 So, we are here uhhh this is the continuation of the previous lecture I am trying to show that uhhh for the uhhh for the origin and a point z0 on the unit disk the geodesic is simply the line segment joining the origin to that point ok. In other words I am just trying to show that any diameter is a geodesic ok. I am just this if I prove this lemma I am just trying I am I am showing that any diameter is a geodesic ok. And uhhh and my point is that uhhh the proof of this theorem will be complete just because of this because uhhh you can move uhhh any geodesic to a diameter by using a conformal automorphism of the unit disk ok. So, essentially the whole proof of the theorem is here ok essentially here ok. So, how do I prove that uhhh the shortest distance from 0 to z0 is uhhh given by this line segment. What I will do is I will I will rotate uhhh like this by this e to the minus i theta0 where theta0 is argument of z0 and I uhhh so this map z the point z0 is mapped to r0 then I will have to show that this is a geodesic ok. And uhhh so this line segment from 0 to this point r0 on the real line that is a geodesic I have to show. And uhhh what therefore what do I have to show I have to show that you see uhhh it is uhhh that is a segment which has least hyperbolic length ok. And uhhh that is because hyperbolic distance is defined as uhhh the least hyperbolic length of an arc between the two given points. So, what I am going to do uhhh to show that this is uhhh this the hyperbolic length along this this line segment is the least I will simply compare it with the hyperbolic length along any path from uhhh 0 to r0. So, so here is gamma which is a path from uhhh uhhh uhhh so let me let me keep this because I need it. So, let me let me estimate the uhhh hyperbolic length along gamma let us do a simple calculation. So, gamma uhhh is from say closed finite interval uhhh on the real line uhhh I am taking values in the unit disc. And uhhh of course gamma of a is uhhh the origin gamma of b is uhhh r0 ok. Gamma is some path uhhh and what is the hyperbolic length of gamma? Hyperbolic length of gamma is well you know it is by definition integral over gamma mod dz by 1 minus mod z the whole squared ok. And uhhh and of course you know we calculated this this is just integral over gamma uhhh mod gamma dash of integral from t equal to a to t equal to b mod gamma dash of t dt by 1 minus uhhh mod gamma t the whole squared ok. So, I have just plugged in z equal to gamma of t and uhhh I have made a change variable uhhh to the real parameter t which varies from a to b ok uhhh. Now you know what you might understand is that uhhh if you give me uhhh uhhh uhhh any path gamma like this ok if I project that path to the uhhh to the real axis I will get a path along the real axis mind you see this path could go out I mean there is no uhhh there is no reason it could go out like this and here also you know it could be even more complicated it could go like this that path gamma could be just just any path inside the unit disc which starts at 0 and ends at r not it could twist and turn I mean there is no uhhh it has just has to start at the origin it has to end at r not uhhh and it has to be in the unit disc that is the only condition ok. So, uhhh but you know give me a path like this if I take the projection of that path on the real axis I will get a path along the real axis which starts with 0 and ends with r not and how do I get that path you know projection on to the real axis is given by simply taking the real path ok. So, what you must understand is that real path of gamma if I take that is also uhhh uhhh is also a path in the unit disc from uhhh real path of gamma of A which is actually because gamma A is 0 real path of gamma A is also 0 to real path of gamma of B and gamma of B is r not that is also real so real path of gamma of B is also equal to r not. So, it is so real path of gamma is also a path from r 0 to r not it is just the projection of that path on to the real axis ok. So, uhhh uhhh you know basically what I am saying is that you give me 2 points on the real axis and then you you give me a path like this ok then as a point travels on on this path ok its projection will travel along this line segment and that will give you a path on the real axis ok. So, uhhh so if so you know if if if gamma is a path like this then real part of gamma will be its projection the fact is real part of gamma is also a path ok and for this path what is the hyperbolic length the hyperbolic length of real part of gamma is what it is by by this definition is integrate over real part of gamma again mod dz by 1 minus mod z the whole square alright. And now if you calculate this is going to be well uhhh here I have to put uhhh uhhh integral over modulus of d of real part of gamma of t divided by 1 minus modulus of real part of gamma of t whole square this is what I will get if I plug it in. So, when I calculate for real part of gamma I will have to again take t equal to a to t equal to b uhhh but I have to plug in instead of z real part of gamma ok and uhhh of course uhhh the what is a derivative of the real part of gamma of t it is a real part of the derivative of gamma of t because gamma of t has a real part and has an imaginary part and its its derivative with respect to t is derivative of the real part with respect to t plus i times derivative of the imaginary part with respect to t. So, this is just