 So, so start with a simply connected domain D which is not the whole complex plane ok, start with a simply connected domain D which is not the whole complex plane and so choose an A which is in the complement of D ok. So, D is not the whole complex plane and it is simply connected and the Riemann mapping theorem is the statement that D is holomorphically isomorphic to the unit disc ok and in fact you can find such a holomorphic isomorphism which carries any fixed point of D to the origin and that such an isomorphism is unique provided you fix the derivative of that map at that point which is being mapped to the origin ok. So, now let me recall something that we already did you choose an point which is outside D ok let so, so f of z so let me not write f of z. So, z-A is non vanishing on D because A is not in D and it is analytic ok and if you have a lower vanishing analytic function on a simply connected domain you can find branch of analytic branch of the logarithm so has an analytic branch of log so D has an analytic branch of log of z-A using which we define an analytic branch of square root of z-A on D ok after all root of z-A is just e to the half log z-A. So, if you use analytic branch of log of z-A then e to the half log z-A is also analytic and that is a analytic branch of root z-A ok and we have seen that this function this root of analytic branch of root of z-A which I am simply writing as root of z-A is actually 1 to 1 is injective and so if you call this as h of z ok and h of D does not intersect minus h of D ok. So, these are all things that we saw few lectures ago ok the image of D under h and minus h these images will be distinct ok. So, there is no intersection between h D and minus h D and in fact and you know it is not very difficult to recall the proof you see if z1 h of z1 is equal to h of z2 will imply that you know z1 which is h of z1 squared plus a will be h of z2 squared plus a which will be z2. So, this will tell you that so this implies h is 1 1 ok and h is of course injective I mean h is of course holomorphic and you know injective holomorphic map is always a isomorphism onto its image because of the inverse mapping theorem right therefore it is injective holomorphic map so it is a holomorphic isomorphism onto its image and so this is the fact that h is 1 to 1 ok that is the injective then so h from D to h of D is a holomorphic isomorphism isomorphism by inverse mapping theorem or yeah inverse function theorem or inverse mapping theorem ok. And the other fact is that there is no there is no point in h D and minus h D and the reason is again the same kind of calculation if w0 belongs to h D and minus h D then w0 is actually h of z1 and it is also equal to minus h of z2 where z1 and z2 are in D remember that minus h is actually the other branch of the square root, h is one branch of the square root of z minus a and minus h is after all the other branch of the square root and all I am trying to say is that two branches of the square root are different ok because every number different from 0 has two distinct square roots that is all that I am saying. So you know I mean that is the that is essentially the reason for all this so you know so if you use this again you will get z1 is equal to h of z1 squared plus a and that is equal to h of z1 I mean sorry h squared of z1 is the same as h squared of z2 so it will be equal to h squared of z2 plus a which is z2 ok and this will tell you that it will give you this funny thing that you know h of z1 is equal to minus h of z1 and that will tell you that h of z1 is 0 and that is a contradiction because you see h is after all an exponential h is h of z is square root of z minus a is e to the something alright and you know the exponential function can never take the value 0 so this is not possible as h is e to the half log z minus a ok therefore this contradiction tells you that there is no point in common with h of d and h of minus d alright. So you know so you know the so that so the diagram is that you have this well I will not draw so this is t simply connected domain and I have this map h then you see if you draw so I will draw d bounded but it need not be bounded ok just because a domain is holomorphically isomorphic to the unit disc does not mean that it is bounded because you should remember that the unit disc is holomorphically isomorphic to any half plane and no half plane is bounded ok so you should not expect a simply connected domain which is I mean a domain which is holomorphically isomorphic to the unit disc do not expect it to be bounded ok that is the only mistake one makes but it will be bounded if you consider it as a domain in the extended Riemann plane ok in the extended complex plane namely on the Riemann sphere the upper half plane will also look like a disc ok so it will be bounded in the external in the extended complex plane ok but not on the complex plane right. So anyway I am drawing d to be bounded just so that you know it is easier to draw and then so you know the picture is like this so you have so you have h of d and then you know you have you have h of minus d which is just its reflection I think it is a pretty bad picture so this is minus h of d and they do not they do not intersect alright this is the situation and now what you do is you know you take a point z0 here and then you take your point w0 there ok and choose a small enough disc surrounding w0 ok. So choose rho so that you know mod w0 so that mod w-w0 less than rho is in h of d mind you d from I mean h from d to h of d is a holomorphic isomorphism because h is injective holomorphic and injective holomorphic map is a holomorphic isomorphism ok. So h is a holomorphic isomorphism of d to hd ok and I am taking a point w0 in hd which is the image of a point z0 in d under h and I am taking the small disc close disc centered at w0 which is inside hd ok then you see then what you will get is you know if I take this I