 Okay, so we have seen in the last lecture the various versions of the monotromy theorem. Now I just want to discuss the application of all this to study the behaviour of an analytic function at a critical point which we had done, which we had begun a few lectures before I started discussing monotromy theory, I want to conclude that now, so you know our situation was like this, we had so let me draw the diagrams so that and remind you of what we have already seen. So you know we had some we had a point z0 in the complex plane and well and you are looking of course this point is in a domain u where an analytic function w equal to f of z is defined and well the analytic function f of z takes z0 to the value w0. So well this is the complex plane, this is the z plane and the target planes of course is the omega plane or the w plane, this is the complex plane which is the w plane and the point z0 goes to w0 which is f of z0 okay and what we have assumed is that we are assuming that z of course you know let me recall everything depends on the vanishing or otherwise of the derivative of f at z0 okay, if f dash the derivative of f at z0 is not equal to 0 then the inverse function theorem will tell you that f is locally 1 to 1 okay, so it means that sufficiently small disc surrounding z0 will be mapped to something that looks conformally like a sufficiently small disc at w0 okay, so what you can expect is a sufficiently small disc will look like something that is conformally equivalent to a disc okay, for example you can get something that looks like a distorted disc okay, for example that is the situation if the derivative of f at z0 does not vanish and that is the consequence of the inverse function theorem right. And this happens when z0 is not a critical point and what happens if z0 is a critical point that is the other case when the derivative of f at z0 vanishes okay and how do we look at that case we look at we try to look at z0 as a 0 of f-w0 okay, so that is the way we think the so you know right from the beginning of this course we have been looking at 0s of analytic functions we have been translating everything to 0s of analytic functions. So what you do is you see z0 is a 0 of f of z-w0 because f of z0 is equal to w0 and you know and f of z-w0 its derivative is the same as f dash of z because after all w0 is a constant alright analytic function minus a constant is also an analytic function because a constant functions analytic for the derivatives of course are the same and suppose inverse function if you know so if f of f dash of z0 is not equal to 0 then inverse function theorem, function theorem implies f is an isomorphism on a small disc centered at z0 this is the inverse function I mean this is the inverse function theorem with applied together with the so called open mapping theorem also see inverse function theorem will tell you that f is at in a neighbourhood of z0 which is not a critical point that is the derivative does not vanish f can be inverted okay open mapping theorem will tell you the image of that neighbourhood is also a neighbourhood and open mapping that the fact that it is a one to one open mapping will tell you that the inverse is also continuous the inverse function theorem will tell you the inverse is actually holomorphic so it is a holomorphic isomorphism. So this is the nice picture that you get the derivative does not vanish at z0 which is the situation when z0 is not a critical point okay but we are want to look at the situation when z0 is a critical point so if z0 is a critical point of order m-1 okay so this is the reason I took m we take m-1 there is because then z0 will become a 0 of order m of f of z-0 okay is it not this is 0 order m of f of z-w0 okay mind you in this case if the derivative is not 0 then z0 becomes a simple 0 of f of z-w0 if f dash of z0 is not 0 then z0 becomes a simple 0 of f of z-w0 okay but if f dash of z0 is equal to 0 that means z0 is a critical point and if it is 0 if it is a critical point of order m-1 then z0 is a 0 of order m of f-w0 okay this is how the definitions are arranged we know that already and well you know so you know what you do is you do you make a you know we factor this through a sequence of transformations so what we did was we first we know we chose first of all we chose a small enough disc surrounding z0 of radius rho where z0 is the only 0 of f of z-w0 and that we can do because you know 0s of an analytic function are isolated okay so choose rho greater than 0 such that f of z-w0 has 1 has is not 0 is non-0 in this derivative neighbourhood which is 0 strictly less than mod z-z0 strictly less than or equal to rho okay I can do this because the 0s after all z0 is a 0 of f of z-w0 which is an analytic function the 0s of an analytic function are isolated of course you know I am assuming the analytic function is non-constant okay and the reason that the analytic function is non-constant is I mean that is because you know throughout I am said I am working with non-constant analytic functions I am not working with constant analytic functions and for analytic functions that are defined on a domain even they are being constant