going to be integral from a to b ok here I am going to get real part of gamma dash of t mod mod dt divided by 1 minus mod real part of gamma of t the whole square this is what I am going to get right because you know if gamma of t you write it as uhhh x of t plus i y of t then gamma dash of t is going to be x dash of t plus i times y dash of t. So, x dash of t which is a real part of gamma dash of t is the derivative of x of t which is a real part of gamma of t. So, real part of gamma of t derivative is the same as real part of derivative of gamma of t that is all I am saying ok. But then if you if you you know the real part of a complex number is always less than uhhh the modulus of the real part of the complex number is always less than the modulus of the complex number. So, you know this real part of mod modulus of real part of gamma dash of t is certainly less than or equal to modulus of gamma dash of t and the denominator 1 minus real part of gamma t whole square that will be greater than 1 minus gamma t the whole square. So, it is reciprocal will be less than therefore you know this quantity is certainly less than this quantity ok. So, what you will get is that you will get that this is less than or equal to hyperbolic length of gamma of gamma. So, what I have proved is you give me any you give me any path gamma alright then it is hyperbolic length is certainly uhhh bigger than the hyperbolic length of its projection to the real axis ok. And therefore you know uhhh the moral of the story is that if you want the hyperbolic distance you have to take a path of minimum hyperbolic length. So, you know obviously I will first of all have to take a path which is equal to its real part ok. Because if if you have a path gamma which is not equal to its real part it means that this integral this length will be bigger than this length you will get equality you will get equality if and only if the imaginary part of gamma is 0 ok. Therefore moral of the story is you give me any path first of all in order to minimize the hyperbolic length it is you will have to take paths which live on the real axis ok ok. So, therefore and among the paths which live on the real axis from 0 to r not which is the path that will give you minimum length minimum hyperbolic length it is the path for which uhhh it is a path which starts at 0 and continuously and it increases to r not that is the only path. Because if the path decreases at some at some sub interval you can cut out that sub interval and get a shorter path. So, for example you know if I have a path which goes like this for example if I take the projection of this path it will start from here it will go up to here then it will come here and then it will slow down then it will go here it will slow down then it will come here slow down a little then it will go all the way there and then it will come back ok that is how the projection of this path will be. But you know if you want uhhh whenever part of the path goes uhhh goes in the reverse direction then there is a sub interval where it is decreasing that sub interval you can cut out by simply cutting out the portion of the path which the moment it starts going backward and until it comes back to the same point that portion of the path you can cut it out because that will unnecessarily give you extra length after all you want to minimize the length. So, all this shows that the if you want a path of shortest length you just have to take this path starting at 0 and ending at r not it should be a increasing function ok. So, this just this calculation and common sense uhhh common mathematical sense of course tells you that the geodesic is simply the straight line segment from 0 to r not ok. So, so let me write that down thus the uhhh geodesic from uhhh from 0 to r not is the straight line segment from 0 to r ok. So, it is very clear. So, the uhhh uhhh uhhh you must remember that uhhh even if gamma even if uhhh even if real part of gamma uhhh decreases ok then this derivative will become negative. But I am taking mod ok therefore uhhh uhhh even if gamma decrease even if you take a path which which goes uhhh which wanders away which which goes starts at 0 then goes back and then comes forward I am going to get more length ok. So, if you want shortest length you have to start at 0 go along this and then straight end at r not ok it should be an increasing function it should not decrease in any sub interval ok. So, that tells you that the so you must you must always remember that there is a mod here there is a mod real gamma dashed of t. So, see the integrand is always uhhh positive quantity ok. So, uhhh so the geodesic from 0 to r not is the straight line segment from 0 to r not and therefore from 0 to z not the geodesic will be just this line segment because uhhh this is the inverse image of this under the inverse map which is rotation by uhhh e power it is which is rotation by theta not this is rotation by minus theta not this inverse map is rotation by theta not and you know it is an automorphism unit disc. So, it will preserve geodesics. So, if this is the if this is the geodesic then this is also geodesic. So, what so the effect of this lemma is that we have proved that all diameters are geodesics ok. So, this lemma actually proves that all the diameters are geodesics ok thus all diameters of delta are geodesics and uhhh as far as this uhhh this expression is concerned uhhh this expression