 Alright so we continue with our discussion of the spherical derivative okay, so there are few things I wanted to point out with regard to the spherical derivative okay, so let me just recall if f of z in a domain d, z varying over capital D which is a domain in C is a meromorphic function that is f belongs to M of D and mind you the set of meromorphic functions on D is considered now as a subset of you know in fact continuous maps from D to the extended complex plane is C union infinity okay, so the script C denotes continuous maps and the point is that you make the meromorphic function continuous even at the poles by defining the function value at the pole to be infinity okay, so you consider it like this and then the spherical derivative of f is f hash of z it is defined to be 2 times mod f dash of z divided by 1 plus mod f z the whole square, this is the definition of the spherical derivative and mind you why did we need the spherical derivative it is because of the following reason, so suppose the you have this is the complex plane with the variable z and you have this suppose this is your this the area inside this dotted region is your domain D and suppose you had an arc gamma inside D now you take the image of this arc gamma under f in the external complex plane okay, so it means that you know you are allowing also the value infinity, so for example you know gamma may pass through a pole of f okay f is a meromorphic function so f is meromorphic on D means f is holomorphic and that is analytic on D except for a subset of isolated points of D where f has poles okay but at the poles also the value of f has been defined to be infinity, so your gamma your curve gamma can pass through the poles and that is a technical thing that I want to explain to you about now you identify this external complex plane via the stereographic projection with the Riemann sphere which I will briefly draw like this, so this is the Riemann sphere which is S2 okay and this isomorphism is actually a homeomorphism given by the stereographic projection, this is stereographic projection with the point infinity going to the north pole which is this point here alright and the fact is that the image of gamma will see gamma will give you you know if you take the image of gamma what will happen is that you will get some curve here on the Riemann sphere okay, see it is a curve in the external complex plane but you know you are thinking of the external complex plane as Riemann sphere so you imagine that the image of gamma is a curve on the Riemann sphere itself okay and what is that curve this is just this is just f of gamma okay this is f of gamma and what is the what is the big deal about this spherical derivative the big deal about the spherical derivative is that you can get the spherical length of f of gamma okay you can calculate the length of that image curve okay and I have put subscript S for spherical length because it is the length you are computing the arc length on the sphere okay and how do you get it you get it in the following way you simply integrate over gamma with the variable see normally you know if you integrate over mod dz if you integrate over mod dz simply on the plane over a curve gamma you will simply get the arc length of the curve okay that is what integrate integrating over mod dz means because mod dz is infinite decimal arc length on the Euclidean plane on on seat on complex plane on the complex plane thought of as a R2 okay it is usual arc length but you know if you put if instead of instead of doing this suppose I put if I add the magnification factor given by the suppose I add the magnification factor given by the spherical derivative so that means I put f hash of z here and do this then what you will get is I will get actually the length of the image curve on the Riemann sphere okay and so this is where the spherical derivative is used okay the spherical derivative will give it is so you know without this if I do not see if I remove this spherical derivative factor okay I will get simply integral over gamma mod dz and that is just length of gamma but if you put the spherical derivative there okay then I will not get the length of gamma but I will get the length of the image of gamma and ref and mind you gamma can pass through here it can pass through a pole the only thing is it means that this image curve will pass through the north pole that is all it is not going to create any problems because if it passes through a pole the function value there is infinity and infinity corresponds to the north pole on the Riemann sphere under the stereographic projection so the point is important the spherical derivative is that it gives you this spherical length okay but there are a few technical things about this there are a few technical things about the spherical derivative which I just indicated towards the end of my last lecture and I want to be more you know elaborate about that so you see so I want to draw your attention to the to this formula which is a formula for the spherical derivative okay there is something that is a little troublesome about this formula see when I define the and you know but before that let me also tell you that here in this formula for length spherical length of f of gamma you know if I replace the spherical derivative if I instead of putting f hash of z suppose I put mod f dash of z suppose I put modulus of the derivative of f then what I will get is actually the and assume that f is you know holomorphic function then I will get the image