mean if I take small enough disc if I take centered at w0 ok with radius epsilon then what will happen is that mod hz plus w0 will always be greater than epsilon for all z in d this happens because you see the distance of hz from w0 I mean this is actually distance of hz this is the distance of minus hhz from w0 this quantity is distance of minus of h of z from w0 because that distance is modulus of minus hz minus w0 which is the same as modulus of hz plus w0 ok and but you know minus h is disjoint from this. So if you take minus if you take any z here ok then hz will lie here so you know minus hz will lie will it be its reflection it will be in minus hd and certainly therefore the distance of minus hz from w0 will be certainly greater than this epsilon ok that is all I have written but now this is what I am one can use to produce holomorphic mapping of the domain into a subdomain of the unit disc so what you do is you put h1 of z to be equal to epsilon by hz plus w0 look at this map this is a map from d to the unit disc why because mod h1 z is mod of this and mod of this is less than 1 because of this inequality and mind you h1 is of course injective why because you see h is injective and hz plus w0 is injective inversion is a mobius transformation that is injective ok and then multiplying by epsilon is also a mobius transformation so I am getting h1 from h by applying a series of mobius transformations. So I first take h I translate by w0 that is a mobius transformation to get hz plus w0 then I apply inversion is zeta going to 1 by zeta that is also a mobius transformation ok and then I multiply by epsilon multiplying by constant is also a mobius transformation therefore I get h1 from h by applying a series of mobius transformations therefore h is injective therefore h1 is also injective so h1 is injective and holomorphic therefore again by inverse mapping theorem h1 will give an isomorphism of d on to h of d which will now be a sub domain of delta ok and it will be of course simply connected because all holomorphic isomorphisms are also homeomorphisms and homeomorphic image of simply connected space is again a simply connected space ok. So h1 is injective and holomorphic so h1 from d to h of d h1 of d inside delta is a holomorphic isomorphism on to the sub domain h1 of d of delta which is of course simply connected and you know now so we have I mean using all this somehow we are able to map this d into the into a sub domain of the unit disk ok ah sorry this should be epsilon correct yeah so first I wrote mod w-w not less than rho but then I change the rho to epsilon so this should be epsilon ok so epsilon sufficiently small. So you know finally we have managed to map d into simply into isomorphically on to holomorphically isomorphically holomorphically isomorphic on to a sub domain of the unit disk but what you want is you want a map that whose image is the whole unit disk ok plus you also want the map with the property that it takes a given point let us say z0 to the origin ok and you also want the derivative to be some fixed the derivative at z0 you want also have the map such that the derivative at z0 is some fixed positive real number ok so these are the things that we want so we want to find holomorphic map holomorphic isomorphism f from d to delta with f of z0 equal to 0 f dash of z0 is equal to lambda which is a positive real number ok you want to find holomorphic isomorphism like this ok and we have already seen that such a holomorphic isomorphism is unique ok such an f we have already seen seen that such an f is unique ok the holomorphic map of the simply connected domain which is not the whole complex plane on to unit disk can be made unique if you specify its value at one point and the derivative at that point so what we do is we fix a point if you fix a point z0 in the domain and specify that the value of the function at z0 which we usually take as 0 and you specify the derivative at that point z0 and usually you take the if you want you can take it 1 you can make the derivative 1 if you want right for that matter you could make it any real number right so if you want you can make it 1 right then such an f is unique because you know if you have 2 such f's f1 and f2 then if you compose if you have 2 f1 and if you have 2 such maps maps f which satisfy these conditions then if you compose f1 inverse and then follow up by f2 you will get an automorphism of delta ok which will take 0 to 0 ok and whose derivative whose derivative you will see that you will get an automorphism of delta and this derivative condition will tell you that it has to be just the identity automorphism and that will tell you that f1 equal to f2 ok this is something that we have already seen so such an f is unique that is something that we have already seen because you know we already know how the automorphisms of delta look like right so I have to find this f's so you know now what we do is we completely change our view point ok you do not try to get this f directly ok what you do is you look at all possible holomorphic maps into the unit disc ok and which take which are injective ok you look at all possible maps like this which map D into a sub domain of the unit disc ok and which are injective plus you also add the condition that Z0 goes to 0 ok you can so you look at that family and then essentially apply Montel's theorem to that family cleverly ok and then you will get this f as an extremal function ok. So now we change the view point so what we will do is let script f so here is the family so here is where I try to bring in something on which I can apply Montel's theorem ok so let script f be the set of all h so let me use small f I hope I am not use small f anywhere yeah so set of all small f from D to delta such that f is analytic injective and f of Z0 equal to 0 you look at you look at this family look at all mappings of the simply connected domain D into the unit disc which