on a locally will also mean that they have to be globally constant that is because of the identity theorem so I cannot even get constancy on a even a small neighbourhood because identity theorem actually says you cannot get even constancy on a subset which has an accumulation point okay so anyway the functions that I am considering are certainly not constant functions so I mean if the function is a constant then its derivative is identically 0 so I will not then that is the only case when the 0s are not isolated because every point it becomes a 0 alright so if a function is constant it has a value 0 at a point then it has to have the same value 0 everywhere so this 0s will be the whole domain alright and then the 0s are not isolated. So whenever you say that the 0s of an analytic function are isolated of course you are worried about non-constant analytic functions okay and we are considering only non-constant analytic functions. So the 0 z0 is isolated so I have a deleted closed neighbourhood of the z0 disc surrounding z0 deleted disc closed disc where there is no other 0 alright and in particular you know if I look at so you know fz-w0 does not vanish even on the boundary circle okay so if I take its modulus on the boundary circle it will have a minimum and I am calling that as delta okay and delta greater than 0 such that delta is lesser than mod of f of z-w0 for mod z-z0 is equal to rho okay. So this is also this is also is true because after all mod z-z0 equal to rho is a circle it is a compact z and mod fz-w0 is a continuous function real value non-negative real value continuous function and you know continuous function on a compact connected set will the image will be an interval and because it is taking non-negative values the left end point of that interval could very well be delta okay and it will be a positive quantity. So you could take delta to be either the minimum value of mod fz-w0 on the circle or anything positive but lesser than that anything will do okay and now what you can do is now you can now look at this delta neighbourhood of w0 okay. So mind you the order of the choices is first I choose z0 I mean I start with this z0 I choose this rho then using rho I choose delta and that gives me the delta neighbourhood here. So the delta neighbourhood here is well I am looking at all w with mod w-w0 less than delta okay then so you have seen then for every w with mod w-w0 less than delta there exist precisely m values z1 of w and so on up to zm of w which are in mod z-z0 lesser than rho such that f applied to z I of w will always end give you w okay. So these are the zi are the so called branches of the inverse of f see when I say inverse of f it is not a set theoretic inverse okay but it is a functional inverse set theoretic inverse will not make sense unless the function is set theoretically injective you cannot talk about the inverse of function unless the function is injective. So the problem is that it is not an injective function alright in fact the point z0 itself is taken with multiplicity m because z0 is a 0 of order m of f of z-w0 therefore the function f of z-w0 has z0 as a 0 not as a simple 0 it is a 0 of order m and m is greater than 1 okay that means that is the same as saying that f takes the value w0 at z0 m times okay alright. So it is not 1 to 1 okay it is a many to one function in fact it is a m is to one function okay and therefore so when you write the inverse functional inverse for every point you will get m values so if you start at the point w here then you know if I try to go back I will end up with m points which are the various zi of m z1 of m z2 of m I am sorry z1 of w z2 of w and so on up to zm of w I will get m of them of course the m of them some of them could coincide but the point is if I count them with multiplicities they have to be distinct and in fact the truth is so long as w is different from w0 they all will be distinctive the truth is they will be distinct if w is different from w0 okay and how do you see that because you see that because you use branches or logarithm so what you do is well you know you have so if you recall what we did in that in our earlier discussion we you know f is we started with the fact that z0 is a 0 of order m of fz-w0 and then you factored out that the term z-z0 power m from the Taylor expansion of fz-w0 centred at z0 okay. So what you write if you take the Taylor expansion that is a power series expansion of fz-w0 at z0 what you will get is fz-w0 is z-z0 to the power of m into some h of z you will get something like this with with you know h analytic at z0 and h of z0 is not 0. So this just happens I mean this is just what happens when I mean this is just a reflection of the fact that z0 is a 0 of order m of f of z-w0 okay so you can factor out the z-z0 to the power of m out we have seen this already that is because of vanishing of the corresponding Taylor the first so many Taylor coefficients right. And then and then now what you do is h is not 0 so since h is not 0 you can find a you can find a branch of the logarithm I mean you can find a branch of the mth root of h in a neighbourhood of z0 okay so we can