the length of the image of gamma under f then f will map only into the complex plane if f is differentiable okay everywhere on gamma okay then it is not meromorphic so the image of gamma will lie in the plane itself it is not going to go to infinity okay because there are no poles okay and because f is if you assume f to be an analytic function and then if you integrate over gamma mod f dash of z into mod dz what you will get is length of the image under f but this will be the Euclidean length it will be just the length on the plane but if you integrate over gamma mod dz with the coefficient f hash of z which is spherical derivative and in addition you allow also f to be meromorphic you will actually get the length of the image curve on the Riemann sphere thought of as the extended plane that is what you have to understand okay fine so you know there is a problem with this at first sight with this definition of f hash because you see there is this f dash here okay f dash is the derivative of f at the point z but the problem is if z is a pole then you are in trouble there is at a pole the function is certainly not differentiable it is a singular point it is a it is a pukka singular point it is a it is a honest singular point it is not a removable singularity okay function is not differentiable alright so you are in trouble so you know when I wrote this definition last time you know I was only you know I was trying to heuristically tell you things but now I am going to tell you things more seriously so let us worry about let us worry about this situation so here is my domain D which is the interior of this dotted line it is an open connected set okay this is inside the complex plane and suppose I have a point z0 and of course you know I have this map f f is a meromorphic function on D and you know of course f is taking values in c union infinity and and z0 is a pole of f of order let us say of order n okay and of course you know z0 will go to f of z0 which is by definition infinity okay this is our definition. Now what about the spherical derivative okay see what is so the question is what is f dash of z0 so this is a this is an issue you see because what is f dash of z0 this we have to worry about this that is the reason is because see suppose I have a gamma suppose I have path gamma passing through a pole okay then my formula for the length of gamma the spherical length of the image of gamma under f on the Riemann sphere which is identified as the extended plane what is the formula it is L spherical of f of gamma you know that is what I have shown in the in the previous slide the spherical length of f of gamma is integral you integrate over gamma so I put if I put just mod dz I will get just the length of gamma but I am put the magnification factor given by the spherical derivative of f with respect to z and what is the spherical derivative of f with respect to z it is well it is f hash of z let me again write it it is 2 times mod f dash of z divided by 1 plus mod f z the whole square this is what is so it was there in the previous slide also okay. Now the point is if I put z equal to z0 z0 see now gamma passes through gamma passes through z0 okay so when I calculate this integral on the right side I have when you do when you when you do an integration you know the variable of integration will lie on the region of integration in this case the region of integration is the path gamma so z will pass through z0 it will vary at some point z will become z0 but when z becomes z0 there is this integrand which is f dash f hash of z the spherical derivative that is in trouble because you know f hash of z depends on f dash of z0 in the numerator f f hash of z0 will be will involve f dash of z0 but f dash of z0 does not make sense why because z0 is a pole I cannot differentiate and at a pole I just cannot find the derivative at a pole so what is a big deal so that is so you see this formula as we have written it last time has this issue that has to be fixed and the reason is because the fact is that as I was telling you last time even at z0 this f dash of z0 is not defined but this spherical derivative is defined as a finite quantity that is the beauty that is the reason why this integral works okay that is what I want to explain to you so you see so let me say that so you see let us assume that z0 is a pole of order n so then what happens is that you know you will get a small disc surrounding z0 so let me use a different colour see I will get a small disc surrounding z0 okay I can find a small disc surrounding z0 where z0 is the only pole okay because you know poles of an analytic function are isolated in any case okay and in fact our meromorphic functions are supposed to be having only pole singularities okay these are the only singularities that are allowed so I can find a small disc surrounding z0 where z0 is the only pole and well you know if you call this if you call the radius of this disc as say epsilon okay then in mod z minus z0 less than epsilon which is the interior of that small disc you see you can write f of z you can write f of z as you know g of z divided by s z minus z0 to the power of n where g of where g of z0 is not 0 okay so you can write it like this and of course g analytic in mod z minus z0 less than epsilon okay so okay I need a little bit more space let me rewrite that let me just write g analytic okay I can do this because this is how the function looks near a pole of order capital N okay now watch carefully let us calculate the derivative of f