 okay. So this is lecture number 35 it is a continuation of the previous lecture uhhh. So what we what we were discussing was the uhhh the proof of the following uhhh statement that the automorphisms the holomorphic automorphisms of the unit disc. So delta is unit disc and uhhh we were trying to show that uhhh set of holomorphic automorphisms of the unit disc uhhh are exactly maps of the form. They are Mobius transformations they are given by maps of the form z going to e to the i alpha uhhh z-z0 by 1-z0 bar z where alpha is of course uhhh an angle uhhh and uhhh and z0 is an element in the unit disc okay and what we uhhh what we saw in the towards the end of the last lecture was that uhhh any uhhh we saw uhhh uhhh map of the form z going to e to the i alpha into z-z0 by 1-z0 bar z is uhhh is uhhh indeed a Mobius transformation. So this Mobius is this o with 2 dots on top of it it is it is called as oum lot and it is supposed to be uhhh the German symbol for oe put together okay. So this is m oe oe uhhh half o and half e okay. So this is Mobius transformation it is of course also called as you know bilinear transformation or linear fractional transformation okay uhhh so it is indeed a Mobius transformation that uhhh maps uhhh the unit disc on to uhhh the unit disc okay. So so the right hand side is of course contained in the left hand side. So an element here is here what I will have to show is a conversely any holomorphic automorphism of the unit disc is of that form okay. So what I am going to do is uhhh so I will conversely let f be a holomorphic automorphism of unit disc. So you know what is happening is see you have you have f here uhhh uhhh all all I know about f is that it is an isomorphism it is holomorphic its inverse is also holomorphic and it goes from the unit disc uhhh to the unit disc and you know uhhh and I have to show that uhhh f is of this form okay. So I have to find an alpha uhhh and uhhh z not such that uhhh f of z is actually e to the alpha into z minus z not by 1 minus z not bar z right. Now what you must understand is that see if f takes 0 to 0 then we are already done because if f takes 0 to 0 then it will be an automorphism of the unit disc which fixes origin and we already seen as a corollary to Schwartz's lemma in earlier lectures that it has to be a rotation. So uhhh you so f will be of the form e power i alpha times z okay which is just this map with z not equal to 0 okay. Therefore uhhh the problem is when 0 does not go to the origin right that is that is the issue. So what we do is uhhh uhhh you know we just take uhhh see we we uhhh well let us look at what 0 goes to alright. So for for so for that matter you know I I can fix any point z not okay and uhhh uhhh suppose z not goes to the point w not. So my map is w equal to f of z I take a point z not it goes to the point w not alright. Now what I am going to do is I am going to write on this side a holomorphic automorphism of the unit disc which takes uhhh 0 to z not okay I am going to do that and you know what I am going to do I am going to use the first statement I am going to use this statement. So what I am going to do is it is it is very very simple you know if I take this map to be z going to uhhh uhhh z minus z not by 1 minus z not bar z suppose I take this map okay. Then I already told you that this map is a mobius transformation that maps the uhhh interior of the unit disc to the unit disc okay and it will take z not to 0. So the point z not will uhhh go to the go to the origin and what I am going to do is this is this map this is going to be the inverse of that map okay. So you know so I am going to call this as g okay and and this is my g inverse and I know g inverse uhhh uhhh is is is a map that takes uhhh the unit disc to the unit disc and it will map z not to 0 because you know in this if you substitute z equal to z not you will get 0 and this is a mobius transformation and I am going to take its inverse. So the inverse map will take 0 to z not okay and then this f will take z not to w not okay because w not is f of z not right and then uhhh you know I am going to then again apply another map here that will move w not to 0 alright and what is that map you know what that map is I will call that is h. So and this map is going to be so I am going to just map w w not to 0 and you know what that map is it is just w going to same formula w minus w not by 1 minus w not bar w okay this map will map w not to 0 and it will be an automorphism unit disc okay. So both for this side and for this side I am using this this fact that I have already proved in the last lecture right and now now watch if I if I take uhhh so this is h okay if I take the composition what will I get see this is also an automorphism unit disc f is also an automorphism unit disc and h is also an automorphism unit disc therefore if you compose you will get the composition of all the three will give an automorphism unit disc unit disc and it will fix the