 So just to recall what we saw in the last lecture we have D inside the complex plane D is simply connected domain and of course D is not the whole complex plane ok. So you know there is a point A in the complex plane which is outside D ok and if you look at the functions Z-A this is analytic and never 0 on D because the point A is outside D alright. And you know if you have a non-zero analytic function on a simply connected domain then you can get an analytic branch of the log of that function ok. So you know so basically if you want you know your domain I am just drawing a picture all that I know about the domain is that it is it is simply connected and it is not the whole complex plane ok. But what I am drawing here is something that is bounded D not be like this right for example it could be the upper half plane right which is unbounded but I am drawing a diagram so that you just for some motivation. So the point A is in the complex plane minus this at D so you know if I take a point is that then I am looking at Z-A which is translation by minus A alright Z-A f of Z equal to Z-A is just translation by minus A so it is a Mobius transformation actually ok. So it translates this whole disc by minus A alright but that is not what I want to say is that Z-A has an analytic branch that is an analytic branch of a log of Z-A here and how do I do that see I take a point fix a point Z not in D alright and you know and you give me any other point Z in D alright take any path gamma alright then you define you know if you integrate along gamma I am trying to find an analytic branch of log of Z-A ok what is its derivative derivative of log of Z-A is 1 by Z-A alright so the integral of that should give me the log right therefore I integrate over gamma ok D by Z-A ok when I do this so gamma is from Z not to Z then what I will get is a logarithm of Z-A ok and you know if I I will get logs see in fact what I will get is log Z-A final point minus log Z-A initial point ok so in fact so you know this is independent of gamma this integral is independent of the path ok and that is because of simple connectedness because you know if I instead of gamma if I put another take another path gamma prime which is also inside D alright then you know since D is simply connected then gamma can be the integral over gamma followed by minus gamma prime which is a loop will be 0 for the function 1 by Z-A because 1 by Z-A will be analytic in D that is because Z-A never vanishes in D and by Cauchy's theorem this integral will be 0 ok so this is independent of gamma and this is what this is just going to be log of Z-A minus log of Z not-A this is up the the the infinite integral is log of Z-A upper limit is Z the lower limit is Z not this is what you are going to get and therefore you know if I if so if I if I take this plus this I will get an analytic branch of log of Z-A so so in other words you know integral over gamma so I will just write this integral over gamma as Z not to Z D Z by Z-A plus log of Z not-A is an analytic branch of log of Z-A in D ok is an analytic branch branch of log Z-A in D so you see I have used I have used you know I have used Cauchy's theorem I have used if you want I have used Marrera's theorem alright I have used everything here ok to say that the integral is independent of the path I need integral over a closed loop is 0 and integral over a closed loop is 0 because the integrand is 1 by Z-A is analytic Z and that is because Z-A never vanishes ok and therefore by Cauchy's theorem integral over gamma is the same as integral over gamma prime therefore this integral is independent of the path alright and then and then of course the derivative of this which follows from you know the proof of Marrera's theorem that if you differentiate this as an independent integral you will get log of Z Z-A alright and therefore this log of Z-A when I write log of Z-A it is some branch of the logarithm which is not just continuous but it is actually even analytic the analyticity comes from observing the proof of Marrera's theorem ok. So what will happen is that this expression here will be an analytic branch of the logarithm and that is the reason why I am writing log but I am not it is some branch you do not know what branch is but it is some branch right but it is analytic that is the most important thing. So I have an analytic branch of the logarithm and once I have analytic branch of the logarithm what I have is I can so let me again tell you I have used the fact that D the domain D is not the whole complex plane because I have chosen a point outside D ok if the domain D is the whole complex plane I do not have a point outside it ok I cannot do this. So I have used the fact that the domain D is smaller than the complex plane then I have used the fact that the domain D is simply connected because if a domain is simply connected then any loop if you take any loop inside the domain the region inside that loop will also be inside that domain because the domain cannot have any holes ok and I need all the region inside this loop formed by gamma and the reverse of gamma, gamma prime to be inside the domain because only then I can apply Cauchy's theorem to say that the integral over gamma is equal to the integral over gamma prime