not at z0 because at z0 you cannot calculate the derivative but in the deleted neighbourhood of z0 let us calculate f of z and here also you know when I write this in mod z minus z0 less than epsilon of course z should not be z0 I mean I cannot literally plug in z equal to z0 because z equal to z0 is a pole and z minus z0 denominator vanishes I cannot write that of course we have agreed to put z equal to z0 and equate it to infinity that is when you consider f to be a function with values in c union infinity but nevertheless if you think of it as a usual function then you do not plug in z equal to z0 because you do not divide by 0 okay and that is the situation you must be if you want if you want really differentiate things okay so you see now let us calculate see let us do this calculation see what is so let me write this in mod z minus z0 less than epsilon z not equal to z0 okay what is f dash of z f dash of z is just d by dz of f of z which is now g of z by z minus z0 to the power of n okay and you can calculate this by if you want quotient rule from basic calculus so it will be I will get z minus z0 to the power of 2n and I will get what will I get here z minus z0 to the power of n g dash of z plus minus g of z n z minus z0 to the power of n minus 1 okay this is what I did if I do this computation and you see now what you do is see this is alright for z0 equal to z0 alright and you will certainly have a problem if you let z tend to z0 okay limit if you actually see in the usual sense if I let limit z tend to z0 then what will happen is that the numerator the first term will go alright the second term well it will go provided n is greater than 1 okay if n equal to 1 I will get g z not okay but the problem will be the denominator as z tends to z not I will end up with essentially I will what will happen is that because f has a pole of order capital N at z not its derivative will have a pole of capital order capital N plus 1 at z not okay that is what could happen so it is going to only get worse limit z tends to z not f dash of z will not exist okay and if at the worst case you want to make it exist you can define it to be infinity by thinking of f dash also as a meromorphic function but now with values in c union infinity you can do that but in any case it is not a finite it is not a it is not a proper limit in the usual sense okay you have to include the value infinity but then so I will say limit z tends to z not f dash of z does not it is not a complex number okay it is if you include c union infinity then you can call it as infinity that is but that is not the case okay we do not want to include the value infinity when we are talking about derivatives but you see but look at what on the other hand you look at what is f hash of z look at the spherical derivative if you look at the spherical derivative what I will get I will get see I will get 2 times modulus of f dash of z okay so I will get 2 times modulus of this whole quantity okay divided by 1 plus mod f the whole squared and mod f the whole squared will be mod f is modulus of this quantity so it will be 1 plus modulus of g of z by z minus z not to the power of n the whole squared this is what I will get okay this is what I will get and now you take limit z tends to z not of f hash of z you take you calculate this limit what will happen is you see the in the denominator you have 1 plus mod g z the whole squared divided by mod z minus z not to the 2n okay so denominator will go to infinity the denominator will go to infinity faster infinity the denominator will go to infinity faster than the numerator so this so the whole quantity will be bounded as z tends to z not that is the whole point so you see if you calculate it okay if you calculate it what will happen is so let so this exists so if you write it out you know I am going to get so I will to simplify things I will multiply both numerator and denominator by z mod of z minus z not to the 2n okay which is what is the common denominator so what I will get up get is that I will get well let me write this here f dash of sorry f hash of z is going to be 2 times I will get the numerator of this which is mod z minus z not to the n so I will I will get this mod z minus z not to the power of n g z minus g z oops I think that must have been that is a g dash of z this is a g dash of g dash of z minus g of z n into z minus z not to the n minus 1 mod divided by okay I have multiplied by I have multiplied by this modulus of this quantity mod z minus z not to the 2n okay so that is gone and the denominator I will get mod z minus z not to the 2n plus mod g z the whole square this is what I will get if I multiply by mod z minus z not to the 2n okay and mind you the spherical derivative is an absolute derivative so it is only absolute value it is a non-negative real value by the way. Now you do if you take limit as z tends to z not what is going to happen you see as z tends to z not this term will vanish because z minus z not power n is there and of course this n is of course greater than or equal to 1 okay it is the order of a pole so it is a pole of order 1 or higher okay so this is going to vanish and this fellow here what will happen here depends on whether n is equal to 1 or n is greater than 1 okay see if n is equal to 1 what is going to happen if n is equal to 1 then this term does not exist okay and as I let z tends to z not I will get 2 times mod g z not okay g is anyway mind you analytic it is continuous so limit z tends to z not g z is g z not and modulus