get branches of so in fact I should say so you know yeah so before I write down let me let me let me tell you what we want to do see we want to take mth root on this side you want to take an mth root on this side so that you know here the z-z0 to power of m when you take mth root you just get z-z0 and then you get h to the power of 1 by m but to write out h to the power of 1 by m you need to know that mth root of h exists okay and mth root of h exists that will happen in a small neighbourhood about z0 because h does not vanish see h does not vanish at z0 and h is analytic by continuity it will not vanish in a small disc surrounding z0 okay so I have a small disc surrounding z0 where h does not vanish in fact h cannot vanish in this whole disc alright because if h vanished on that disc then fz will also vanish on that disc but I mean fz-w0 will also vanish on that disc but I know the only place where it vanishes on that disc is at the centre okay so h really cannot vanish anywhere alright so h is nonzero on that disc and the disc is simply connected and I told you that whenever you have a nonzero analytic function analytic function that never vanishes on a simply connected domain there is always a logarithm there is a there is a there is an analytic branch of the logarithm of that analytic function defined on that domain okay so I have a branch of log h and once I have a branch of log h it is very easy to define h power 1 by m because it is just e power 1 by m log h h power 1 by m is e power 1 by m log h so just to define h power 1 by m all I need is an analytic branch of the logarithm of h and that analytic branch of the logarithm of h does exist in this neighbourhood because it is simply connected and h does not vanish okay so let me write that since h does not vanish on mod z-z0 less than rho which is simply connected there exists an analytic branch branch of log hz on mod z-z0 less than rho there is an analytic branch so we can define h as the analytic branch analytic branch of hz to the power of 1 by m to be exponential of 1 by m log hz on mod z-z0 so I can so I can make sense of h to the 1 by m and once I can make sense of h to the 1 by m I can multiply this h to the 1 by m with z-z0 I will get something whose mth power is fz-w0 now you put so you know g of z which is defined to be this thing z-z0 times hz to the 1 by m by which I mean z-z0 times exponential of 1 by m into log hz okay is such that is analytic in mod z-z0 less than rho and fz-w0 is just gz to the power of m okay. So you know therefore you know the whole mapping that goes from that goes from z to fz can be you know factorized into a series of transformations so what you do is first you apply g so I will write neta is equal to g of z okay so what I do is I get so I go to the c plane which is a neta plane okay now you see what will happen is that if I plug in z equal to z0 then this will go to 0 g of z0 is 0 because you see g of z0 is what g of z0 is I plug in z0 here and there is a z-z0 factor so it will become 0 so g of z0 is 0 therefore what will happen is and mind you note also that g dash is non-zero okay note also that g dash is non-zero that is very important because why do I need it I need it to say that g is actually 1 to 1 in a neighbourhood of see g dash of z is g dash of z0 is not equal to 0 see what I want to say is that at least this argument tells you that g dash is not 0 okay that is all I want okay and the whole point is that g dash is not 0 that will tell you that g is 1 to 1 on a inverse function theorem will tell you that g is going to be 1 to 1 on a small enough neighbourhood of z0 okay so if necessary I even further shrink row I can do that always alright so that is good enough for our discussion so well let me say the following thing so by taking row smaller if needed perhaps you do not need it need to do that we can we can we can we can ensure g is 1 1 on a neighbourhood on mod z-z0 is less than row okay this you can do because you know so I am using the inverse function theorem using the inverse function theorem the inverse function theorem says that whenever you have analytic function which is whose derivative does not vanish at a point then there is a sufficiently small disc surrounding that point where the analytic function is 1 to 1 okay so if necessary I take row smaller I can make g 1 to 1 okay now since and g so you know g becomes a 1 to 1 analytic function on this disc okay and it will take z0 to 0 so the image of and g is 1 to 1 mind you so it is a holomorphic isomorphism because I told you a 1 to 1 analytic function is a isomorphism onto its image the inverse function is also analytic it is a holomorphic isomorphism so if I take the image of this disc I will get some disc like neighbourhood of the origin okay z0 will go to the origin alright so the image of this disc will be another disc like neighbourhood of the origin which you can think of as a distorted disc okay so this is the effect of applying g then now what you do is now from here to here you apply you take the map zeta equal to neta to the power of m okay and you know what this map is this is the power map this power map will you know