origin because it will take 0 to 0 that is the whole idea of putting this g on this side and h on that side. So it will be an automorphism unit disc which fixes the origin which by an earlier corollary of the Schwarz lemma we have seen is a rotation. So what you will get is you will get that uhhh should take uhhh first apply g then apply f then apply h is uhhh an automorphism a uhhh a holomorphic automorphism of the unit disc which fixes the origin okay. But you know any automorphism holomorphic automorphism the unit disc which fixes the origin is rotation that is the first corollary one of the corollaries of the that is the corollary of the Schwarz lemma that we have seen okay. So uhhh by a corollary to the Schwarz lemma what you will get is that h circle f circle g of uhhh uhhh w is just going to be a rotation and therefore it is of the firm e power i alpha w okay. Since you know this is something that we have already seen the set of automorphisms holomorphic automorphisms of the unit disc which preserve the origin. So this is this is a subset of this set these are all holomorphic automorphisms bijective holomorphic maps of the unit disc onto itself and these are the subset of maps which which send 0 to 0 and this is this is the same as this is identified with uhhh this is equal to rotations the set of all z going to e to the i alpha z where alpha is varying from 0 to up to 2 pi okay and this is identified with you know S1 which is uhhh the boundary of the unit disc we have already seen this and this is what I am using here okay. So so h circle f circle g is an element here therefore it is a rotation. So it is just the variable multiplied by uhhh an e power i alpha for suitable for some alpha okay for some uhhh for some alpha 0 less than or equal to alpha less than 2 pi of course alpha is just an angle and you have to read it mod 2 pi okay. So I think I will have to worry about my variables here. So here this is the z plane this is z plane and this is omega plane and uhhh let me call this as eta plane. So this is eta so eta is h of w okay and uhhh w is f z and here let me use zeta and uhhh zeta yeah z is g of zeta. So this is zeta uhhh zeta goes to z under z goes to zeta is g inverse and zeta goes to z is g okay. So so here you have variables so if I if I use these variables then I should let me change this variable to eta okay because uhhh uhhh not eta it should be zeta. So you know uhhh uhhh you will get uhhh uhhh so g zeta g of zeta is z okay. So I will get h of uhhh f of z see h of f of g of zeta is h of f of g zeta but g zeta is z so I will get f of z. So I will get h of f of z is equal to uhhh e power i alpha into zeta and what is the formula for zeta? The formula for zeta is z minus z naught by 1 minus z naught bar uhhh z okay. So I will get so f of z will be h inverse of this okay and uhhh uhhh and what is what is the formula for h inverse? It is going to be uhhh it is going to be the map in this direction uhhh. So it will be h inverse of eta will be what uhhh uhhh you have to solve this for w omega okay. So you have to solve this for omega and what will you get? Well if I make that calculation somewhere uhhh let me let me make it somewhere here say I will get uhhh neta is equal to w minus w naught by 1 minus w naught bar w you cross multiply you get neta minus w naught bar w neta will w minus w naught. So I am solving for uhhh uhhh so it is neta going to w I want write w in terms of neta that is h inverse h inverse of neta is w once I have a formula for h inverse neta I will put I will apply it to this okay. So uhhh so I have to calculate uhhh w from this or omega from this. So I will get omega into uhhh so I will get neta plus w naught is equal to omega plus uhhh w plus w naught bar w neta which is w into 1 plus w naught bar neta. So you will get w equal to neta plus w naught by 1 plus w naught bar neta this is the formula for w. So this is neta plus w naught by 1 plus w naught bar w of w naught bar neta. So this is the formula for h inverse of neta right and apply it to this apply it to this and you will get again an expression of this form okay you uhhh I mean if you are you must have probably done this such exercises when you have uhhh taken a first course in complex analysis but it is very easy to do it even if you have not done it. So I will I just have to apply so this is my neta okay. So I substitute it in here I will get e to the i alpha into z minus z naught uhhh by 1 minus z naught bar z uhhh plus w naught by 1 plus w naught bar into neta which is this e to the i alpha into z minus z naught by 1 minus z naught bar z this is what I get okay. In in this whole calculation I have chosen z naught I have chosen some z naught in the unit disc and I have taken w naught to be the image of z naught okay but you know I can make special choices for example I can take w naught equal to 0 if you want I I can reduce the calculation by I can make the calculation simple by simply taking w naught equal to 