are injective and which take Z0 to 0 see this family is non-empty the reason is because see h I already I have already constructed a injective holomorphic map from D into the unit disc but I can adjust it so that I can make Z0 go to 0 ok so f is non-empty see if you put f let me put f0 to be you know you put f0 of Z to be some you know let me use delta times h1 of z-h1 of z0 for delta small ok see h1 so I have this h1 ok which maps D into the unit disc alright and it is injective now what you do is you define f0 to be delta times h1z-h1z0 you know why I am putting this h1z-h1z0 that is because if I plug in z equal to z0 this will become 0 so it will map z0 to 0 because that is a condition for functions in f but the other thing is I want this function also to go into the unit disc and for that it is enough to take I think it is enough to take just enough to take delta less than half probably see mod f0 so then see I do not have to take delta very small probably you see because f0 of z0 is 0 ok and f0 is of course f0 is injective analytic it is injective analytic because you know h1 is injective so h1-constant is injective and epsilon I mean some constant multiplied by an injective function is also an injective function so long as a constant is not 0 alright therefore it is an injective analytic function there is no problem about that and the only thing you have to worry about is whether it takes values in the unit disc so you know if you calculate mod f0 of z0 fz you know by triangle inequality this will be less than or equal to mod delta into modulus of disc minus this which is less than or equal to modulus of this plus modulus of this which is 2 ok because h1 takes values in the unit disc so any value of h1 has modulus less than 1 so triangle inequality will give this so if you choose delta less than half I mean less than half then you are done so this implies f0 is in your family f if delta is less than half because if delta is strictly less than half then mod f0 is strictly less than 1 which means that f0 lines inside the unit disc and that means f0 is in this family script f so therefore this family is non-empty I have a non-empty family and now comes the now you know the whole purpose of defining this family is because you know Montef or Montel's applying Montel's theorem your family should be family of analytic functions which are uniformly bounded that is the only condition you want ok the point is this is automatically uniformly bounded because you see they are all functions here into the unit disc therefore modulus of the functions always bounded by 1 ok whenever you consider a family of analytic functions taking values in a bounded domain ok it is automatically uniformly bounded so this script f is automatically uniformly bounded that is the whole point ok and you can apply Montel's theorem ok for all functions small f in script f mod f is less than or equal to 1 implies this script f is uniformly bounded ok in fact I in the in Montel's theorem I need only uniformly bounded on compact subsets I need only uniformly bounded on compact subsets but here I am getting uniformly bounded throughout ok and that is because the target domain is a bounded domain ok and so you know if I so the moral of the story is that Montel's theorem applies and you give me any sequence of functions in script f ok there will always be a subsequence which converges uniformly on compact subsets of t ok. So Montel's theorem applies and given any sequence fn in f there exists a subsequence fnk which converges uniformly on compact subsets of d I have to keep saying compact subsets of d because I do not know d is bounded d need not be bounded ok f domain that is holomorphically isomorphic the unit disc need not be bounded for example it could be the half plane which is not bounded ok. So whenever you are working with an unbounded domain all these uniform convergence etc you should only expect on compact subsets ok. So now so the question is to which sequence am I going to apply this my question is to which sequence am I going to apply it. So you know what I am going to do is I am going to take so look at the set of all derivatives at z0 and their moduli for f in small f in the family f look at this set. So I am taking all these functions these functions are all holomorphic maps that map z0 to the origin and the map d holomorphically on to a simply connected sub domain of the unit disc I am looking at all their derivatives at the origin at z0 at that fixed point z0 ok. What you must understand is that you know we have already seen during the course of the proof of Montel's theorem that you know if you look at a family if you look at a uniformly bounded family of analytic functions ok then and you take a point in the domain then you know in a sufficiently small neighbourhood of that point all or even at that point that given point the derivatives are all uniformly bounded because of Cauchy's estimate ok. So you know so note that if mod z-z0 is equal to rho less than or equal to rho is in d ok then mod f dash of z0 is going to be I am just repeating what I wrote I am just applying Cauchy's integral formula and this is less than or equal to 1 by rho where that is because mod f is always less than or equal to 1 ok. So normally when you apply this Cauchy's estimate what you will get is you will get m by you will get the bound for the derivative at the centre of the circle to be equal to m by the radius of the circle ok where m is the maximum it is a bound for the mod of the function on the circle and of course the bound on the circle will also be a bound inside the circle because of the maximum principle ok. So it will be a uniform bound but for the functions f in f in for the function small f in script f the bound is 1 so I will get this but what does this tell you this tells you that this will tell you that this set is bounded above this is for all this is for all f in script f this is true