alright. Therefore the moral of the story is that I have used simply connectedness of the domain I have used the fact that the domain is not the whole complex plane I have used both I have used both alright ok so now this is so I have an analytic branch of log z-a here alright now I can define h of z to be an analytic branch of root of z-a ok. An analytic branch of root of z-a and how do I define it is very simple this is just exponential of half log z-a ok. So I already have an analytic branch of log z-a I have multiplied by half and take e to that raise it raise e to that power and I will get an analytic function and this analytic function is just root of z-a ok because you know this if you think of it this half will go to the power here and the e and log will cancel and I will get z-a to the half ok. So this is an analytic branch of the square root function ok and now I want to I want you to understand that this has some properties so first thing I want to say is that you see since it is an analytic branch of z-a you will have z h square of z will be z-a so in other words you will have z equal to h square of z plus a I will get this alright and from this I want to derive two properties of h namely the first property is that h is one to one and the second property is that the image of h the image of d under h and minus of h they are the images are disjoint ok so but before I continue let me try to say what I am trying to do I see what is a Rayman mapping theorem Rayman mapping theorem is you know give me a domain d which is simply connected which is not the whole complex plane then it is conformally equal to the unit disc that is the Rayman mapping theorem. So I have to find from the domain d I have to find Rayman map f ok I have to find Rayman map f which goes into the unit disc delta this is a set of all complex numbers with modulus less than 1 ok so I have to find I have to find a map into this ok from this domain into this I have to find a map f which is analytic which is one to one ok and whose for which you also have an inverse ok which is also analytic in other words I want to find an analytic isomorphism of this domain d into the on to the unit disc that is a Rayman mapping theorem ok that is what I want to do but what I am trying to do now in the first step is try to at least map this disc into a smaller possibly smaller domain inside the unit disc that is what I am going to do in the first step. First I will show that you can map this d into a smaller domain inside the unit disc ok and then I am going to use that and some more techniques to show that you know you can from that and using some analysis you can you can get a map which maps this not just inside the unit disc but on to the unit disc ok so you can you can use some analytic techniques to show that you can this instead of this map you can find another map for which I mean which which covers the whole unit disc ok. So the first step is you are just mapping this into a smaller domain inside the unit disc like this ok and what is that map and that map is going to be cooked up using h ok that is the importance of h. So let me so let me make these claims that I said some time ago h is 1 to 1, h is 1 to 1 on d why is it 1 to 1 on d because you know h of z 1 is equal to h of z 2 on d means well you put it here you will get z 1 is equal to h squared z 1 plus a but h z 1 is h z 2 so it is h squared z 2 plus a but that is equal to z 2 ok. So h z 1 equal to h z 2 means z 1 equal to z 2 which means it is 1 to 1 ok. So your map h is analytic in 1 to 1 ok but what you know about in 1 to 1 analytic map it is an isomorphism ok. We have seen this we have seen this inverse function theorem if you have 1 to 1 analytic mapping it is already an isomorphism on to the image ok and analytic and non-constant analytic mapping is always an open map mind you if you take a 1 to 1 analytic map then its image is open and on the image you can define the inverse you can define the inverse because it is a 1 to 1 function but the inverse function theorem will tell you that the inverse will also be analytic therefore 1 to 1 analytic map is an analytic isomorphism so holomorphic isomorphism or conformal isomorphism therefore what this will tell you is that h from d to h of d h from d to h of d will be a holomorphic isomorphism ok. So thus h from d to h of d is a holomorphic isomorphism of course the other words that are used are well in the literature instead of saying holomorphic isomorphism people say analytic isomorphism or they use conformal isomorphism or some people sometimes use the word conformal conformally equivalent ok so this is this is because of the open mapping theorem and inverse function theorem ok. Now so you know so the point is well here is my d and then I have this h so of course whether there exist an f like this is the question that is the Rayman mapping theorem but I am looking at d and here is h d this is some domain some other domain it is another domain ok and then I can also look at minus h d ok minus h d is consist of all those complex numbers whose negatives are in h d ok. So well in other in other words you know minus h d will be just reflection of h d so if I draw a diagram well I need to draw let me do this so here is my point a