 And how do you how do you prove this so it is in this proof that I need this hyperbolic geometry okay so what you do is you know you first so here is you choose a point you choose a point okay let me use B you choose sorry I will tell you what we are going to do it is very very simple choose B in delta minus D0 okay choose a point B which is outside D0 but inside delta and this is possible because D0 is not the whole unit disc okay. And what you do is you take a mobius transformation that will map which is an automorphism unit disc and that will map the point B to the origin okay so what you do is so you know I am going to take a map like this so I am going to take a h small h so let h of zeta P zeta minus B by 1 minus B bar zeta okay you know that this is an automorphism of the unit disc that will map B to 0 okay and it is a it is an automorphism so what this will tell you it will tell you that you know h of B is 0 okay h of B is 0 and that will imply that h of D0 will not contain 0 because B is not in D0 h of B will not be in h of D0 because h is a mind you it is a bijective map okay since B goes to 0 and B is not in D0 okay h of B will not be in h of D0 that means 0 will not be in h of D0 so you know the effect of h on this thing will be something like this now you can think of I am just drawing something figuratively you know this h will map D0 into something this is h of D0 it will it will push B to 0 and you know it will map h of D0 D0 onto h of D0 and this h of D0 will not be it will not contain 0 alright and since h of D0 does not contain 0 I can find see whenever a domain a simply connected domain does not contain the 0 I can always find an analytic branch of the square root okay this you can the problem with finding branch of the square root is that the square root of the function you are trying to find that should not vanish and the domain where you are trying to define the square root it should be simply connected okay only then you will get a square root. So h of D0 is a simply connected domain because it is the image under h of the simply connected domain D0 and h is an automorphism h is a holomorphic automorphism so it is a homomorphism okay so h is an isomorphism topological isomorphism also from D0 to h of D0 and since D0 is simply connected h of D0 is also simply connected and h of D0 does not contain the origin so there is a branch of the I mean there is a branch of the logarithm and you can use that you can find a branch of the branch of log neta and you can therefore find a analytic branch of root neta where you know if you want to call neta as the variable here neta is h of zeta okay there exist analytic branch of log neta on h of D0 why because h of D0 does not contain 0 and it is simply connected okay so there is an analytic branch of log of neta okay and so and hence an analytic branch of the square root of zeta so let me use capital G neta of root neta and you know root neta is e to the half log neta I mean the moment you get an analytic branch of log e to the half log will give you a square root right and and then so you know now you know therefore I have this G this G is a G is defined on this and G is a branch of the square root okay and if I take its image of course this does not contain 0 so if I take its image under G I will again get something that is not 0 so you know if I apply G I will probably get you know so I do not have enough space to draw this but anyway I will draw it so this is what will happen if I apply G okay I mean my diagrams are not accurate but they just they will just help you to visualize what is going on so this is what happens if I apply G so G will move this HD0 into GHD0 and that will be a of course the image of the image under G of HD0 will be again an open set because again open mapping theorem will always tell you the image of a non-constant analytic map will always be open so you know H so what I will get is I will get this domain which is G of HD0 H of D0 I will get this domain right I will get this domain and well all I can say about it is but there is even more see the fact is see the square root if you take the square root function okay the square root function is injective okay the square root function is injective because you have chosen an analytic branch choosing an analytic branch means you are choosing a continuous branch that means you are choosing one square root only one of the square roots and that too you are choosing them choosing it in an analytic way so the square root is always injective the square root function when you take the analytic branch of the square root that will always be injective okay therefore the moral of the story is G is in fact holomorphic isomorphism because it is an injective analytic map okay so G you see G from H of D0 to D0 to G of H of D0 is a holomorphic isomorphism holomorphic