for all script f in script f oh sorry yeah this is right this is square in the denominator but you will get only a rho there yeah thank you yeah that is correct because if you look at nth derivative you should get zeta minus z0 to the n plus 1 ok. So if you look at first derivative it is zeta minus z0 to the square that is correct that is right so you get this but now so what this tells you this implies that if you take supremum of this of all this derivatives at z0 where small f is in script f this is if you call this as a this is finite because you know this is a set of non-negative real numbers it is bounded above by this 1 by rho ok and therefore it is therefore this supremum exists and it is finite alright and of course you know the point is that the same rho works for all f ok. So all the derivatives at z0 are uniformly bounded the family of derivatives of functions in f all those derivatives are uniformly bounded at z0 ok and therefore this a is a finite quantity supremum is a finite quantity I said that I mean so this is standard property of the subsets of the real line you know that if you have a subset of the real line which is bounded above then it has a supremum ok and that is equivalent to actually the completeness of the real line ok. So this a is finite alright and now you also notice that so one of the things that I want to tell you is that see this you see this a is so the thing I want to say is that this a is positive ok I want to say that a is positive ok that is one fact this a is a positive number that is something that I want to say and that is because you see so the claim I want to say is that a is positive and I think that is more or less clear because you know if a is 0 then the supremum these are all non-negative numbers the supremum is 0 then each is 0 ok then all these derivatives at z0 have to vanish right because each of these numbers is less than or equal to supremum is supposed to be an upper bound it is a least upper bound. So each mod f dash of z0 is greater than or equal to 0 and it is less than or equal to a so if a is 0 then all the derivatives at z0 vanish and that is not possible that is because you know all the see all the f's I am considering they are all injective ok for an injective you know an injective analytic function is an isomorphism onto its image. So derivative cannot vanish ok derivative cannot vanish because there is an inverse function so they just the injectivity of all these f's will force that you know a cannot be 0 a has to be positive ok since f each since any f in script f is 1 1 ok. So a is a positive number a cannot be 0 alright so a is some positive number and now you know I am going let me go back to your good old real analysis where you know the supremum has a has a approximating property. So the supremum of a set of numbers is can be obtained as the accumulation point of that set of numbers ok. In other words what is the property of supremum it is the supremum is called the least upper bound so it is an upper bound and any number less than that is not an upper bound ok. So which means that you take any number less than that you can find a member of this set which exceeds that and that number less than that you take could be epsilon less than that and in this way you can cook up a sequence of elements from this set which tends to a ok and it is that sequence of f that to which I am going to apply Montel's theorem ok. So we can we can find f n such that f n inside script f such that f n dash of z0 more in modulus density you can do this because of the approximation property of the supremum that is because the supremum is a is an accumulation point ok and I can therefore get a sequence of elements here which tends to the supremum and that sequence corresponds will give rise to a sequence that will correspond to a sequence of functions applied to the point z0 whose derivatives are applied to the point z0 and that is the sequence I want that is the sequence to which I am going to apply Montel's theorem ok. Now apply Montel's theorem by Montel's theorem f n converges to a function f ok my claim is that this is my function ok probably maybe I may have to modify it so let me call it as f0 ok. So by Montel's theorem so what is Montel's theorem it tells you that whenever you have family of analytic functions which is which is uniformly bounded on compact subsets then given any sequence in the in this collection there is always a subsequence which will converge uniformly on compact subsets so it means there is a I will find a subsequence fnk of fn which converges uniformly on compact subsets so basically it converges to a function that function will be analytic and the analyticity is because the convergence is uniform on compact subsets ok. So you know that if you have a uniform limit of analytic functions it is analytic ok and more generally if you do not have uniform limit on the whole domain if you have uniform limit only on compact subsets it is enough because that allows you to check within a close disc which is compact ok. So uniform limit of a normal limit of analytic functions is analytic so this f0 is analytic clearly you see clearly you can see that f0 mod f0 is less than or equal to 1 because each of the functions here are into the unit disc the limit function will also be into the unit disc ok and the derivative of f0 at z0 will be this a ok because you know under normal convergence not only does the sequence of functions converge to a given function which is analytic the derivatives will converge to the derivative of the limit function and this will happen for any nth derivative ok because normal convergence is a very powerful thing right therefore what will happen is this converges to this so its derivative will converge to its derivative but its derivative at z0 converges to a so its derivative at z0 is a ok. So you know what we have so more or less what we have done is we have found a function