same Ki okay this Ki that I got for this may be different from that Ki okay so in fact I should call this as if you want Ki and Ki but then it holds for all values beyond certain states then I can take the maximum of those two and call that as Ki okay so that is the adjustment I make. So what you must understand is that this row in principle I should write as row 1 and that I should write as row 2 okay and this I should write as Ki and that I should write as Ki prime okay but I can choose the maximum of Ki and Ki prime and replace that Ki call that as Ki okay and I can take the minimum of row 1 and row 2 and call that as row okay and then you know what you can do is once you have done it for row you can next do it so you know maybe I will first call it as let me first call this as let me do it for okay so let me do the following thing here instead of row let me put row by i okay instead of row let me put row by i right. So the point is that the reason why I am putting row by i is that you know the distance between zeta i and z1 is less than row by i which as i tends to infinity goes to 0 which tells you that the zeta i will converge to z1 and the eta i will converge to z2 okay. So I can do I can take this these row by i's okay and you must think that as I increase this i okay then the for example if I put i equal to 1 it is just row if I put i equal to 2 it is row by 2 okay then it becomes row by 3 you get smaller and smaller and smaller discs okay so with this kind of thing what you get is the following you get that okay so I should I should try to rewrite this as a zeta i. So you see zeta i converges to z1 eta i also converges to it converges to z2 fk i of zeta i is actually w0 which is equal to fk i of eta i because zeta i is a 0 of fk i of z-w0 and is also 0 eta i is also 0 of fk i of z-w0 and but then fk i is given to be 1 to 1 so this will tell you that zeta i is equal to eta i eta i since fk i univalent and this implies taking limits that limit zeta i is limit eta i but that means you will get z1 equal to z2 and that finishes the proof okay so what I want to tell you is that you have to be a little careful in choosing the zeta i and eta i okay and so basically you should choose a set of sequence of zeta i which converges to z1 and the sequence of eta i which converges to z2 as i tends to infinity okay so that is the proof. Now what I want to discuss next is another proof of Hurwitz's theorem which actually uses Roush's theorem okay so recall what was Roush's theorem see Roush's theorem basically says that if you have the number of zeros of an analytic function in a simple closed inside any simple closed curve is not going to change if you add to the analytic function another analytic function which is a smaller function on the boundary okay so let me write that so let L of z and B of z be analytic on D union dou D where D is bounded and dou D is a contour okay so it is a piecewise smooth contour and the function the both functions so L is supposed to be thought of as the little function B is thought of to be thought of as a bigger function okay. Suppose that the little function is lesser than the bigger function in modulus strictly lesser than on the boundary okay then B of z and B of z plus L of z have the same number of zeros inside D okay this is the Roush's theorem where you think of so what it says is the number of zeros of B z is the same as number of zeros of B z plus L z now that B z plus L z is thought of as a small perturbation of B because you have added the error term that you have added is L of z which is analytic of course but the point is that L of z is strictly smaller than B of z in magnitude on the boundary okay so this is as if you remember we proved this very easily using the argument principle but the point is this also yields a beautiful proof of Hurwitz's theorem okay so how so the reason why I am doing this is this discussion of you know zeros of analytic functions essentially uses residue theorem I mean argument principle and all these ideas are interrelated okay so you should understand how each idea you know is kind of connected to another okay. So you see so proof of Hurwitz's theorem using Roush's theorem so you see it is a suppose I want to prove this using Roush's theorem then it is very easy to guess what you have to do you see what is Roush's theorem or what is Hurwitz's theorem actually want to say it wants to say that you know in a disk like this f and fk have the same number of zeros that is what you want to say so you see so it is very clear that you know you have to take one of the big functions the big function has to be f okay and this small function should be chosen so that when you add it to the big function you get fk see the Roush's theorem says there is a big function and the big function plus a smaller function they have the same number of zeros okay now if you want to get this from that then but here I want f and fk to be the two functions for which the number of zeros are the same beyond a certain stage so the big function has to be f and the big function plus a small perturbation must be fk okay so the small perturbation has to be fk-f it is very simple to see that so what you do is put take row so that mod so that there is no zero of f of z other than z0 in well z-z0 less than or equal to row choose such a row of course as I told you this is possible because you are assuming that f is analytic I mean you have that f is analytic okay that is because f is a normal limit of analytic functions right and then put big function to be f of z little function to be fk of z-f of z okay then you know of course if I add the two I will get fk okay and of course to apply Roush's theorem I am the domain on which I am applying Roush's theorem is this disk is this disk and the boundary is just the boundary circle okay so I am applying Roush's theorem here okay I am just applying Roush's theorem here alright to the little function and the big function and I will get that the big function which is f and the sum of the big function and the little function which is fk they will have the same number of zeros okay provided the little function is really little than the big on the boundary okay but you see it is a that is something that you can that is something that you can easily see because you see see fk of z fk converges to f normally this implies that you know fk my fk-f goes to 0 okay fk-f goes to 0 that is what it means and that also normal okay and of course in this case normally means that it will be uniformly in this region so I should say so in fact I can rub off this normally here and simply write uniformly in this and this is also uniformly goes to 0 uniformly in