somewhere outside let me save some space here and here I will draw the unit disc and well so if I draw this a little bigger my situation is like this here is a complex target complex plane and well this complex plane this complex plane here is the w plane is the z plane and here is a w plane and I am using the map w is equal to h of z where h is a square analytic branch of this root of z minus a and you know well the image of d under h is going to be something well so let me draw something here so this is h d ok it is some domain in the complex plane ok mind you h d is isomorphic to d therefore it will also have the properties of d namely it will be a domain and it will be simply connected so h of d will also be a simply connected domain alright and what is minus h d it is just the reflection of h d so I will have this minus of h d alright I have the map minus h see if h is analytic then minus h is also analytic h if h is analytic then minus h is also analytic and in fact minus h will give you the other branch ok if h is one branch of root z minus a minus h will give you the other branch because you will always get two branches for the square root you will have two branches for the nth root you will have nth branches so if you take minus h it will be the other branch analytic branch of the square root and h and minus h will be mirror reflections of each other under the real axis and the claim is that h and minus h cannot intersect ok h no point of h can belong to minus h I need this fact so here is one more claim h of d intersection minus h of d is the null set there is no point in h and minus h ok where minus h means the reflection of h so minus h if you want to write it as sets minus h of d is a set of all minus of h of z where z belongs to d ok you just apply minus h instead of h and you know if h is analytic and d minus h is also analytic function defined by I mean analytic function multiplied by a constant is also analytic function after all minus h is h multiplied by minus 1 minus 1 is a constant therefore minus h is also analytic function right. Now you see what I want to say is that there is no point common to both of this ok so you know if there is a point common to both of this let me try to get a contradiction then it means that w not is h of z not and it is also equal to h of well let me use z 1 and is also equal to h of z 2 minus h of z 2 for z 1 and z 2 in d this is what it means to say that there is a point common to h of d and minus h of d so this there is so if you call that point as w not then there is a z 1 in d which is mapped by h into w not and there is a z 2 in d which is mapped by minus h into w not ok and well this cannot happen why this cannot happen is just a simple calculation because you know now I can use I can I can start I can use the same kind of calculation here I will get z 1 is equal to h square z 1 plus a and but you see h z 1 is minus h z 2 therefore h square z 1 is the same as h square z 2 so this is equal to h square z 2 plus a and that is equal to z that is equal to z 2 ok see whether you take h or minus h whether you take h or minus h h square is always z minus a because both h and minus h are two branches of the square root both h and minus h are branches of root of z minus a so h square is always z minus a ok. So I am using that here so I will get z 1 equal to z 2 ok but if z 1 is equal to z 2 then it means h of z 1 is equal to h of z 2 ok so this implies h of z 1 is equal to h of z 2 but h of z 1 is also equal to minus of h of z 2 so what this will tell you is that h of z 2 is 0 ok but what is h of z 2 after all is h of z 2 is minus w naught so minus w naught is 0 this will tell you that w naught is 0 ok but then that will mean that h of c but this cannot happen you know because see w naught cannot be 0 that is a contradiction because w naught is a value of h or minus a and also it is also a value of h and it is also a value of minus h but just use the fact that it is the value of h w naught is the value of h but h is what h is an exponential the exponential function can never take the value 0. Therefore it is a contradiction so contradiction as h is not 0 that is because h is exponential the exponential function that can never take the value 0 ok. Therefore if you assume that there is a point in the intersection of HD and minus HD you get a contradiction ok. So this implies that HD and minus HD they do not intersect their intersection is null side ok. So the diagram is the diagram is really like this HD and reflection they do not meet each other they do not intersect. And now why do I need this I need this for the following reason. See if I take a point W not let me choose a point W not in the image and choose a small disc centred at W not which is in the image. So this is a disc mod W minus W not less than or equal to epsilon I take a small disc close disc in the image ok. Then let us see what happens if mod W minus W not less than or equal to epsilon is in HD ok then you see where of course you know W not is H of Z not ok. So I mean I take this point Z not and it goes to a point W not and I take a small enough disc close disc which is in HD ok centred at W not radius epsilon ok. Then the distance of H of Z then the distance of minus H Z from W not is greater