 So the next thing that I want to discuss is Schottky's theorem okay which is now very easy to prove okay. So let me write it down Schottky's theorem so this is let alpha be a positive constant and beta be a fraction and a positive real number between 0 and 1 okay. Then there is a constant C alpha beta such that if script f is a family analytic functions on the open unit disk omits the values 0 and 1 and satisfies f of 0 is bounded above by alpha for all functions small f in the family f then the mod fz is less than C alpha beta for all fz with mod z less than beta okay this is Schottky's theorem and the point is that this constant C alpha beta it depends only on alpha and beta and it does not have anything to do with what the family is it works for any family and the such theorems when they were first proved they were they were pretty they were considered pretty difficult but then because of because we have Montel's you know theorem on normality and Montel's test for normality it is easy to reduce this theorem okay. So let me tell you the proof see you can see immediately that you want the family to be arbitrary therefore you consider the biggest possible family namely you take all analytic functions on the unit disk okay satisfying the condition that the you know value at 0 is bounded by alpha okay so you apply it to the largest possible family that you can think of okay and you know see the moment you are given that these functions omit the value 0 and 1 okay it means that the family is normal see in fact you see what is Montel's theorem on normality otherwise called the fundamental normality test see if you want to decide a family of meromorphic functions on a domain is normal then you need to know that it omits three values but three values in the extended complex plane so one of them could be infinity right but then if you are working only with analytic functions you already know infinity is not going to be taken okay so you have to only ensure that for a family to be normal to be able to apply the normality test you have to only ensure that the family does not every function the family does not take two values so here it is given that the all these functions they do not take the values 0 and 1 so you know if you apply the fundamental normality test that is Montel's theorem it will follow that if you take the family of all analytic functions if you take the family of all analytic functions on the unit disc which omit the value 0 and 1 that will be normal okay so this is Montel's theorem on normality right but then we also saw another theorem of Montel okay which was a translation or an improvement of the Arzela-Ascoli theorem which said that for a family of analytic functions to be normal on a domain you need that the family is normally uniformly bounded that is it is uniformly bounded on compact subsets so if you see mod z less than so you know so let me write let me consider mod z less than or equal to beta okay if you look at mod z less than or equal to beta so I will change this here to mod z less than or equal to beta which is what I meant to write but I did but if you take mod z less than or equal to beta that is a compact subset of the unit disc okay because it is closed and bounded and therefore by the other Montel theorem which is improved version of the Arzela-Ascoli theorem the normality of the family will tell you that the family is going to be normally uniformly bounded so it is uniformly bounded on any compact subset so on this compact subset it has to all the functions should have a bound and call that bound as c alpha beta it is a simple as that okay so what you must remember is that we have applied two Montel theorems one Montel theorem which equates the normality of a family that is a normal sequential compactness of a family with the uniform boundedness of the derivative uniform boundedness of the original functions in the family on compact subsets normal uniform boundedness and that is mind you that is a improved version of the Arzela-Ascoli theorem and in fact it used Arzela-Ascoli theorem plus a diagonalization argument alright and then we apply the more serious Montel's theorem on normality the fundamental normality test or fundamental criterion for normality which is a very deep theorem mind you that was the key to prove proving Picard's theorem okay that the movement of family of meromorphic functions omits three values it is normal the movement of family of analytic functions omits two values it is normal okay. So you apply those two theorems then Schottky's theorem is simple corollary alright so it happens that there is a paper of Salkman in the building of the American mathematical society where several where he explains how several problems in function theory have easy solutions by use of this Salkman lemma. So in fact that is what is called the there is a very deep theorem called Bloch's theorem okay and it involves roughly it is trying to estimate the size of the largest size of a disk under the image of you univalent or one to one analytic function you take a one to one analytic function okay and then you know try to estimate you take the image and then you try to see what is the largest disk radius of the largest disk that is contained in the image okay there are theorems of this type and there is a particular theorem called Bloch's theorem which is very deep okay and this can be proved by using this so called Bloch's Salkman principle which is also called as Bloch's principle okay and the whole point is Salkman's lemma is very powerful it gives you proofs of easy proofs of very deep reserves right. So it is not a surprise that you get Schottky's theorem okay has a simple corollary