mod z-z0 that is because the convergence is uniform on compact subsets okay and mod z-z0 less than or equal to rho is a compact subsets it is closed and bounded. So but what does this mean this means that the modulus of this can be made lesser than any small quantity that is what it means and you see f note that f mod f is greater than or equal to delta on mod z-z0 is equal to rho okay this is the fact that we also use during the proof of Fourier system because you see mod f is a continuous function okay it is a continuous real valued function and when defy and when you restrict it to mod z-z0 equal to rho mod z-z0 equal to rho is a circle centred at z0 radius rho that is compact because it is closed and bounded okay. So we have this fact from analysis you take a real valued continuous real valued function if you restrict it to a compact set then it will be uniformly continuous and it will attain its bounds. So in particular mod f will have lower bound it will have an upper bound and it will take the lower value and it will take the upper value also okay and delta is the lower value okay on on on this compact set the circle boundary circle okay and of course this delta is positive that is because mod f is positive mod f vanishes only at the centre and it does not vanish anywhere else so it is on the boundary it is positive therefore the minimum value is also positive okay and that is because minimum value is taken by mod f okay and mod f cannot be 0 if mod f is 0 then f is 0 and f is not supposed to be f is not supposed to vanish anywhere in that closed disc except at the centre that is the choice of rho okay. So mod f is greater than or equal to delta but then so you know fk-f converges to 0 uniformly means that I can choose a you know a index large enough index n such that for k greater than or equal to n fk-f in modulus can be made less than delta I can do that okay since fk-f converges to 0 uniformly in mod z-z0 less than or equal to rho we can choose so let me continue here we can choose we can find n such that k greater than or equal to n implies that mod fk-f can be made less than delta I mean this is just uniform converges okay and this is I am not writing fk of z-f of z because all this is done independent of z and this independence of z is exactly the uniformness of convergence okay. So this n does not depend on z it does not depend on what value of z you plug in where z is in this closed disc okay that is the uniformness that I am using there alright but you see this is but delta is less than or equal to mod f. So what you get is you get the modulus of the little function is strictly less than modulus of the big function by our choices and that is precisely what you need to apply Roush's theorem okay so by Roush's theorem f of z which is bz and fk of z which is b of z plus l of z have the same number of zeros in mod z-z0 strictly less than for k greater and that is exactly Kurwitz's theorem okay. So you see you get Kurwitz's theorem as a consequence of Roush's theorem right fine so having done this what I want to do next is I want to go to a topic which is called as I want to go to the topic of open mappings okay so I want to prove the very important open mapping theorem the open mapping theorem says that any non-constant analytic function maps open sets to open sets okay it is a very deep theorem but the point is that somehow the proof of the theorem also involves ideas of this type it just involves it is again about zeros of analytic functions okay and it also again literally involves the if you want you know the residue theorem in the form of the argument principle okay. So here is the open mapping theorem it is a very deep theorem very important theorem if f is a non-constant analytic function on domain t then f is an open map that is for any open set u in d f of u is open okay so this is the open mapping theorem it says that a non-constant analytic function if you take the under a non-constant analytic function if you take the image of an open set you will again get an open set okay this is a very deep theorem because you see you cannot find any counterpart for this in for example functions of one real variable okay it is rather I mean it is God given and it is beautiful normally you cannot expect a map to take open sets to open sets which is a very important condition it is an important condition because along with this if you put the condition that f is 1 to 1 okay then it means that since f is 1 to 1 f inverse makes sense set theoretically and saying that f is open will tell you that f inverse is a homeomorphism okay and what it will tell you is that is what we are going to see after this there is an inverse function theorem which will tell you that f inverse itself is analytic okay that is the next step okay so put so for all these things so the final statement is that if you have an injective analytic map then the image of the source domain will be an open set and f inverse on that open set will again be analytic so that means that f is an analytic isomorphism okay what it tells you is that an injective analytic map is a isomorphism onto it is image which is open analytic isomorphism and inverse has an inverse which is also analytic okay so the starting point is this for even for the inverse to be even continuous you need the fact that f is open helps okay so yeah so how does one prove this so basically you know we do the idea of proof is to count the number of times f takes a value omega not okay so it is again the counting principle the argument principle okay so in fact so not only count the number of times f takes a value omega not in fact you also let this omega not to vary okay so let me explain that so you see so here is so let me draw a diagram so here is my source complex plane and well here is some domain and here is a point z not and here is my function f, f is all constant and this is of course the this is the z plane and the target plane is also complex numbers but it is the omega plane where omega is equal to f of z okay if it is one real variable you write y equal to f of x okay since it is one complex variable you write now we now write omega is equal to f of z and suppose you take a value omega not which is f of z not okay now what one does is how will you count the number of times f takes the value omega not okay so that means you know you have to look you have to think of z not as a 0 of f of z minus omega not okay you think of z not as a 0 of f of z minus omega not you see that is the idea that we have been using all the time right so think of omega z not as a 0 of f of z minus omega not okay and notice that f of z minus omega not is also non constant analytic function