is also continuous function so I can push the limit inside the variable okay inside the argument of the function and then that is what I will get in the numerator so this is if n is 1 okay this is if n is 1 and in the denominator what I am going to get this term is going to vanish as it tends to z not I am going to simply get mod g again I will get mod g z not the whole square I will get divided by mod g z not the whole square which is just 2 by mod g z not this is what I will get and mind you g z not is not 0 because g z not is g is the you know if you want g is analytic function divided by z minus z not power n which is equal to f in the in the neighbourhood of f in fact you know g z not is if you check very carefully g z not is the coefficient of the of 1 by z minus z not power capital N if you write out the Laurent expansion okay and that is not supposed to be 0 okay because g f has a pole of order capital N right so this is what you will get and you see and mind you in the case that n equal to 1 g z not is actually the coefficient of 1 by z minus z not power N which is 1 by z minus z not but you know what is the coefficient of 1 by z minus z not called it is called the residue so actually this is 2 divided by the residue of f at z not that is what it is this is nothing but 2 in 2 divided by modulus of residue of f at z not this is what happens if you get if f is a simple pole any capital N equal to 1 alright and the point is that if now if n is greater than 1 everything is gone because you see if n is greater than 1 there is no problem with the denominator I will get mod g z not the whole squared this term is anyway going to vanish and the numerator will also go now numerator has a z minus z not term common so it is going to go so I will get 0 if n if n is greater than 1 so here is the so here is the so of course this is on the left I forgot I have forgotten to write f hash of z so here is the nice thing f hash of z not you can now call it see you can define f hash of z not by continuity to be equal to limit z tends to z not f hash of z okay if you want to think of f hash as a continuous function okay if you want to think of the spherical derivative as a continuous function then it is natural to define f hash at z not to be the limit as z tends to z not of f hash of z okay and you see this what this does is that it makes the spherical derivative continuous even at z not and mind you z not is a pole so what this tells you is that the spherical derivative f hash of z is continuous at all poles so it is continuous throughout the domain and therefore you know because it is continuous at all throughout the domain this formula is valid okay what I really meant here was what is f hash of z not of course f dash of z not does not make sense so the question is what is f hash of z not alright so now you know f hash makes sense even at poles so this integral is well defined there is no issue as if you are you can blindly integrate f hash you cannot blindly integrate f dash because f dash will not exist at a pole you can integrate f dash mod f dash only where so long as you are on a path which is not going through any poles but if it is going through a pole you cannot integrate f dash but you can integrate f hash always even if you are passing through a pole that is a big deal that is a big deal so that is the reason why this formula works and what this calculation we did just tells you is that the spherical derivative is actually 2 divided by modulus of the residue of f at the simple pole z not if z not is a simple pole and it is 0 if not this is if not means I mean pole of higher order so the moral of the story is that you know you are in you are in good shape f hash the spherical derivative is a very nice thing okay and therefore whenever you want to find the arc length you can integrate mod dz over multiplied with with the you know integrand as f hash and that is pretty important okay now and you know again I will tell you why we are doing all this we are doing all this because you know somehow the kind of analysis that is required to prove Picard's theorem is involves Montel's theorem okay and this I will tell you roughly the idea is that you know there are these there is a very close relationship as I told you between compactness and sequential compactness and equic continuity and normal normal convergence okay and boundedness of the derivatives okay so this is a this is a bunch of results in analysis which is usually covered by the Arzela Ascoli theorem okay and that is a that is the Montel's theorem is a is an is something that comes out of that okay and why we are doing all this is because you know you the basically you know the idea is that you want to look at a space of meromorphic functions on a domain okay so you have some domain alright this is a domain in extended plane it could include infinity also okay on that domain you are looking at meromorphic functions alright and you are looking you want to think of them at least as continuous functions so you are allowing the value infinity at a pole okay so you are looking at that space of meromorphic functions and the and you see the the convergence that you are worried about is normal convergence okay it is not uniform convergence everywhere it is only uniform convergence restricted to compact sets which is called normal convergence and with this convergence idea you want to study topology of this space of functions and explicitly what kind of topology you want to study compactness okay you want to study