so let me write down proof is uses so let us consider the family of all analytic functions on the open unit disk that we meet 0 and 1 this is normal this is normal by Montel's theorem on normality otherwise called the fundamental normality test or sometimes it is also called fundamental normality criterion again by another theorem of Montel along the lines of Arsila Ascoli the family is normally uniformly bounded hence bounded uniformly by c alpha beta on mod z less than or equal to beta okay see the point is I did not even use the fact that the functions at the origin are bounded by alpha okay I just I know that there is a uniform bound alright and I simply call that uniform bound c alpha beta actually I need not put that alpha there but I can call it c alpha beta the point is that I am I have to put in beta because I am looking at the bound on mod z less than or equal to beta which is a sub disk of the unit disk close sub disk of the open unit disk okay fine. So what you must understand that is that this easy proof is because you have the strong Montel theorem on normality which is a fundamental normality test okay so see alright so this is one thing then I would like to I would like to discuss the I would like to discuss the solution to the first assignment that I gave okay so here is a problem that I gave earlier let D be a domain in the complex plane and f from D to C be continuous such that for some in positive integer n integer f power n is analytic then f is analytic of course you know I need to take n greater than 1 because otherwise it is trivial okay because we put n equal to 1 f power n is just f power n is just f power 1 which is f okay and what is the solution to this well so the first thing is that so f power n first of all f power n means f of z whole power n which means f of z into f of z multiplied n times okay. The first thing I want to tell you is that we will use the fact that the 0s of an analytic function are isolated so since f power n is analytic okay the 0s of f power n are isolated but then the 0s of f power n are the same as the 0s of f therefore the 0s of f are isolated okay and therefore what we will do is we will first throw away the 0s and look at the complement of 0s in the domain which is an which is a sub domain okay alright and what we will do is that on that sub domain we will first prove that f is analytical alright and then we will have to worry only about these points where f becomes 0 alright but then we can apply Rayman's removable singularity theorem because each of these points will be isolated points in a neighbourhood of which f is continuous therefore they will be analytic even at those points and that is the proof okay. So let me write this down since the 0s of an analytic function are isolated the 0s of f power n the set of 0s is isolated but the set of 0s of f power n is the same as the set of 0s of f so the set of 0s of f in D is isolated of course you know when you want to say the 0s of an analytic function are isolated you must make sure that the analytic function is not identically you know constant so this is the only case where this will fail is when the analytic function is identically 0 if the analytic function is identically 0 then this 0 set is the whole domain alright that is the only extreme case but of course if f power n is identically if f is 0 then f power n is identically 0 and if f power n is identically 0 f is 0 so let us assume that f power n is not 0 assume that f is not 0 okay so there is nothing to prove if f or f power n is identically 0 okay so that has to be we assume that f is not identically 0 on D right so that is the only thing that we will have to worry about whenever you want to apply this result that the 0s of an analytic function are isolated you better make sure that the function is not identically 0 okay and usually we are not interested in that function fine so now what you do is in any case you look at consider D minus ZF okay throw away the 0s of f okay it is an isolated set of points so you get a domain again so D minus ZF is also a domain right and it will still be open okay it will still be an open set and because you are throwing away some isolated subset and it will also be it cannot get disconnected okay so because you are just throwing isolated points away it cannot get disconnected so D minus ZF is also it is also a domain okay and now we are going to look at this domain the advantage with this domain is that f power n does not vanish f does not vanish because the 0s have been thrown away so f power n does not vanish and f power n is an analytic function okay so you have a non-vanishing analytic function on a domain now you know if you take any point in the domain if you take a sub then there is a sufficiently small disc around that point which is in that domain okay and the point is that if you have a non-vanishing analytic function on a simply connected region on a simply connected domain then you can find an analytic branch of the logarithm of that function and in particular you can find n th roots of that function for any n the point is you can find n th roots which are analytic that is the whole point so if you want to find an n th root of a function which is analytic okay then the function should not vanish and the region and the set on which you want to find it must be simply connected alright so the point is that if you take f power n which is analytic and f power n does not vanish on d minus z f so if you take any point in d minus z f and you take a small disc surrounding that point in that small disc I can find an n th root of f power n and what do you expect it to be it has to be f okay but this n th root is