because if f of z minus omega not is constant that will tell you that f of z is constant but I have assumed f is non constant and so after all f of z minus omega not is the analytic function f with minus omega not added to it minus omega not is just a constant you have added adding a constant to analytic function continues to keep it analytic okay. So if you want because a constant function is trivially analytic okay and the sum of analytic functions is again analytic so again f of z minus omega not is a non constant analytic function and z not is a 0 so the number of times it assumes the value z not is given by the argument principle in a disc surrounding z not where there are no more zeros other than z not. So what you do is that you choose you choose a disc you choose a disc of radius rho okay choose rho so in this case okay so choose rho so that z not is the only 0 of f of z minus omega not okay in mod z minus z not less than or equal to rho in this disc centered at z not radius rho so z not is the only 0 of f of z minus omega not this you can do because f of z minus omega not is a non constant analytic function and the zeros of a non constant analytic function are isolated so the 0 z not is isolated so you can find a small disc surrounding z not where there are no other zeros okay even on and you can choose the disc small enough so that there are no zeros on the boundary as well on the boundary circle as well okay. Now so you see so what is the number of times f assumes the value not assumes the value omega not in mod z minus z not less than rho how is this how is this given by the let me call this n sub w not this will be 1 by 2 pi i integral over mod z minus z not equal to rho d log f of z minus w this is just the argument principle the argument principle tells you that d log of something if you take and then you integrate over a simple closed curve and divide by 2 pi i you will get the number of zeros of that inside the closed curve. So I will get this will actually give me the number of zeros of f of z minus w not and they will be exactly the number of points inside this region enclosed by this circle namely the disc centred in z not radius rho where f takes the value w not okay this is just again by the argument principle or so this is a counting principle okay. So now what you do is you see note that mod f of z minus w minus omega not is say again greater than or equal to delta greater than 0 on mod z minus z not less on z mod minus z minus z not equal to rho. So this is again the same kind of argument that we used earlier namely f z minus w not does not have any zeros on the boundary circle because in this closed disc the only 0 of f z minus w not is at z not at the centre. So there are no zeros on the boundary circle and the boundary circle is closed in bounded so it is compact and mod f z minus w not is a continuous function when recited to this compact set it has it is uniformly continuous and it will have a minimum and a maximum value and delta is the minimum value and the minimum value is positive because it does not vanish okay. Now you see the trick is what you can do is consider any omega such that mod omega minus omega not is less than delta okay so you see it is the same delta I am using so what you do is you now take a disc centre at omega not and radius delta okay then you know I can completely replace omega not in this equation by omega and that will give me the number of times f takes the value omega in the disc mod z minus z not less than rho. So define n of omega to be 1 by 2 pi I integral over mod z minus z not equal to rho d log f of z minus omega okay mind you this makes sense because you see what is d log f of z what is this this is actually f derivative of this which is f dash of z divided by f z minus w d z this is what it is okay and mind you you see the f z minus w cannot vanish on the boundary okay f z minus w cannot vanish on the boundary why is that so that is because of this choice of w okay the choice of w is that see the choice of w tells you that the distance from w to w not is less than delta whereas the distance of f z from w not is greater than or equal to delta okay so this will tell you that f z f z the modulus of f z minus w cannot be 0 therefore this is well defined this integral is well defined and what does it give you it gives you the number of times the function f assumes the value w or omega in the unit disc I mean in the disc centered at z not radius rho. So this is number of times f assumes the value w in mod z minus z not less than rho okay this makes sense now after having written all this let me tell you that the whole point is that you see if you think of w as now a complex variable okay then this is a function n of w is a function of w okay the amazing fact is but it is amazing but it is very easy to prove the amazing fact is that n of w is actually an analytic function of w okay it will turn out that n of w is an analytic function of w okay and that will mean that n of w is constant because you see it is an analytic function but it is values are in integers okay and you know the image of the if you have an analytic function if you take the values of analytic function okay if for example when I say n of w is an analytic function of w in mod in this disc then this disc is of course connected. So if I take the image of this disc I should get a connected set but on the other hand the values are integers so I should get a connected set of integers okay but what is the connected set of integers it has to be only a single integer so what it will tell you is that n of w is a single integer and that is irrespective of w so it will be the same integer as n of w not okay but then what does that tell you it tells you that if f assumes a value w not n w not times then f assumes every other value w the same n w not times in this in this mod z minus z not less than rho. What this tells you therefore is that this whole disc is in the image and that is a proof that the image contains an open disc centered at z not so if you take a point centered at w not so if you take a point w not in the image then you get a whole disc centered w not in the image and that is exactly saying that every point in the image is an interior point of the image and that means that the image is open and that is the proof of open mapping theorem. So the technical point is to show that this is an analytic function okay and everything follows from that okay and mind you the idea is very simple we are just using the counting principle the argument principle okay so I will expand upon this in my next lecture I will explain how to show n of w is an analytic function okay so I will stop here.