compactness and you know if you have studied for example in Euclidean space compactness is is as good as sequential compactness which is the same as saying that you know every sequence if you have an infinite sequence there is always a convergence of sequence alright and so you have compactness is somehow strongly related to sequential compactness okay so basically you want so basically given a given a sequence you always want to have a convergence of sequence alright and now you want this also to happen for meromorphic functions that is the that is the that is the central idea the central idea is give me a bunch of give me a sequence of meromorphic functions on a domain and now you try to find conditions topological conditions that will tell you topological of course includes also analytic conditions that will tell you that I always will be able to find from this sequence I can find a subsequence which converges but mind you now it is not just convergence is normal convergence because in the context of complex analysis in the context of holomorphic functions uniform convergence will not work you will get only uniform convergence on restricted subsets namely only on compact subsets that is called normal convergence okay so you have to worry about this and in order to do all this see I need to see the boundedness of the derivatives is for example something that is strongly related to all this so I want to be able to work with derivatives but the problem is the functions I am trying to work with are all what they are meromorphic functions and meromorphic functions are not differentiable at the poles they are not differentiable so what do I do at a pole with a meromorphic function what I do is the clever thing is that I do not look at the ordinary derivative I look at the spherical derivative the spherical derivative makes sense even at a pole that is where a spherical derivative comes in okay that is what you have to understand all this is required for me to do analysis on a space of meromorphic functions and that is the kind of see that is the kind of you know analysis you have to do to prove the Picard theorems okay fine so okay so now what I want to do next is I want to tell you well that this other fact that I was telling you last time that you know the spherical derivative has another important advantage the spherical derivative allows you to you know forget meromorphicity so it is a very clever trick you see on the you introduce a spherical derivative because you want to look at a derivative of an function which is meromorphic at a pole for example okay that is why you introduce a spherical derivative but I the it is beautiful that once you introduce this notion you can in most cases you can even forget the pole which means you can reduce everything to just studying analytic functions so that is a beauty the reason is the spherical derivative of meromorphic function is the same as a spherical derivative of its reciprocal okay and what is the advantage of passing through the reciprocal passing to the reciprocal the advantage is that a pole becomes a 0 okay and a 0 is a very nice thing the function is differentiable there alright so that is the advantage so that is a that is a that is an added advantage you get free okay and why is this true this is true because the spherical length that is invariant under inversion okay so that is what I want to tell you about next. So you see so recall that the spherical distance d sub s is invariant under inversion so you have the spherical distance between Z1 and Z2 is the same as the spherical distance between 1 by Z1 and 1 by Z2 where you know Z1 and Z2 are now taken to be in the external complex plane and in the external complex plane mind you 1 by 0 is defined to be infinity and 1 by infinity is defined to be 0 this is the convention so you know the spherical distance is the is actually the spherical distance on the Riemann sphere okay which means the distance between 2 points on the Riemann sphere is given by the length of the minor arc of the biggest bigger circle that passes through those 2 points and which lies on the Riemann sphere okay and that length how do you get it and that length you get it by for example using basic analytic geometry okay it is after all length of arc of a circle and you can always derive its formula okay and that length you transport it by the stereographic projection to the extended complex plane. So the advantage of that is that I can measure for example distance between a point in the complex plane and the point at infinity okay I can do which I cannot do with the usual Euclidean distance because Euclidean distance becomes unbounded as the points as one of the points is fixed and the other goes to infinity alright so and I told you in an earlier lecture that you know the mapping z going to 1 over z which is the inversion mapping that is a map of C union infinity onto itself it is a in fact it is a homeomorphism and the fact is that under that homeomorphism see what it does is that it just maps the extended complex plane back to the extended complex plane it exchanges 0 and infinity infinity goes to 0 0 goes to infinity okay but on the other hand since it is a it is a self homeomorphism of the extended complex plane it will also induce a self homeomorphism of the Riemann sphere because after all the Riemann sphere is homeomorphic to the extended complex plane see whenever in mathematics whenever one object has an isomorphism and you take another isomorphic object then an isomorphism on the