supposed to be analytic therefore it will prove that f is analytic okay so but a little bit of a little bit more has to be written down okay so let us do that f power n is analytic and non-zero on d minus z f for z not in d minus z f that exists a small disc small open disc mod z minus z not lesser than epsilon in d minus z f since this is simply connected and f power n does not vanish there exists an analytic branch log f power n in mod z minus z not less than epsilon okay consider the analytic function so let me call this analytic branches g okay consider the analytic function e power 1 by n g okay which is actually see it is e power 1 by n g is actually a log of f power n okay and you know this must be f alright you should expect this to be equal to f right now you see we will use the we will use the following how will you show that this is the claim is that e power 1 by n g is actually equal to f okay the claim is e power 1 by n g is actually equal to f once you once that is proved it means f is analytic because e power 1 by n g is already analytic and e power 1 by n g is analytic because g is analytic and why is g analytic because g is an analytic branch of the of the logarithm okay so I just have to we just have to prove that e power 1 by n g is equal to f in the small disc okay and this will show f is analytic in a small disc but then the point z not was arbitrary so it will show that f is analytic on d minus z f okay and then to at points of z f you can apply Rayman's removable singularity and conclude that f is analytic on the whole of t alright so the only issue is now that I will have to show that e power 1 by n g is f okay now what is common to e power 1 by n g and f they are both n th roots of f power n f is an n th root of f power n by definition alright and e power n 1 by n g is also an n th root of f power n because if I take e power 1 by n g and raise it to the power n I will get f power n okay I will get e power g if I take e power 1 by n g and raise it to the power of n I will get e power g but g is log f power n so I will get e power log f power n which is f power n okay so both e power 1 by n g and f are n th roots of f power n alright and the point is you see you take if you see the so now we have to use the following property if you take the take any two logarithms of a complex number okay they will differ by a constant they will differ by a constant multiple of 2 pi i you see if you take if you calculate the logarithm of a complex number of course it is only defined for a complex number which is different from 0 there is no logarithm for 0 okay so if you take a non-zero complex number and calculate its logarithm then you know different logarithms you know logarithm is a multi valued function okay and the point is that the real part is the real logarithm of the modulus of the number which is non-zero since the number is non-zero and the imaginary part is the argument of that number of that complex number and the argument can be the argument is defined up to a multiple of 2 n pi therefore the imaginary part of the logarithm can be changed by 2 n pi i I mean by 2 n pi okay so any two logarithms of a number will differ by 2 n pi i we have to use that for each z in mod z minus z0 less than epsilon we have to do a little bit of thinking see you look at the function f of z times e power minus 1 by n g of z you look at this function okay see look at this function this function if I raise this to the power of n I will get 1 because you see if I raise this power n f of z will give me f of z power n and e power minus 1 g z if I raise it to the power of n I will get e power minus g z okay but e power minus g z is 1 by f power n because g z is a log of f power n alright so this is equal to 1 so this means that you know f of z into e power minus 1 by n g of z is an n it is an nth root of unity of unity okay and this nth root of unity but the point is it is an nth root of unity and this nth root of unity in principle could change if you change the z so I will call it as omega z because it depends on z seemingly okay for every z if you take f of z times e power minus 1 by n g of z its power n is equal to 1 so it is an nth root of unity so for every point z you are getting an nth root of unity call that function as w of z so w of z is that function okay but you see what is this so this so I am just calling this function as w of z so what is w of z? w of z is just f of z times e power minus 1 by n g z okay but notice here is where I will use the fact that f is continuous see I have been given that f is continuous I have to so f is continuous and e power minus 1 by n g z is also continuous so it is so the power is continuous so w becomes a continuous function w or omega of z is a continuous function so it is a continuous function from a disk and what is the image set it is the nth roots of unity that is a discrete set okay therefore the image has to be constant okay the image of disk has to be connected under a continuous function so you must get a connected subset of the set of nth roots of unity it has to be only it can be only a constant okay can be only a single term so that means that this omega of z is a constant it is one you get the same nth root of unity for all z okay you get the same nth root of unity for all z so that is where you are using the continuity of f okay since f is continuous w is continuous on mod z minus z not lesser than epsilon which is connected so w of z is equal to a constant nth root of unity okay so what you get is you get f of z easy times e power minus 1 by n g of z is equal to constant so which means tell you which tells you that f of z is equal to the constant times e power 1 by n g