is something that uhhh I I I calculated in the uhhh the last uhhh in the in the lecture before the last lecture ok uhhh. In fact uhhh I can do that for you uhhh hyperbolic length length uhhh from uhhh 0 to r not along geodesic which in this case is the right is the radial line from 0 to r not line segment. What is the hyperbolic length you have to just integrate over this line segment uhhh uhhh. So, you know uhhh uhhh how do you parameterize this uhhh line segment from 0 to r not you simply parameterize by the interval 0 to r not itself ok. So, you you take gamma from 0 r not to delta given by identity map gamma of t equal to t ok. This is simply this is just the inclusion of uhhh the interval 0 r not as this line segment this is after all the interval. So, the parameter is just given by t ok and uhhh if you take this as the path ok then uhhh integral over gamma what is the hyperbolic length you have integrate over gamma mod dz by 1 minus mod z mod z the whole square you will simply get integral from t equal to 0 to r not uhhh you will get uhhh uhhh uhhh gamma dash of t is 1. So, you will get mod dt. So, in this case it will be just dt because t is increasing by 1 minus t square and if you integrate this you will get half long 1 plus r not by 1 minus which is this expression ok. So, uhhh so that finishes uhhh that finishes the proof of lemma 1 ok this finishes and mind you uhhh uhhh this geodesic uhhh length is the same as this geodesic length and therefore this geodesic length is also given by the same quantity mind you this is also this is also equal to half long 1 plus mod z not by 1 minus mod z because after all mod z not is r not z not is r not e to the i theta not ok. So, so that proves uhhh uhhh that proves lemma uhhh that proves lemma 1 now uhhh after I have uhhh after I do this I have to now show that uhhh I still have to prove the theorem the theorem is that given any 2 point z not and z 1 on the unit disc the geodesic is given by the unique circular path of the circle which passes through z not and z 1 and which is orthogonal to the unit circle ok. So, uhhh so here is uhhh uhhh lemma 2 so lemma 2 is uhhh is something that you know uhhh is something that you know already uhhh given it is more of a recalling uhhh uhhh lemma that you already seen given 2 points z not, z 1 the unit disc we can find uhhh uhhh uhhh an automorphism h of the unit disc such that h takes uhhh z not to 0 and then h takes z 1 to uhhh the real axis. So, this is uhhh this is just playing around with uhhh Mobius transformations of the unit disc onto itself. So, you know this is something that uhhh uhhh we have already seen you see. So, so let me draw this diagram now. So, I am all I am saying is well take this uhhh uhhh lemma 2 uhhh it is it is more of uhhh it is more of recalling a statement that we already seen you see you just take uhhh so you know so put h of z is equal to z minus z not by 1 minus z not bar z you know that such a map uhhh is a automorphism the unit disc and it will map z not to 0 ok. It will map z not to 0 because if you put z equal to z not I will get 0 alright and uhhh what will it take uhhh z 1 2. So, it will take z 1 2 h of z 1 will be z 1 minus z not by 1 minus z not bar z 1 that will be some complex number ok. You see it will have some argument alright how will you how will you bring that complex number to the to the real axis by simply rotating by the negative of that argument. So, what you do is you put e to the i alpha into this where alpha is minus of principle argument of h of h of z 1 because if I put it like this then the argument of uhhh h of z 1 ok will be argument of this which is alpha which is minus argument h z 1 uhhh no I should not put h z 1 I should be careful uhhh I should put not h z 1 I should put h without this. So, uhhh e z 1 minus z not by 1 minus z not bar z 1 this is what I should ok you put you put this then you know then what is argument of h of z 1 the principle argument of h of z 1 is 0 because the argument of h of z 1 will be argument of this which is alpha plus argument of this evaluated at z 1 which is minus alpha. So, you will get alpha plus minus alpha you will get z ok if you do not want to use principle argument you can use ordinary argument and read model of 2 pi ok. So, uhhh this is a of course a Mobius transformation which takes the unit disc to the unit disc it maps z not to 0 and it maps z 1 to a point on the real axis ok that is what we wanted and uhhh uhhh in fact positive real axis right. Now I am done now now I can give the proof of the theorem. So, uhhh so, so this is end of proof of lemma 2 there is nothing to prove except that you should write this formula properly we have already seen that any automorphism of the unit disc looks like this for some alpha for some real alpha and for some z not a point inside the unit disc that we have already seen ok we are I am just using that fact here right. So, now is we come to the proof of the theorem we come to the proof of the theorem. So, what is the proof of the theorem? So, you know what do I have to prove in theorem? So, here is unit disc and here are 2 points z not and z 1 ok and what am I supposed to prove I am supposed to prove that if I take this circle that passes through z not and z 1 and which is orthogonal to the boundary of the unit disc which is s 1 the set of unimodular complex numbers I have to show that this is the uhhh this circle if you take this