or analytic isomorphism because you know it is one to one it is injective analytic it is an injective analytic function and inverse function theorem will tell you an injective analytic function is a holomorphic isomorphism image will be open because of the open mapping theorem because it is a non-constant function in fact it will be isomorphic to the image so G is a holomorphic isomorphism alright okay G is a holomorphic isomorphism. Now you see so you know so you know you have this H that moves this to this so you know that 0 is was in D0 okay 0 was in D0 therefore you know I will get the image of 0 under H it will be H of 0 that will be a point in H of D0 okay and then I will get its further image under G I will get a point here and this point will be H of G what was it I use capital G of H of 0 okay so this is what I am getting so let me get rid of this this arrow is confusing so let me remove it okay so I have this D0 which is moved by small H onto H of D0 which is and that is an isomorphism because small H is an automorphism unit disk then from H of D0 to G of H of D0 is also an isomorphism because G is an isomorphism on H of D0 okay but there is something funny about G see this G is not defined on the whole unit disk this G is not defined on the whole unit disk it is only defined on this simply connected domain okay and therefore you know what this G as far as it is see the pix lemma and hyperbolic geometry only tell you that if you have an automorphism of the unit disk then it will be an isometry for the hyperbolic metric but G is of course an isomorphism from H D0 to G H D0 but G will not be an isometry with respect to the hyperbolic in fact G will be an expansion G will become a strict expansion why because you see we have already seen that G is the inverse of this function the squaring function which is a strict contraction therefore G will be a strict expansion okay alright so the fact that the squaring the square function is is a strict contraction will tell you that it is inverse which will be a branch of the square root that will be a strict expansion okay so you know neta going to neta squared is a strict contraction implies that G is a strict expansion so what does this mean this means that is you see G of neta so you know strict expansion with respect to what of course strict contractions strict expansion etc are with respect to the hyperbolic metric so you know if you take the image under G of 2 points neta 0, neta 1 and take the you take 2 points neta 0 and neta 1 in the in H D0 okay and you take their images under G there will be 2 points in G of H of D0 and you take the hyperbolic distance that will be you know that will be greater than or equal to C times a constant times the hyperbolic distance between the 2 original points for suitable C greater than 1 this will happen see this is expansion see expansion means this expansion means that the distance between the image points hyperbolic distance between the image points is greater than the hyperbolic distance between the source points and there is a constant which will appear and the constant will be greater than 1 you know what is that constant that constant is actually the reciprocal of this constant with R replaced by root R where root R is a root R is such that mod neta less than or equal to root R contains H of D0 okay. So in fact in fact C is equal to 1 plus R by 2 root R where mod neta less than or equal to R mod neta less than or equal to root R contains H of D0 that is in fact that will come from here okay that will come from here note that after all small g capital G is an inverse for small g small g is a square function alright capital G is a square root capital G is an inverse for small g okay. So you know if you in this in this expression you call this as neta not call this as neta 1 okay then this will be G neta not and this will be G neta 1 okay and you will get this expression where you will have to put G neta to be inside this okay and if you want the image energy to be in the disk bounded by R then the source this should be bounded by root R okay. So from this you will automatically get this okay. So the moral of the story is that you know we need to use this now so alright so now we have that G is a strict expansion alright and then what you do is see now I still you know my originally 0 was there in my D0 okay and then I translated I mean I used I mapped this small b to 0 therefore when I took H of D0 0 was not there okay and then I am using the square root function so 0 will not continue to be there alright but then I would still like to go back to the origin so what I will do is I will apply another mobius transformation that will map gh0 to the origin okay. So I will bring this fellow back to this to the origin okay by applying a suitable map and let me call that map as I have too many arrows