of z but of course the right side is analytic so f is analytic so which is analytic on mod z minus z not lesser than epsilon okay so the moral of the story is that since z not was arbitrary you get that f is analytic on d minus z f okay on d minus z of f now I will have to only vary at points of z of f points at which f becomes 0 you take a point where f is 0 that is of course an isolated point we have already seen that so it is an isolated similarity for f alright but f is continuous there therefore by Rayman's removable similarity f is analytic at those points as well therefore f is analytic on all of t okay so it is an application of Rayman's removable similarity here at each point of z f we have an isolated similarity of f but f is continuous there so by Rayman's removable similarity is theorem f is analytic thus f is analytic in t okay so that is the proof that f is analytic okay so you should see that the point is that you are bringing in you are using isolatedness of zeros of an analytic function okay you are using the existence of an analytic branch of logarithm you are using Rayman's removable similarity theorem you have question no it is what we have proved is around that point f is analytic you have proved so it becomes if around a point a function is analytic that point is automatically by definition it is a similarity it is an isolated similarity and Rayman's removable similarity theorem applies what is the singular point of a function it is a point which can be approached by points of the where the function is analytic and what is an isolated similarity it is a point where in a deleted neighborhood the function is analytic so if you take any point of z f it is an isolated point and you can find the deleted neighborhood of that point where the function is analytic because I have already we have already shown that the function is analytic outside the zeros of f so that point becomes an isolated similarity and then the question is what kind of isolated similarity is it and you know Rayman's removable similarity theorem says that if the function is has a limit at that point or is continuous at that point or is bounded in a deleted neighborhood of that point all these things are equivalent to the function being analytic at that point you can extend the function to that point you can define the redefine the function value at that point if it is not already defined and make it analytic but in our case the function value at the points of z f is 0 by our own definition okay and the point is again it is given you are again using importantly the hypothesis that the function is continuous even at points of z f that is the importantly used you are given the function is continuous everywhere so in particular the function is continuous at each point of z f and you now apply Rayman's removable similarity theorem okay that is one thing and then of course I also wanted to discuss this problem namely that the only 1 1 onto maps from the complex plane to the complex plane are of the form z going to A z plus B where A is not 0 okay so these are the only automorphisms of the complex plane okay. So let me do that also because it is an application of the idea of singularities so here is another problem the only bijective the only bijective holomorphic maps f from C to C are those of the form f of z is equal to A z plus B where A and B are complex numbers and A is not 0 and what is the solution to this well the point is that if f from T C to C is bijective then f inverse from C to C is defined and bijective okay and mind you that see f is analytic and there is the inverse function theorem which will tell you that f inverse will also be analytic because f inverse will be locally analytic okay so by the inverse function theorem f inverse is analytic okay and now we use the following thing you know you treat so the whole point is to treat infinity as an isolated singularity of f okay you treat infinity as an isolated singularity of f inverse okay so f you look at f inverse okay and look at infinity alright infinity is an isolated singularity okay and or you can also take f it really does not matter now what kind of singularity is isolated singularity is infinity it can be either removable or it can be pole or it can be essential okay if it is removable since f is entire it will by Lewis theorem f will become a constant okay so certainly f is not a constant function because it is bijective okay it is surjective so infinity is not a removable singularity the other possibility is infinity is a pole if infinity is a pole then f has to be a polynomial okay but if it has to be bijective in particular if it has to be injective it should be a polynomial of degree 1 so it is it has to be of the form A Z plus B alright and the only other possibility is that f is the infinity is an isolated essential singularity but if infinity is an isolated essential singularity then in every neighbourhood of infinity f will take every complex value except one several times in fact infinitely many times and that will contradict the injectivity of f so it cannot be an essential singularity so here you are using Picard's theorem alright so the moral of the story is that because of Picard's theorem you are forced to conclude that f is of the form f of Z is of the form A Z plus B alright and A cannot be 0 okay so this is an application of Picard's theorem that is why I wanted to mention it by look at Z equal to infinity as an isolated singularity of f or f inverse it cannot be removal by Liouville it cannot be essential by Picard so it has to be a pole hence a polynomial this must be of degree 1 by injectivity and that finishes the proof okay so I will stop here.