circle uhhh it is this arc which is a geodesic I have to show that ok. Now what I will do is well I will simply map I will use this h I will put w is equal to h z uhhh with h as in lemma uhhh lemma 2 ok. Then what will happen is that you see uhhh after all this h is uhhh you know an automorphism unit disc. So, it will take the unit disc to the unit disc and what it will do is that it will map z not to 0 ok. It will map z 1 to h z 1 but h z 1 will be a point on the positive real axis. So, h z 1 will be here this 0 will be h of z not and I will get a point here which is h z 1 this is what the Mobius transformation will do it will map z not to 0 ok and it will map z 1 to a point on the real axis. And because this map is a Mobius transformation it is conformal. Therefore, if you take this circle suppose I call this circle as uhhh C if this circle you know Mobius transformation will always map circles onto circles or straight lines ok. So, the image of this circle at least the portion of the circle inside the unit disc ok will be it has to be a circle or a straight line inside the unit disc. But it has to pass through these 2 points then what else can it be it has to only be the diameter it it has to only be this diameter. So, it is clear that h of C intersection delta is equal to the diameter uhhh on the real axis right that is just just because of property of Mobius transformation right. And but then what does a lemma 1 tell you lemma 1 tells you that the geodesic from z from 0 from h of z not to h of z 1 is this line segment ok. Lemma 1 tells you that the geodesic from h of z not to h of z 1 is just that line segment. Therefore it is inverse image under h ok which will be this arc will be the geodesic between z not and z 1 and that finishes the proof of the theorem. I am just using the fact that uhhh you know uhhh mobius transformations are conformal maps they have to map straight lines and circles on to straight lines and circles. And they are conformal therefore any circle perpendicular to the unit circle has to go to a circle or a straight line perpendicular to the unit circle which passes through these 2 points and that has to be only a diameter it it has to be a straight line right. Therefore the this is a geodesic it is inverse image mind you h inverse is also an automorphism of the unit disc. The image of a geodesic under an automorphism is again a geodesic because geodesics are isometric I mean uhhh because automorphism unit disc they are isometrics they preserve distance ok. So uhhh therefore so what this so what this tells you is that it gives you the proof of the theorem namely uhhh the the circle the arc of the circle uhhh from C uhhh from z not to z 1 is the unique geodesic geodesic the the as it is the inverse image as it is the inverse image as it is the image of h inverse the image of the line segment from h of z not equal to 0 to h of z 1 under h inverse which is an automorphism of delta ok. So that tells you that that proves the theorem ok. So you know it is uhhh it is just uhhh clever play of Mobius transformations nothing I mean all this is all this uhhh geometry is pretty easy to work out but uhhh the results you get are very nice ok. So that proves the theorem that tells you how geodesics actually look like uhhh for the hyperbolic uhhh geometry. Now what I need to do next is uhhh now I come to something more serious I come to this fact now you know uhhh now I am going to change the point of view and uhhh try to prove this uhhh prove another statement uhhh which is what I want for the Riemann mapping theorem. So for that you know I am going to look at Schwarz lemma again I am going to look at these 3 statements Schwarz lemma infinitesimal inversion of the Schwarz lemma and Picks lemma but I am going to look at them uhhh I am going to look at the non-equality case ok. So look at the Schwarz lemma what does it say if f is an analytic map from the unit disc to unit disc which which preserves the origin then mod fz is less than or equal to mod z ok for all z in the unit disc and equality happens only if f is an automorphism. So if f is not an automorphism you have strict inequality and what is the strict inequality the strict inequality will say mod fz is strictly less than mod z for all z in the unit disc what does it say it says the the distance from the origin of the image fz the Euclidean distance ok is certainly is strictly less than the Euclidean distance of z from the origin. So with respect to the Euclidean distance what is happening it is it is a contraction if f is not an automorphism of the unit disc onto itself the effect of f is like a contraction that is what it says mod fz will be strictly less than mod z because even for a single z0 not equal to 0 if you get mod fz0 is equal to mod z0 then f has to be an automorphism that is these that is a strong implication of the equality which that that part of the shorts lemma. So so long as f is not an automorphism ok and you know the condition that f is not an automorphism is equivalent to saying that f is not 1 to 1 1 to 1 and 1 to f is not bijective because you know injective a bijective holomorphic map is certainly a injective holomorphic map and therefore it is an automorphism right. So the condition is either f is not injective or f is not surjective if you take such an f then such an f is not an automorphism and then it will surely be a contraction ok and you get the same thing here