than epsilon you see all Z in D. See try to understand the statement see you take any Z in D ok then H Z is going to be lie in HD ok. But if I take minus H Z it is going to lie in minus HD ok. So you know H Z is going to lie here minus H Z is going to lie here therefore the distance of minus H Z from W not has to be certainly greater than epsilon ok. Because if the distance of minus H Z for some Z if the distance of minus H Z to W not is lesser than epsilon then that point will be an intersection of HD and minus HD which is not possible ok. Therefore you get this fact ok and this that is in other words what you will get is if you write it the distance between two complex numbers is given by taking the modulus of the difference. So it will give you that minus H Z minus W not is greater than epsilon for all Z in D. So you get modulus of H Z plus W not is greater than epsilon for all Z in D you get this ok. Now you know now you put F of Z to be epsilon by H Z plus W not put F of Z equal to epsilon by H Z plus W not. If you do this then this is analytic which is analytic on D. See this is analytic on D because you know H Z plus W not is always greater than epsilon ok therefore it never vanishes. So 1 by H Z plus sorry H Z plus W not is always greater than epsilon and epsilon is positive so modulus of H Z plus W not is greater than epsilon which is positive and does not vanish so that means H Z plus W not never vanishes. So 1 by H Z plus W not is an analytic function and I have multiplied it by an epsilon on the numerator. So this is an analytic function and what is more this is actually this is even 1 to 1 analytic function because after all what I have done is I have taken the 1 to 1 function H I have translated by W not because translation is also a bilinear transformation it is analytic and it is 1 to 1 and then I have then I have inverted it then I have multiplied by epsilon ok. So I have applied a series of Mobius transformations to H ok to get from how do I get F from H first I take H and translate by W not that is a translation ok. So I translate by W not then I invert ok I apply the transformation W going to 1 by W an inversion which is also Mobius transformation ok then I multiplied by epsilon and multiplying by complex number is also Mobius transformation. So I get H F from H by applying 3 by composing with 3 Mobius transformations therefore since H is 1 to 1 F will also be 1 to 1 so F will also be a 1 to 1 analytic map on D but the beautiful thing is now F will land inside the unit disc ok clear clearly F is analytic and 1 1 1 to 1 on the unit disc I mean on D and what is mod F set mod F set is going to be epsilon by mod of H set plus W not which is less than 1 because of this inequality this inequality tells you that epsilon by mod of H set plus W not is equal to 1 is strictly less than 1. So I will get mod F set less than 1 what does it mean it means that F takes values in the unit disc ok thus F gives a holomorphic isomorphism isomorphism from D on to F D which is contained in the unit disc delta which is a set of all sets is at mod z less than 1 if you want I can put W because there ok. So what we have done so far is we have just use the existence of square root of z minus a to derive that you can find an analytic map which goes which maps the domain D isomorphically on to a sub domain of the unit disc so it lands here this is the unit disc ok. So there exists an analytic isomorphism of the domain D on to a sub domain of the unit disc the sub domain is F D this is the unit disc delta and this contains the sub domain which is F of D and F from D to F of D is an analytic isomorphism because it is analytic and 1 to 1 alright. So this is the first step this is the first step that using which you can map any simply connected domain which is not the whole complex domain on to a smaller domain in the unit disc what we want is we want a map that will fill the whole unit disc ok that is the main thing. So the moral of the story is that somehow you have to show that if you give me a map if you give me a simply connected domain inside the unit disc ok you have to somehow show that that is also equivalent to the unit disc ok. I have to somehow show that a simply connected sub domain of the unit disc is also conformally or holomorphically equivalent to the unit disc that is what I have to do. So what we have done is we have translated the problem for an arbitrary simply connected domain which is not the whole complex plane to a problem of looking at a simply connected sub domain of the unit disc which is not the whole unit disc ok. So this step reduces to studying everything inside the unit disc ok so what it tells you is that you have to study the unit disc carefully and that is what we are going to do in the following discussion what we are going to do is we are going to study geometry on the unit disc a very special geometry called the hyperbolic geometry on the unit disc ok. So if this leads us to study the hyperbolic geometry on the unit disc so let me write that down hyperbolic geometry on the unit disc. So you know we already know we already know some facts about the unit disc so what are the couple of facts that we know about the unit disc we know one fact that we know is Schwarz's lemma ok