0. W naught equal to 0 means I am taking z naught so I am taking the unique point z naught that goes to 0 under f okay. So you make that assumption uhhh choose choose the unique z naught in delta with f of z naught is equal to w naught is equal to 0 this has to happen because after all f is a f is given to be a holomorphic automorphism of delta therefore uhhh there is something is mapped to 0 okay whatever is point is mapped to 0 you call that as z naught okay. So if you do this then you are actually putting w naught equal to 0 and if you put w naught equal to 0 you get f of z in the form that you want okay. So w naught equal to 0 gives f of z to be simply you put uhhh w naught equal to 0 you see this is gone this is gone you will simply get e power i alpha z minus z naught by 1 minus z naught bar z okay immediately and that is a proof that is the end of the proof. So what you have done is you started with an element here f here and showed that it is of this form okay. So these two are one and the same okay and uhhh so in fact there is uhhh uhhh uhhh so in fact the truth is that this is also a group the holomorphic automorphisms of the unit disc that is also a group because this is subgroup this is also uhhh certain subgroup of mobius transformations mobius transformations mind you form a group because you can compose mobius the the the group law the group multiplication is just composition okay. So under composition mobius transformations form a group and this is a subgroup of mobius transformations okay these are the uhhh these are certainly uhhh I mean these are mobius transformations of this particular form and they are all they are exactly the automorphism unit disc okay and it is a subgroup under composition of mappings alright and there is a much smaller subgroup among these you have the smaller subgroup which are further those which fix the origin and they are the rotations okay these are not rotations there is this part is a rotation there is something more here okay. So the fact is just like the this subgroup of mobius transformations that fix the origin or the rotations it it can be identified with the boundary okay that also can be identified in a nice way okay it can be identified with delta it can be identified with delta cross the boundary okay. So you see what will happen is that if you take the the holomorphic automorphisms of delta uhhh that is that I have written down a set of all maps of the form z going to e to the i alpha into z minus z not by 1 minus z not bar z where alpha is an angle trade mod 2 pi and uhhh z not is a point with uhhh point in the unit disc okay. So from here I can uhhh you know I can give you a map into delta cross dole delta which is delta cross s1 s1 is the unit circle okay set of all complex numbers with model s1. So it is numbers of the form e power i alpha okay and you know what is this map this map is you know you send the map you send the mobius transformation z going to e to the i alpha z minus z not by 1 minus z not bar z you simply send this to a pair where this first member is this z not because you see z not is an element of the unit disc and the second member is that e power i alpha and because e power i alpha will be on the boundary of the unit disc which is unit circle okay and what will happen is that this will be also an isomorphism of groups this will be an isomorphism of groups alright. So and what will happen is that this uhhh this will contain uhhh properly the automorphic holomorphisms I mean holomorphic automorphisms of uhhh uhhh of the unit disc which preserve the origin okay which are only the rotations and only rotations means that it will correspond to z not equal to 0. So only rotations okay and what you will get is the same map will give you this isomorphism it will you are going to send z going to e to the i alpha z to simply s1 you are going to send this to the element e e power i alpha of s1 okay and this is identified as uhhh uhhh 0 cross s1 which is sitting inside s1 okay. So the the holomorphic automorphism unit disc are identified with delta cross s1 okay whereas the subgroup of holomorphic automorphisms which fix the origin are simply identified with s1. So s1 so you have an identification of uhhh uhhh as a group uhhh of both the automorphisms of delta and the automorphisms of delta which fix the origin alright. So this is a remark now uhhh of course this diagram commutes okay and these vertical arrows are both group homomorphisms okay they are not just bijective maps they are group homomorphisms right. So uhhh now you see we come to uhhh uhhh uhhh uhhh so you know see what we have just looked at are the automorphisms of the unit disc okay the question is what can you say about a general analytic map of the unit disc onto itself what are what are the most general statements you can make. So I mean what can you say about uhhh maps which are not automorphisms of the unit disc but which are maps analytic maps of the unit disc into