here so you know I have so I will apply a map like this so consider so put so I have used small h and let me use h1 so h1 of omega to be you know I will put omega minus gh0 by 1 minus gh0 bar omega I will do this okay what this will do is it is an automorphism unit disk that will map gh of h of 0 capital G of small h of 0 to the origin okay so now and what I will do is now I will consider this composite from the so I have D0 I first apply h okay and I land inside h D0 that is an isomorphism then from h D0 I apply G and I land in gh D0 this is also mind you an automorphism I mean this is also an isomorphism because G is an analytic branch of the square root which is injective and an injective colomorphic map is an isomorphism onto its image okay and then now and then I have applied this h1 that goes to h1 of G of h of D0 and mind you h1 is also an isomorphism because h1 is a automorphism unit disk it is also injective so it gives an isomorphism from this to this okay now what you do and notice that 0 goes back to 0 so you know 0 goes to h0 which goes to G of h of 0 and that goes back to 0 so 0 goes to 0 right and now the big deal is that you look at the derivative of this function so you know so let me continue from here so now look at put psi to be this composition that is first apply small h then apply G then capital G and then apply h1 put psi1 to be this okay then psi then psi of 0 put psi to be equal to this then psi of 0 equal to 0 alright of course and now calculate calculate the derivative psi dash of 0 okay you know I am trying to look at the lemma says that if you have smaller simply connected subdomain than the unit disk then I can then I can find an analytic map which maps it isomorphically onto a subdomain of the unit disk which which takes 0 to 0 and whose derivative the origin is greater than 1 okay that is the lemma right and I am trying to prove that lemma I claim that this psi is the psi that I need so it is certainly a map it is so it is of course psi is of course an isomorphism because it is a it is injective it is a composition of three injective maps so it is an injective map so it is an isomorphism it takes 0 to 0 okay calculate the derivative of psi at the origin let me use the correct variables my D not what was my variable on D not on D not it was zeta and h of zeta I call it as neta and neta goes to g neta which I have called as g neta is called as w and then I have h1 of w which I will have to give another name so I am short of so let me use gamma okay normally gamma is used to denote a path okay but I am short of symbols so let me use gamma so you know the variables on D not it is zeta on h D not it is h zeta which is neta okay on g h D not it is g neta which is w and again on h of g of h1 of g of h of D not it is the target variable is gamma which is h1 of w okay so let me write one more step the way I have written it is a function of this starting variable which is zeta and the ending variable is gamma okay is how it is right and what is what is psi dash what is modulus of psi dash of at the origin this is limit zeta tends to 0 psi zeta by zeta it psi zeta minus psi 0 by zeta minus 0 right and but psi of 0 is 0 and so I will get limit zeta tends to 0 psi zeta by zeta okay I will get this okay certainly I can write this as limit zeta tends to 0 mod psi zeta by zeta I can write this because after all the map psi at the origin is analytic so the derivative exists so this limit does exist and if the limit exists I can take this the modulus is a continuous function so I can write this but my aim is I want to show that this is greater than 1 because that is the purpose of the lemma purpose of the lemma is to show that I can map a smaller simply connected sub domain of the unit disc I can map that isomorphically on to another similar smaller simply connected sub domain of the unit disc which contains the origin but with the extra condition that the derivative of the origin can exceed 1 the fact that I can the fact that I can make the derivative of the origin in modulus exceed 1 is exactly the due to the fact that I am working on a strictly smaller domain than the unit disc because if had it been the unit disc Schwarz lemma will tell you that the differential version of Schwarz lemma will prohibit this from happening it will make the derivative at the origin to be less than or equal to 1 but the fact is because I am mapping my domain is not the whole unit disc but a smaller simply connected domain smaller than the unit disc I can make the derivative at the origin greater than 1 I can exceed the bound of the Schwarz lemma that is the whole point. So somehow I have to make this greater than 1 ok now the fact is you know I have to somehow use this estimate ok this estimate this estimate is connected with the mapping in between which is capital G ok the mapping in