in the infinitesimal version also if f is not an automorphism then the derivative at the origin if you take the models it has to be strictly less than 1 ok you will get equal to 1 if and only if f is an automorphism and look at pix lemma pix lemma also says that if you take any analytic map from the unit disk to the unit disk which need not preserve the origin ok. So in these two cases you need origin to be preserved but here you do not need origin to be preserved that is f of 0 need not be equal to 0 but it tells you that the modulus of the derivative is bounded by this quantity and you will get an equality if and only if f is an automorphism if so if f is not an automorphism this is strictly in equality ok. Now what I want to tell you is that the same statement this in the same philosophy we can make a statement using the hyperbolic metric so there is a version of the Schwarz lemma for the hyperbolic metric what it says is if you take any analytic map from the unit disk then it will be either an isometry in which case it will be an automorphism of the unit disk and if it is not an automorphism unit disk it will be a contraction with respect to the hyperbolic metric ok. So there are only two possible analytic maps from the unit disk to the unit disk one are isometrics which are given by holomorphic automorphisms ok the others are non-automorphisms they will strictly be contractions ok this is the fact that we need for the Riemann proof of proceeding with the proof of the Riemann mapping theorem ok. So let me state that so here is the theorem so so let me state it elegantly in words the the analytic maps of delta into delta are either automorphisms which are isometrics that is preserve the hyperbolic metric or they are non isomorphisms which are necessarily contractions for the hyperbolic metric. So this is the theorem this is the theorem elegantly stated in words but how do you state it in symbols in symbols if f from delta to delta unit disk unit disk is analytic the hyperbolic distance from between fz0 and fz1 which are the images of z0 and z1 the unit disk under f is less than or equal to the hyperbolic distance between z0 and z1 for all z0 z1 in the unit disk with equality occurring for a single pair z0 not equal to z1 if and only if f is an automorphism so thus rho h of fz0 fz1 is equal to rho h of z0 z1 for all z0 z1 unit disk if f is an automorphism of delta and rho h of fz0 fz1 is strictly less than rho h of z0 z1 for all z0 z1 in delta if f is not an automorphism this is the statement in symbols ok. So with respect to hyperbolic distances with respect to the hyperbolic metric a map of the unit disk is going to itself either it preserves the metric in which case it is an automorphism if it is not an automorphism it will just shrink it ok it will be its effect will be like contraction you need to give a proof of this proof is pretty easy I mean proof is pretty easy because we have built upon we have built lot of simple lemmas and theorems but even otherwise all this all of this argument is very very simple just involves movies transformations nothing more than that so you know so here is a proof so you know for z0 z1 in delta let gamma be a geodesic from z0 to z1 in delta for the hyperbolic metric. So as I have drawn in the diagram here z0 the geodesic is given by a circle which passes through z0 and z1 and is orthogonal to the unit circle and the geodesic is the portion of this arc of the circle between z0 and z1 ok now what is the hyperbolic distance between let us calculate the hyperbolic distance between fz0 and fz1 ok. So you know so you know my diagram is very much like that but anyway let me draw it so I have diagram like this this is unit disc so I have this z0 I have this z1 and here is my geodesic ok this is my geodesic path gamma and I have applied this map f which is an analytic map from unit circle to unit circle I mean from unit disc to unit disc and so I am going to get I am going to get these points fz0 and I am going to get a point fz1 and of course the image under f of gamma is going to give me a path from fz0 to fz1 so this is a geodesic but then I will get so let me draw something like this so this is the path which is first apply gamma then apply f just fz in gamma ok. So the image of a path from the image of any path from z0 to z1 ok I have drawn the arrows in the wrong direction should be from z0 to z1 this is z1 so gamma is a geodesic from z0 to z1 but in any case it is a path from z0 to z1 alright and its image under f is going to give me a path from fz0 to fz1 because I do not know that f is an automorphism I cannot claim that the image path is also a geodesic I do not I cannot say that so but in any case you know if I measure the length of this the hyperbolic length of this arc that has to be greater than or equal to the hyperbolic distance between these two paths these two points because hyperbolic distance is a minimum length of a hyperbolic path. So what I can say this is less than or equal to integral over f circle gamma of so I should say it is less than or equal to hyperbolic length of f circle gamma this is correct right because the hyperbolic metric distance is supposed to be the minimum of such hyperbolic lengths of paths from fz0 to fz1 and f circle gamma is one such path so the hyperbolic length of f circle gamma is certainly going to be greater than or equal to this it will be equal to this if and only if f circle