itself okay what can you say about such maps. So I will tell you what the main result is the main result is that on the unit disc okay there is a there is a metric called the hyperbolic metric okay it is a way uhhh of defining the distance between two points in the unit disc of course you know metric is something that gives you the distance between two points in a space okay. So on the unit disc there is a special metric it is called the hyperbolic metric and we will we will see what that metric is later but the point is what uhhh the Schwarz lemma actually says is that if you take an analytic function from the unit disc to itself which is not an automorphism then with respect to this hyperbolic metric it is a contraction mapping okay. So this is the this is the this is the philosophical statement I mean this is the ideological statement the ideological statement is I mean this is the statement that we need this is this statement is needed for proceeding with the proof of the Riemann mapping theorem okay you see last in the last lecture what we did was we proved that if you give me a simply connected domain which is not the whole complex plane then I can conformally map it to a subdomain of the unit disc that is what we proved okay but then somehow you have to a alter the map so that the image fills out the whole unit disc because finally the Riemann mapping theorem requires I mean the proof of the theorem requires you to find an isomorphism holomorphic isomorphism of the given simply connected domain which is not the whole complex plane with the whole unit disc okay. So that is the reason we have to study the hyperbolic geometry on the unit disc right and the point is that the Schwarz lemma actually in a in a disguised avatar of itself actually tells you that with respect to this hyperbolic metric on the unit disc if you take an arbitrary analytic function which is not an isomorphism of the unit disc onto itself then it has to only contract so the moral of the story is if either it is an isomorphism of the unit disc onto itself okay if it is not an isomorphism unit disc onto itself it will be a contraction mapping which means you take two points okay and you take their images the distance between the images will be smaller in the hyperbolic metric than the distance between the original points that is what a contraction mapping is a contraction mapping between two metric spaces is a map that reduces the distance okay. The distance between two source points is greater than the distance between the image points that is what a contraction mapping is and what the statement of the result that we need is that if you take an analytic map of the unit disc into itself which is not an isomorphism which is not a member here then it is certainly a contraction map for the hyperbolic metric alright so for that I will have to introduce the hyperbolic metric okay and then of course there is also the question that what will happen to these maps what will happen to automorphic I mean holomorphic automorphisms of the unit disc under the hyperbolic metric the answer is that they will preserve the metric they will be isometrics okay a map between two metric spaces is called an isometry if it preserves distances that is you take two points in the source metric space and you take their images under this map in the distance between the images is the same as the distance between the source points if the distance is preserved the mapping is called isometric okay. So what happens is whenever you take a map from the unit disc to itself which is analytic either it is an isomorphism and it preserves the hyperbolic metric or it is a contraction so it will shrink the image of the domain okay so this is a geometric fact in hyperbolic geometry which we need okay and that is what I am trying to explain alright so you know everything comes from Schwarz lemma you know so the fact is that Schwarz lemma gives you a description of the holomorphic automorphisms of the unit disc which fix the origin and also the general holomorphic automorphism of the unit disc not only that it also gives you information about any other analytic map of the unit disc onto itself okay so the first thing is let us recall the Schwarz lemma so here is the Schwarz lemma the Schwarz lemma is well uhhh uhhh f is defined from the unit disc and taking values in the closed unit disc uhhh analytic f takes 0 to 0 okay then mod f z is less than equal to mod z for all z in delta this is the Schwarz lemma which already tells you that you know any analytic mapping of the unit disc is is a contraction okay in the sense that if you if you measure the distance from the origin then it is a contraction okay the the mod f z is just the distance from the origin of the image f z of the point z under f and the distance from the origin of the image cannot exceed the distance from the origin of the source point that is what it says. So the fact that there is a contraction uhhh is already there