which is mapping into the unit disc because mod f0 is less than or equal to 1 but mind you if it is less than or equal to 1 it is not exactly into the unit disc it is into the close unit disc ok it is a function which maps into the close unit disc and so fine so I have this function which maps the domain D into delta bar ok because the see each for each each function you know in fact each function maps into the unit disc ok in fact the strict inequality here alright and therefore here also I will have strict inequality alright but here I cannot claim the limit I cannot probably claim strict inequality alright but more or less I will get strict inequality simply because of the open mapping theorem ok anyway but let it be as it is let me not worry about it the first thing I want to fix is that you know I want to say this f0 is I want to say this first of all I want to say this f0 is a non-constant function it is a non-constant analytic function and I want to say f0 is also injective now why is f0 non-constant because its derivative is at z0 is non-zero a is positive so this implies that f0 is non-constant ok and now comes another beautiful thing see the fact that f0 is injective comes from Hurwitz's theorem ok see Hurwitz's theorem there is one version of Hurwitz's theorem which says that you know if you have a sequence of analytic functions which are you which are univalent which are injective if that sequence converges normally ok then the limit function is either constant or it is again injective ok it the original Hurwitz's theorem the standard version of the Hurwitz's theorem will say that you know if you have a sequence of analytic functions converging normally to a given to a certain function then that if you take a 0 of that limit function of multiplicity m then you know the zeros for beyond a certain stage all the sequence of in the sequence of analytic functions that you are considering beyond a certain stage all the functions in your sequence will have zeros of the total number of zeros in a small neighborhood of the zero of the limit function of multiplicity m will be again multiplicity m that is what the Hurwitz's theorem will say it will say that if the limit function has zero of it has a zero of order m at z0 then beyond a certain stage all the functions in the sequence will also have m zeros in a neighborhood of z0 sufficiently small neighborhood of z0 and all these zeros will accumulate at the zero of the limit function and that is the original version of the Hurwitz's theorem but another version of the Hurwitz's theorem is if you apply this original version carefully you get the other version of Hurwitz's theorem which says that if you have a sequence of univalent that is injective analytic functions okay if it converges normally to a limit function then either the limit function is constant and if it is not constant the limit function continues to be injected. I mean the injectivity is just the fact that it is just the fact that it takes every value once so the multiplicity is one so basically Hurwitz's theorem says that multiplicity is will be preserved okay the multiplicity will be preserved so if all the functions in the sequence are injective their multiplicities are all one okay therefore the limit function will also have multiplicity one if it is not constant okay that is Hurwitz's theorem. So if you apply it here you see f this f0 is actually uniform limit it is a normal limit of these functions and these functions are all injective okay that and the limit function is non constant therefore by Hurwitz's theorem f0 is injective okay so by Hurwitz's theorem f0 is injective okay now you know we are we are more or less done we are more or less done okay and you see the there is only one more claim left and the claim is the following that the image of f0 is a whole unit disc okay so this is the final claim see you know f0 is a non constant analytic function and it is injective so it is a holomorphic isomorphism so it is a holomorphic isomorphism of D into the closed unit disc alright but the image has to be open because image of a holomorphic map is open so it is an open set inside the closed unit disc okay and the claim is that open set is the open unit disc in other words f0 achieves the job of mapping the given simply connected domain which is not the whole complex plane on to the whole unit disc and proves Rayman mapping theorem okay and then of course the derivative at z0 if I want it to be 1 I have it as a so if you take instead of f0 if you take 1 by a times f0 that will have derivative 1 so you can always adjust the derivative to be 1 at z0 that is not a big deal. The big deal is to show that this f0 fills out the image fills out the whole unit disc open unit disc okay and here is where the big deal comes the big deal comes from hyperbolic geometry on the unit disc that is what helps us to ensure that this f0 has to fill out the whole unit disc okay and that is because this f0 is an extremal function it is extremal because you know it achieves this limit so whenever you have family of functions and you know if you look at you look at some numbers defined based on that family okay and you take a limit okay then if you are able to find a function in that family which achieves this number it is called an extremal function okay. So I am able to find a member of f script f okay namely f0 f0 is a member of script f such that the derivative at z0 is exactly this extremal value this limiting value which is a supremum mind you a supremum is also a limit okay so and the point is because of this extremal property of f0 it has to fill the whole unit disc that is the and that is where hyperbolic geometry comes so hyperbolic geometry comes in to actually tell you that the image of f0 is exactly the unit disc so let me write that f0 hence use Riemann mapping theorem the claim uses rather the proof of the claim uses hyperbolic geometry. So let me stop here.