gamma is a geodesic alright. So let us calculate that what is that I mean so you know but what is the hyperbolic length of gamma it is integral over f circle gamma of mod d you know so if I call this as z and call here suppose there is this z plane this is the w plane then my function is w equal to fz so I will get mod dw by 1 minus mod w whole squared alright this is a hyperbolic length and if I compute it what will I get well I get integral f circle gamma so let me rewrite all that here the hyperbolic distance from fz0 to fz1 is less than or equal to integral over f circle gamma of mod dw by 1 minus mod w whole squared and that is you know if I make a change of variable I will get integral over gamma f dash of z mod mod dz by 1 minus mod f dash f of z whole squared I will get this is what I get if I put w equal to fz alright but then by pix lemma this is less than or equal to integral over gamma mod dz by 1 minus mod z the whole squared why this is because of pix lemma which tells you that mod f dash of z is less than or equal to 1 minus mod fz the whole squared by 1 minus mod z the whole squared so if I use pix lemma I will get this but what is this this hyperbolic length of gamma but gamma is a geodesic therefore this is the distance from z0 to z1 hyperbolic distance so this is this is rhoh z0 to z1 because it is the length of the hyperbolic it is the length of the geodesic from z0 to z1 because I started with gamma to be a geodesic so therefore I am getting the fact that the hyperbolic distance between the image points is less than or equal to the hyperbolic distance between the source points which is essentially the statement of this theorem okay and then I will tell you that you will get equality only if it is an automorphism and if it is not an automorphism you will get a strict inequality so for that what I wanted to understand is that of course you know if it is an automorphism then if f is an automorphism then of course since gamma is a geodesic f circle gamma will also be a geodesic so I will get equality here okay and if f is an automorphism I will get an equality in pix lemma so I will simply get this hyperbolic distance is equal to this hyperbolic distance okay I will get equality if f is a automorphism of delta okay but even otherwise I mean we have already noted similarly that automorphisms do preserve the hyperbolic distance because they are isometrics okay but the more serious assertion is suppose you have a pair of points z0 and z1 which are distinct points for which this equality occurs for even one pair for even one pair if it occurs okay then the assertion of the lemma is very strong it says that f has to be an automorphism okay if it preserves the distance between even a single pair of distinct points then it has to preserve all the distances see it is such a very powerful condition that is what you must understand okay so how do we look at that situation see for a particular z0 and z1 if this is equal to this okay what you will get is that you will get this integral is equal to this integral alright and that means you are getting equality in pix lemma for all points on the contour okay and but pix lemma tells you that if you get equality even at one point it has to be an automorphism okay so you will get equality here if and only if f is an automorphism even for a single point and if you get equality for a single pair of distinct points you will get equality everywhere because it is an automorphism okay so that finishes the proof of this theorem so let me write that last line and we are done and what it also tells you is that therefore if f is not an automorphism then you have strict inequality and strict inequality means it is a contraction with respect to the hyperbolic matrix okay it is not an isometry it is a contraction with respect to hyperbolic matrix that is exactly what this theorem says. So we get equality for even a single pair is not a z1 of distinct points if and only if integral over gamma mod f dash of z dz mod dz by 1 minus mod fz the whole squared is equal to integral over gamma mod dz by 1 minus mod z the whole squared which forces which forces equality in the conclusion of pix lemma for points on gamma which means f is an automorphism okay. See because you know if you get this integral is equal to this integral then you can bring it to one side and you will get and you know this is always greater than or equal to this okay. So if you bring it to this side I will get integral of a non-negative object equal to 0 and if you have a real integral okay with the integrand non-negative if it is 0 then the integrand has to identically vanish. So you will get equality of this integrand with this integrand at all points on the boundary I mean on the domain of integration which is the path gamma which is the geodesic okay but then pix lemma tells you that you will get equality union at one point it has to be an automorphism but now you are able to get it for all points on gamma therefore it is certainly an automorphism okay. And if it is an automorphism then you always get equality because it is an isometric okay. So if it is not an automorphism then it will be a contraction because it will be a restricted inequality. So if f does not belong to automorphism delta then f is a contraction. So that finishes the proof of this theorem which we will need to proceed with the proof of Riemann mapping theorem okay. So I will stop here.