in Schwarz lemma okay but what we are going to do is we are going to uhhh improve upon this okay and of course what is the other statement in Schwarz lemma if you get equality here for even a single z not then it is a rotation okay equality occurs for some z not in delta z not not equal to 0 if and only if f is a rotation okay. So this is the uhhh this is the Schwarz lemma from this version of the Schwarz lemma I am going to I am going to get another version of the Schwarz lemma which is called the differential version of the Schwarz lemma okay. So let me write this uhhh differential version of the Schwarz lemma so what is the differential version of the Schwarz lemma uhhh you see again differential version is that uhhh you have uhhh the the result in terms of derivative at the origin okay. So again you take f uhhh defined of the unit disc and taking values from the closed unit disc analytic and f of 0 is equal to 0 that is it takes the origin to the origin okay. Then the differential version says you take the derivative at the origin of f that cannot exceed 1 the derivative at the origin of any analytic map of the unit disc into the closed unit disc the derivative cannot exceed 1 okay and again equality occurs if and only if it is a rotation if and only if f is a rotation okay. So this is the differential version of uhhh Schwarz lemma or infinitesimal version of Schwarz lemma because it involves the original statement in uhhh the original Schwarz lemma involves uhhh f okay whereas the differential version will involve its derivative it gives you a bound for its derivative. The derivative of uhhh any analytic map from the unit disc to the unit disc at the origin cannot exceed 1 and it will be 1 only if it is a rotation if it is not a rotation then it will be strictly less than 1 that is what it says okay. Now uhhh so what is the proof? Proof of the differential version you can get the proof of the differential version from the original version okay and uhhh what is the proof for that? So you know you see uhhh see you we have uhhh mod f of z is less than or equal to mod z this is already there by the Schwarz lemma alright. So what you do is for z not equal to 0 you divide and you take a limit and you will get this because after all limit z tends to 0 f of z by z is just f dash because f of 0 equal to 0 okay. So you so you get it so you know uhhh so so so this implies so mod modulus of f dash of 0 is modulus of limit z tends to 0 uhhh uhhh f of z minus f of 0 by z minus 0 which is modulus of limit z tends to 0 f of z by z but you know if the limit of some expression exists okay then uhhh modulus is a continuous function so I can remove the limit outside okay. So I can write this as limit z tends to 0 mod f z by z but mod f z by z uhhh of course when I say limit z tends to 0 I am assuming z is not equal to 0 okay. So if z is not equal to 0 mod f z by z is less than or equal to 1 and this is this inequality is because of the original Schwarz lemma okay. Therefore so this so this is this is the quantity less than or equal to 1 and you are taking the limit so this is also going to be less than or equal to 1 okay. So you get this this is a differential form of infinitesimal version of Schwarz lemma okay then we have to say one more thing we we we must say that the derivative is 1 if and only if it is a rotation okay and why is that true uhhh if uhhh the derivative is equal to 1 okay then uhhh the function the analytic function g of z is equal to f z by z okay. So uhhh if you remember the uhhh in the last the last lecture uhhh or the lecture before that where we proved Schwarz lemma what we did was we defined this function g of z to be f z by z and applied the maximum principle to it okay because it was analytic it was not defined at z equal to 0 but then at z equal to 0 it had only a removable singularity okay which you can see because the z will cancel off if you write out a Taylor expansion of f at 0 namely if you write a McLauren expansion of f at 0 the first uhhh constant term will be 0 that is because f of 0 equal to 0 alright and the z will cancel out and therefore g of z will be represented by a convergent power series at the origin therefore it is also analytic at the origin therefore this is actually though I am writing f z by z it extends to an analytic function at z equal to 0 it is just like sine z by z if I take limit z tends to 0 I will just get g of 0 is equal to f dash of uhhh uhhh 0 okay and uhhh what this will tell you uhhh so so modulus of g dash of modulus of g of 0 will be uhhh will be 1 if mod f dash of 0 is 1 then mod g 0 will be 1 okay and you see uhhh now go back to the uhhh again to the proof of the Schwartz lemma in the Schwartz lemma what we did was we applied the maximum principle to g which is f by f of z by z and showed that uhhh uhhh the maximum value of uhhh uhhh the maximum value of mod g is always less than or equal to 1 I mean actually we proved mod g is less than or equal to 