first object will automatically induce an isomorphism on the second object which is transported by this isomorphism between them so the inversion will also induce an isomorphism of the Riemann sphere and what is it it is nothing I have asked you to check this you should do it I hope you have done it so it is just rotation of the Riemann sphere about the x axis by 180 degrees that is all it that is what it is okay and you know if you take 2 points on a sphere okay and you take the spherical distance between them that Rql distance now if you rotate the sphere that is not going to change okay so it is invariant under that rotation alright and therefore the net effect is that the spherical distance is invariant under inversion okay now this implies that the spherical derivative of f is the same as the spherical derivative of 1 by f okay and so why is this for this is for f in meromorphic f a meromorphic function on D okay and why is this true because you see I will tell you if you want you can try to do direct calculation okay you can do a direct calculation so you know so what is f hash f hash is just 2 mod f dash z by 1 plus mod f z the whole squared this is what it is alright now this is okay where f dash exists this formula is correct where f dash exists let us assume that for simplicity let us assume that f dash is not 0 at a point suppose the derivative does not vanish at a point 1 by f also makes sense at that point the usual derivative of 1 by f also makes sense at that point so if you calculate 1 by f hash what it will what will it be it will be 2 times you know I will get modulus of 1 by f dash of z divided by 1 plus mod 1 by f of z the whole squared now if you if you calculate this what will I get I will get 1 by f hash is well you know 2 times if you take the derivative of 1 by f I am going to get minus 1 by f squared into f dash this is what I will get okay if I use a chain rule and then I have to divide by 1 plus 1 by f of z the whole squared mod f okay and now if I simplify it you see I will again end up with 2 times mod f dash of z divided by 1 plus mod f z the whole squared if I simply multiply numerator and denominator by mod f is at the whole squared I will end up with this which is the same as f hash okay so this is a very heurist I mean it is a simple calculation the only thing the only problem with this calculation is that you know so I am I am cancelling of mod f dash I am cancelling of mod f okay and to cancel of mod f f should not vanish okay otherwise I cannot cancel mod f in the numerator and denominator so this is okay if f is f of z this is okay at a point where f is not 0 okay so let me write that valid if f is not 0 that is one thing then the second thing is f dash should exist okay derivative should exist otherwise I cannot write so and f dash exist so what I am saying is that the spherical derivative of f and the spherical derivative of 1 by f they are the same you can verify it at all points of f which are different from zeros and poles okay now what you do is you check it at a pole using the same calculation that we did last time alright and you will see by last time I mean just some time ago we did this calculation to calculate the f hash of z0 at a pole z0 that is by basically writing out f locally at the point z0 which is a pole in this form f is equal to g by z minus z0 to the n alright and you use the same calculation if you use the same calculation what will happen is that you can see that when f is 0 at a point that is at a point where f has a 0 or at a point where f has a pole the same calculation is correct what you do is you calculate in the deleted neighbourhood and then you let limit z tends to z0 you do it both at a pole and at a 0 and you will see that the limit will exist and whether you calculate the limit for 1 by f or whether you calculate the limit for f you will get the same thing okay so that is that I will leave it to you as an exercise okay so one of the important things is that see one of the important things is that you know if you take the spherical length of f of gamma okay this is going to be by definition integral over gamma f hash of z mod dz and since f hash is 1 by f hash this is also integral over gamma 1 by f hash of z mod dz and this is by definition the spherical length of 1 by f of gamma and f of gamma and 1 by f of gamma they only differ by an inversion and under an inversion the spherical length should not change so it is correct okay. So this way also you see that the you know you can when you calculate the spherical derivative whether you calculate for f or whether you calculate for 1 by f there is no difference okay so what is the advantage suppose you are proving something involving spherical derivative and suppose you have to deal with a point which is a pole okay suppose I have to deal with function f at a point which is a pole and suppose I am working with a spherical derivative without loss of generality I can replace f by 1 by f because by replacing f by 1 by f my spherical derivative does not change but my pole for f becomes a 0 for 1 by f and 1 by f becomes analytic so I am dealing with a nice analytic function okay so that is the advantage of having the spherical derivative okay. So now what we will do is in the forthcoming classes we will use all this background that we have developed so far to you know in a series of lemmas and propositions and finally theorems we will prove the Picard theorems okay and on the way we get the very important Montel theorem okay so I will stop here.