between which is capital G which is the square root function so analytic branch of the square root is expanding ok and I have to use that estimate ok to show that you know this happens right. So you know let me do something let us keep this as this let us try to apply that let me try to apply this. So you know let us calculate rho h of psi of zeta, psi of 0 by rho h of zeta, 0 let us calculate this ok let us calculate this well you see this is the same as rho h of psi of zeta, psi of 0 is just 0 divided by rho h of zeta, 0 ok. But you see now but what is psi of zeta psi of zeta is h1 of g of h of zeta, so this is rho h of h1 of g of h of zeta, let me keep let me write this here h1 of g of h of 0 by rho h of zeta, 0 ok so I get this alright and now see I can knock off the h1 because h1 is an automorphism of the unit disc ok and therefore it is an isometry with respect to the hyperbolic metric. So you know I can throw out that h1 so you know that will be equal to rho h of of course you know please do not confuse this subscript h with the h with this h because this h subscript of rho is supposed to signify that I am taking hyperbolic distance ok. So do not confuse the subscript h of rho with the map h ok. So I can knock this h1 out because h1 is an automorphism of the unit disc and it is an isometry with respect to the hyperbolic metric that is part of Picks lemma alright or Schwarz lemma. So I can simply write as rho h of g of h of zeta, g of h of 0 divided by rho h of zeta, 0 ok I can write this since h1 is an isometry of delta with respect to the hyperbolic metric rho h ok. So I can do this and then you know but g is I already have an expression for rho h of g of something divided by rho of that thing alright. So you know in this expression what you do is you put neta0 is equal to h of zeta and you put neta1 is equal to h of 0 ok. So I will end up with so you see this will be greater than or equal to c times by using this estimate I will get c times rho h of what should I put instead of neta0 it is h zeta and instead of neta1 I have to put h0 ok. So here is where I am using this estimate where c is greater than 1 alright I have used so this is very important I have to use this estimate which is reflection of the fact that the square root function is expanding and that is because it is inverse function which is a square function is contracting and why it is contracting is because it is defined on the whole unit disc and it is not an isomorphism and any analytic function on the whole unit disc taking values in the unit disc which is analytic and which is not an isomorphism is necessarily strictly contracting ok that is where see here is where I am using hyperbolic geometry throughout the proof ok. So I will get this but you know if you look at small h small h is also an automorphism of the unit disc so small h is also an isometry with respect to the hyperbolic metric so in the numerator I can knock off that h. So finally this see this ratio turns out to be this ratio turns out to be greater than I mean it is greater than or equal to C which is greater than 1 ok. So now you know then I will have to compare this you know so you know as zeta tends to 0 ok what will happen is this ratio rho h of psi zeta psi of 0 by rho h of zeta 0 this is equal this is greater than or equal to C which is greater than 1 ok you have this ok if you combine all this together this works if you let zeta tend to 0 alright. On the other hand my claim is as zeta tends to 0 this is exactly psi zeta by zeta this is this behaves like psi zeta by zeta therefore in other words as zeta tends to 0 if you take mod psi zeta by zeta it is a quantity which is greater than 1 and therefore the derivative at 0 is greater than 1 that is the claim. So what I what we need to understand is that this as zeta tends to 0 behaves like psi zeta by zeta ok and I think that that is should be alright because you see rho if you calculate rho h of psi zeta, psi 0 is rho h of psi zeta, 0 ok and we know what this is this is half we have seen a formula for this is half long 1 plus mod psi zeta by 1 minus mod psi zeta this is what rho h is ok right and rho h of rho h of zeta, 0 will similarly be half long 1 plus mod zeta by 1 minus mod zeta ok we know this formula ok and now again you know if you actually divide and let zeta go close to 0 then the ratio will be psi zeta by zeta will behave like psi zeta by zeta just because of again because of if you want the Lopitas rho ok. So you know if you calculate half long of 1 plus mod psi zeta by 1 minus mod psi zeta divided by half long 1 plus mod zeta by 1 minus mod zeta we calculate this see what I want you to understand is half long 1 plus x by 1 minus x is approximated by x for x sufficiently small. See see if I expand half long 1 plus x by 1 minus x where x is a small quantity ok then you know