and the other thing is the corollary of the Schwarz's lemma which says that any automorphism of the unit disc a holomorphic automorphism the unit disc that fixes the origin has to be a rotation ok. So let us recall that Schwarz's lemma so what is Schwarz's lemma f from delta to delta bar analytic and f is defined as a unit disc and takes values inside the closed unit disc f takes 0 to 0 ok then Schwarz's lemma says that the disc the length of fz cannot exceed the length of z for all z in the unit disc for all z in the unit disc this is Schwarz's lemma and it tells you that you get equality for a single non-zero member of the unit disc if and only if it is a rotation ok further equality mod fz0 is equal to mod z0 for even a single z0 not equal to 0 with mod z0 less than 1 implies f of z is a rotation a rotation and it is not only that it implies it is if and only if. So you have equality even for a single vector if the image of that vector I mean if you think of the point as a vector in the complex plane the vector joining with the origin to that point the position vector of that point ok then if the length of the vector is equal to length of its image even for one point then your analytic map has to be a rotation there is no other choice ok. So this is a Schwarz's lemma and here is a corollary and the corollary is a set of automorphisms the holomorphic automorphisms of delta so this is a set of all maps from delta to delta which are holomorphic isomorphisms they are in other words they are injective holomorphic maps they are bijective holomorphic maps from delta to delta and you know bijective holomorphic map is automatically a holomorphic isomorphism because of the inverse function theorem and the open mapping theorem therefore this is a set of automorphisms of the unit disc and what are they they are just rotations ok. So in fact the point is I come back to it soon but here we are only looking at automorphisms which fix the origin ok so let me put let me write delta, 0 this is the set of all these are all the rotations so this is just a set of all z going to e to the i alpha okay and so and that is that can be identified with the each rotation therefore is given by this angle alpha ok. So you can identify it with just the unit circle S1 which is the boundary of the unit disc every point of S1 has a unique angle alpha namely the angle joining that point to the origin made with the x axis the real axis and therefore the set of holomorphic automorphisms is just S1 and this so what is the map you send any holomorphic automorphism is of the form z going to e to the i alpha z and you send z going to e you send this map z going to e to the i alpha z to the element e to the i alpha which is on the unit circle which is the boundary of the unit disc ok that gives you a bijection and this is not just a bijection ok it is actually it is actually a group isomorphism because if you compose two rotations the angles will get added ok and therefore on the circle also the you have the set of all complex numbers of modulus 1 is also a group e power i alpha and e power i beta when you multiply you will get e power i alpha plus beta so this bijection is not just a bijection of sets it is even a bijection of groups ok so the group of what holomorphic automorphism of the unit disc which fix the origin can be identified with the unit circle as a group under multiplication ok this we have seen that this is a corollary of the short slimmer right so these are the first two facts now the whole point about hyperbolic geometry comes from what is called hyperbolic the hyperbolic metric on the on the unit disc and the hyperbolic metric that you have a nice hyperbolic metric which is preserved by all these holomorphic automorphisms of the unit disc ok now that itself comes from something that is called pic slimmer which is a kind of a nice version of short slimmer ok so so let me make one more statement here I have looked at all the automorphisms holomorphic automorphisms of the unit disc with the origin fixed ok the origin going to the origin but what about any automorphism of the unit disc what are what is a general automorphism of the unit disc so let me write that so here is a lemma so the lemma is automorphisms any holomorphic automorphism of the unit disc what is it it is of the form z going to e power i alpha into z minus z naught by 1 minus z naught bar z where of course alpha is a where alpha is an angle between 0 to 2 pi and z naught is a complex number with modulus less than 1 so here is a lemma this lemma tells you what are all the holomorphic automorphisms of the unit disc what are all the one to one analytic maps of the unit disc onto itself a general map is of this type and you get this case when you put z naught equal to 0 you take you put z naught equal to 0 this expression will become z and I will get the map z equal z going to e to the i alpha z which is just a rotation about the r more generally if you if your automorphism of the unit disc does not fix the origin it may map origin to something else then it has to look like this ok this is the statement and what is the proof of this well the proof of this is that well see if you look at if you look at the map z going to z minus z naught by 1 minus z naught bar