1 which translated to mod f z less than or equal to z which is the conclusion of the Schwartz lemma okay. So and what is the maximum principle what does it say if the maximum modulus if the maximum value is uhhh is it occurs at an interior point then the function is a constant okay so by the maximum principle for g principle for g g is a constant on the unit disc okay you you will you will get g is a constant for g is a constant on the unit disc and uhhh uhhh and that constant is uhhh and and uhhh mod g will be mod of that constant on delta which will be which will be mod g0 which will be 1. So this implies that this constant has uhhh modulus 1 okay so mod g is mod of that constant and mod g is mod of that constant it has to be equal to mod g0 but mod g0 is 1 so modulus of that constant is 1 okay it is a complex number with constant modulus 1 therefore it is of the form e power i alpha. So g of z is of the form e power i alpha it is a constant function but what is g of z g of z is f of z by z. So this will tell you that f of z is a rotation so in other words you know you have proved the differential form of Schwartz lemma the derivative the derivative at the origin of an analytic map of the unit disc on to itself cannot exceed in modulus 1 okay its modulus cannot exceed 1 and it is equal to 1 if and only if it is an automorphism and that automorphism has to be a rotation. So if it is not a rotation the derivative of the origin is strictly less than 1 that is the infinitesimal version of the Schwartz lemma okay so that finishes uhhh the proof of this. So this is the differential version of the Schwartz lemma and now uhhh we come to uhhh you know uhhh this question of uhhh trying to generalize this differential version of the Schwartz lemma. So you can ask what will happen if I put instead of 0 is z here okay. So take an analytic function from the unit disc to the unit disc okay I know f dash 0 in modulus is less than or equal to 1 but my question is what will happen if you take f dash of z for z an arbitrary point on the unit disc not necessarily the point 0 okay. Then also you get a bound and that is uhhh grander version of the Schwartz lemma and that is called Picks lemma okay and which is a which is a fundamental lemma that is required for hyperbolic geometry okay. So I will I will just state uhhh Picks lemma uhhh somewhere I think uhhh so here is Picks lemma. So again same assumptions f from delta to delta bar analytic uhhh f takes 0 to 0 okay uhhh no I do not think I need f takes 0 to 0 I do not need that this is very general f is any analytic map f is any analytic map from the unit disc taking values in the closed unit disc okay and uhhh then you see mod f dash of z okay is less than or equal to uhhh 1 minus uhhh mod f z squared uhhh by 1 minus mod z squared for all z in delta. You see if you uhhh in this you know if I take in particular an analytic function which takes 0 to 0 okay then if I substitute z equal to 0 what I will get is mod f dash of 0 is less than or equal to 1 minus 0 by 1 minus 0 because I am putting z equal to 0 and f of 0 is also 0 I must if I assume 0 goes to 0 then I will get mod f dash of 0 is less than or equal to 1 which is the differential form of the uhhh Schwartz lemma. So this is a this Picks lemma is a generalization of the Schwartz lemma okay the only point in Schwartz lemma is that uhhh you are looking at the derivative at the origin the modulus of the derivative at the origin and you are also assuming that the map takes 0 to 0. Now you are relaxing both the things you are taking any analytic map of the unit disc into itself and you are saying that the derivative modulus of the derivative at any point has a bound and this is the bound okay and again just like that case you will get equality if and only if f is a automorphism okay otherwise you will have strict inequality right. So uhhh further equality occurs uhhh for a point of uhhh delta if and only if f is a holomorphic automorphism of delta so this is Picks lemma okay. So you you see uhhh uhhh you you can see the flavour of our arguments we are more and more worried about the mappings of the unit disc into itself. The Schwartz lemma itself is a statement about the mapping of the unit disc into itself. So we are worried about the mappings of the unit disc into itself we are worried about which of these mappings are automorphisms and we are worried about properties of such mappings okay and this con and this focusing on the unit disc is actually uhhh leading us to studying hyperbolic geometry on the unit disc which is uhhh essentially given by defining the so called hyperbolic metric on the unit disc okay. So so this is Picks lemma uhhh and what do you uhhh what do you do to prove Picks lemma uhhh the nice I mean I it is a pity I erase this diagram here I have to use the same diagram to prove Picks lemma but I will have to finally apply the infinitesimal or differential version of Schwartz's lemma. So let me remove I I