what I am going to get is I am going to get half into long of 1 plus x minus long of 1 minus x ok and this will see this will be x plus terms involving x square and so on because long 1 plus x is x minus you know x square by 2 plus x cube by 3 and so on right. So you know if I expand both in power series for x sufficiently small alright if I expand and subtract what I will get is I will get x plus something and then here I will get minus of minus x and so on. So I will get 2x and then there is a half outside so I will get x plus terms involving x square and higher powers ok. So as x becomes small this quantity is so as x tends to 0 half long 1 plus x by 1 minus x is behaves like x ok I need this fact so as x becomes very small ok half long 1 plus x by 1 minus x can be approximated to x. So this can be approximated to the numerator it will be mod zeta psi of zeta by mod zeta as zeta mod zeta tends to 0 ok this is an approximation this comes very close to the behavior of this alright and therefore the moral of the story is that if I now calculate if I like take the limit as zeta tends to 0 of mod psi zeta by zeta the behavior is like taking the limit as zeta tends to 0 of this mod of this quantity and that is always greater than 1. So you know therefore what you will get is mod psi dash of 0 is greater than 1 and that finishes the lemma ok that finishes this lemma which you see is highly technical it is not very difficult but the point is the ideas involved they use lot of hyperbolic geometry ok. Now you know now I can go back and say why the image of f0 is delta ok why is the image of f0 is less than delta I mean is equal to delta because if it is less than delta I can apply the lemma ok if so you know now it is now the proof is just one line if f0 of t is proper properly contained in delta by the lemma we have an injective analytic map psi from D0 which is f0 of D to delta such that psi dash psi of 0 is 0 and mod psi dash of 0 is 1 I mean sorry is greater than 1 ok. Now I can apply this lemma but the beautiful thing is if I combine if I apply this f0 and then follow it by this psi the resulting thing is continuous to be in this family that is the big deal ok note that first apply f0 then apply psi this is also in the family because after all it is also a map you first apply f0 and then you follow it by psi that continuous to be a map from D to the unit disc ok because psi is takes from this to the unit disc ok so this composition is also a map from D to the unit disc first point it is also analytic because it is composition of analytic functions second point third point is injective because both are injective psi is injective and f0 is also injective therefore composition is injective but now comes a big deal what is the derivative of psi circle f0 at z0 by the chain rule this is derivative of psi at f0 of z0 modulus times derivative at f0 I mean derivative at z0 of f0 ok and this is and you know derivative f0 of z0 is 0 and psi dash of 0 modulus is greater than 1 this is a so this will be greater than a ok because this is greater than 1 this quantity is greater than 1 by the lemma and this quantity is equal to a ok so the product is greater than a now that is a contradiction because a was supposed to be a was supposed to be supremum of all the derivatives of all the functions in the family f the modulus of supremum of the modulus of the derivatives at the origin at z0 at the point z0 you cannot find a for any member of the family if you take the derivative at the at z0 and take the modulus it cannot it has it cannot exceed a is a supremum but I have found something that exceeds a that is a contradiction it is a contradiction to the fact that this psi circle f0 is in the family script f so this contradiction proves that f0 of d is delta in other words the extremal function f0 fills out the whole unit disc ok contradiction to definition of a ok so this proves the important fact that f0 does map this domain on to the unit disc isomorphically ok and of course it we have already adjusted it to make 0 z0 go to 0 you can make the derivative at z0 to be positive by using a suitable rotation ok so you can also make the derivative positive right and and that finishes the proof of the Riemann mapping theorem ok of course one has to also think about whether you can make the derivative at z0 equal to 1 ok but that one that one has to ponder about but what we have is that given any simply connected domain d which is not the whole complex plane you can find an you can find a holomorphic isomorphism of that of on to the unit disc which maps any given point chosen point z0 of d on to the origin and you can make the derivative at the origin I mean you can make the derivative at that point also to be a positive real number you can do this much ok so that finishes the proof of Riemann mapping theorem ok.