z if you look at this map ok then this map mind you this map is a mobius transformation ok this map is a mobius transformation because you see it is of the form see it is of the form z going to a z plus b by c z plus b where a is 1 b is minus z naught c is 1 c is minus z naught bar and d is 1 and if you calculate ad minus bc you are going to get 1 minus bc is going to be z naught z naught bar so I am going to get 1 minus mod mod z naught squared and that is in fact that is that is greater than 0 because mod z naught is less than 1 since mod z naught is less than 1 mod z naught squared is also less than 1 and therefore this is a positive number so this is of the form z going to a z plus b by c z plus d with ad minus bc non-zero in fact ad minus bc is positive so it is a mobius transformation and you know a mobius transformation is a it is a bijective map of the extended complex plane to the extended complex plane it will always map you know the properties of mobius transformations namely it will map straight lines in circles onto straight lines in circles on the complex plane but even a straight line on the complex plane can be thought of as a circle on the extended complex plane therefore if you make the statement for the extended complex plane it will always map circles to circles okay and it will map the interior of the circle the interiors and exteriors it will preserve because of continuity okay. So now what you must understand is that if I if I calculate if I take the image of if I take z equal to e power i theta okay and calculate mod z minus z naught I will get mod e power i theta minus z naught but this is equal to it is the modulus of it is conjugate okay which is equal to e power minus i theta minus z naught bar alright and that is equal to if I multiply throughout by e power i theta which I can do because of because the modulus of e power i theta is always 1 what I will end up with is modulus of 1 minus z naught bar e power i theta okay right and that is just 1 minus z naught bar z mod. So what this tells you is that mod z is equal to 1 is mapped onto mod omega equal to 1 so it takes unit circle to the unit circle see if I if you call this as omega as z minus z naught by 1 minus z naught bar z if you put z equal to e to the i theta then mod omega turns out to be 1. So in other words mod z equal to 1 goes to mod omega equal to 1 so this Mobius transformation maps the unit circle to the unit circle after all it is a Mobius transformation so it has to map a circle to a line or a circle but here what it is doing this calculation actually shows that it is mapping the unit circle to the unit circle okay and what about the point 0, 0 goes to when I put z equal to if I put let me not put 0 let me put z naught where does z naught go to z naught will go to 0 if I substitute z naught I will get 0. So but where is z naught z naught is inside the unit circle okay this z naught is inside the source unit circle and 0 is inside the target unit circle therefore this will tell you that the interior of the unit circle will also go to the interior of the unit circle. See what is happening is that you have on the z plane you have the unit circle this is z plane you have the unit circle and on the omega plane also you have the unit circle and you know the unit circle on the z plane goes to the unit circle on the omega plane because of this calculation under the map z going to w or omega given by z minus z naught by 1 minus z naught per circle. Once a mobius transformation maps a circle onto a circle then there are only two choices the interior of the circle can either go to the interior of this circle or the interior of the circle will go to the exterior of that circle that is the only possibility but what I am saying is there is if you take the point z naught here which is inside the circle that is going to the origin so the point in the interior of the circle is going to a point in the interior of this circle therefore what this will tell you by property of mobius transformations is that this whole circle unit disc is going to be mapped in the unit disc. So what this will tell you is that z going to w equal to z minus z naught bar by 1 minus z naught bar z maps the unit disc onto the unit disc ok so it is a holomorphic automorphism of unit disc it is a holomorphic automorphism of unit disc and then you know if I take if I take if I multiply it by e power i alpha e power i alpha is only a rotation ok and a rotation will just map the unit disc back onto itself alright therefore if I take this map of this form it is certainly a holomorphic automorphism of the unit disc ok implies z going to e power i alpha z minus z naught bar by 1 minus z naught z minus z naught sorry this should be z minus z naught by 1 minus z naught bar z is an element of the holomorphic automorphism of the unit disc. So what I have proved is that this is contained in this I have proved that any element like this is certainly a holomorphic automorphism unit disc conversely I have to show any holomorphic automorphism unit disc is like that and for that I will use this corollary ok I will just use this corollary to show that I will continue later because I just have to compose by suitable transformation of this type and apply Schwarz lemma ok.