do not want it to take values on the unit circle alright. So let me remove this let me just assume that I am taking the map is from the unit disc to the unit disc okay. Let me not think the let me not take make the assumption that f takes a boundary value alright. So map from the unit disc into the unit disc right. So uhhh so you know the diagram is the same so I have I have this uhhh unit disc I have the map f and uhhh it goes from the unit disc to the unit disc okay and this is w equal to f z this is z plane and this is the w plane okay and well if you take a point z0 it will go to a point w0 okay and I want this w0 to lie inside the unit disc that is the reason why I did not take f to be a mapping from the unit disc to its closure okay. So uhhh so I want w0 to lie inside the unit disc right and uhhh well if you take any point in the unit disc it is going to go to a point inside the unit disc because I have I have I have removed the bar there okay and what I am going to do is as I did before I am going to put here this map which uhhh which is an automorphism unit disc which takes z0 to 0 okay and you know what is that map. So you know if you remember I call this the zeta plane earlier in this lecture and then I I I wrote a map like this I wrote a map like this which is g which is an isomorphism it is an automorphism unit disc which takes 0 to z0 and its inverse is an automorphism unit disc which takes z0 to 0 and you know what that map is it is simply z going to z-z0 by 1-z0 bar z this is the map and this is this is this is our zeta okay and this is g inverse okay and I am taking g to be this map so that it takes 0 to z0 because g inverse takes z0 to 0 g will take 0 to z0 alright and I have f here mind you f is not f is not an automorphism okay f is not an isomorphism it is not even given to be 1 to 1 it is only given to be analytic nothing more is given about f okay and then here what I do is I take this map from here a neta is equal to h of w this goes lands in the neta plane and I will take an automorphism unit disc that maps w0 to 0 okay and you know what that map is what is that map it is w going to w-w0 by 1-w0 bar w which is your neta okay. So, I have this set of maps alright this is an isomorphism the last one is an isomorphism the central one is just an analytic map the central one is analytic I do not know it is 1 to 1 I do not know it is 1 to I do not know anything about it now what I do is I compose if I compose what I get is g followed by f followed by h if I compose I get h circle f circle g it is a map from the unit disc to the unit disc okay I cannot again say anything about this map except that I can only say 0 goes to 0 okay because 0 goes to z0 z0 goes to w0 w0 goes to 0 so 0 goes to 0 but that is good enough for me to apply the Schwarz's lemma the inferential version of the Schwarz's lemma which will essentially give me the proof as you as you as you will see. So, apply differential version version of Schwarz's lemma this composite function h circle f circle g what will you get you will get h circle f circle g derivative at 0 modulus is less than or equal to 1 that is what the infinitesimal version of Schwarz's lemma says infinitesimal version of the Schwarz's lemma says whenever you have an analytic function unit disc to unit disc which takes the origin to the origin okay which this is the origin then the derivative of the origin has modulus less than or equal to 1. So, h circle f circle g is an analytic function from the unit disc to unit disc which takes 0 to 0 so its derivative modulus of the derivative at the origin cannot exceed 1 okay but what is this this is if you see this is g dash you see see I will have g dash of 0 okay. So, if I differentiate it using chain rule I will get h dash of f of g of 0 into f dash of g of 0 into g dash of 0 this is what I will get I will get this because you know if I differentiate this by chain rule I differentiate the outermost function then the next inner function and then the next inner function this is just the chain rule okay and you know but this is what you see this is mod g dash of 0 mod f dash of g of 0 but what is g of what is g of 0 it is z not so g of 0 is z not and what is and I will have mod h dash of f of g of 0 which is w not this is h not equal to 1 I will get this alright so I will tell you what to do what you do is you calculate g write out g alright g will be just zeta going to zeta plus z not by 1 plus z not bar zeta you just have to change minus to plus that will give you the inverse map okay you differentiate it and substitute 0 okay you literally calculate it and you will get mod f dash of z not is less than or equal to 1 minus f of z not the whole squared by 1 minus z not the whole square you will directly get it it is direct calculation ordinary calculation nothing to do nothing complicated and since this is true for any z not you get that estimate alright it is just a straight calculation ordinary differentiation right.