this will take you on to it will map this on to again a disc like neighbourhood alright and you know it is a m is to 1 map you know that okay so the combination of these two will be z going to g power mz okay but what is g power mz g power mz is just fz-w0 so when you go from here to here you have already got z going to fz-w0 so if you want further to get z going to fz I have to add w0 so that is this final map so this map is zeta going to zeta plus w0 this is the final map it is a translation so this is of course this is 0, 0 goes to 0 and now the 0 is going to go to w0 okay so finally this is how the map factors this is how the map w equal to fz factors alright so I should tell you that this map is going to be this map is just going to be z going to g power mz gz to the power of m which is fz-w0 and then if you followed by this which is translation by w0 you will get fz okay that is how this map factors alright. Now the fact is that I can write out the I can write out how I go from this starting from this w how I got these m values okay how did we get these m values that is because of this you know it is because of that function n of w which we defined and used to prove the open mapping theorem the inverse function theorem and so on you know so you know you define this n of w what is this n of w? n of w is 1 by 2 pi i integral over mod z-z0 equal to rho of the log f of z-w you know this integral this is for mod w-w0 lesser than delta this integral will give you the number of times the function actually it will give you the number of zeros minus number of poles okay but there are no poles because everything is analytic so actually it will give you the number of zeros of fz-w and number of zeros of fz-w is actually equal to the number of values z at which f takes the value w so you are counting the number of times f takes the value w okay so that is this n of w alright and by of course you know this is a logarithmic derivative it is f dash of z divided by f of z-w that is what this integrand is that is logarithmic derivative of fz-w alright and you know that this n w we have already seen this n w is an analytic function and it is an analytic function which is defined on this disk okay and it is but it is integer value because of the residue theorem or the argument principle whichever way you want to see it. So an integer value analytic function has to be a constant if it is defined on a domain okay therefore n of w is a constant and therefore n of w is actually equal to n of w not which is equal to m so you will get that this is equal to m okay this is equal to n of w not which is actually equal to m. So what this tells you is that you take any w with in this disk mod w minus w not less than delta then there are precisely n values of z z1 of w z2 of w etc up to zm of w where f takes the value w okay so you get precisely m values z1 of w through zm of w at which f of that each of those values is w you get these distinct ones and the fact is that all these are distinct there you only this n of w is just the number of points z counted with multiplicity okay but it is just n of w just tells you how many times fz minus w has zeros okay but those it is only counts with multiplicity but the analysis actually shows is what it shows is that all these all these for w different from w not all these z's are different okay why are they different that is because you see you can write down you can write down the inverse function okay so how will you get the inverse function when you write down write down the inverse function what you will do is well the of course here I can write g inverse because g is 1 to 1 okay inverting this is very easy this is just g inverse because it is a 1 to 1 function translation is also 1 to 1 so the inverse function is translation by minus w not okay the translation by w not is undone it is reversed by translating by minus w not so this inverse function is also easy this is a 1 to 1 function so holomorphic map the Mobius transformation so you can easily reverse this so only thing that gives you to gives all the branches is all these branches of the inverse function of this which are the mth root functions okay so here you get all these so many branches of zeta to the 1 by m you get m branches now what you do is that you combine this with each of these m branches and then you follow by g inverse you will get all these m branches which are which are given by w going to z1 w going to z2w etc okay that is how you get all these m branches and all these m branches are different for w different from w not why because these m are different the m branches of the m the m branches of the mth root function are all different because there are m different mth roots for any non-zero complex number that is the reason why for w not equal to w not the zi's you get the zi of w you get they are all different okay so you get exactly m of those values for w not equal to w not of course when w is equal to w not you get z not okay but you get z not with multiplicity m if w is not equal to w not you get various zi's and there are m distinct zi's that is how it behaves so what happens is that you get branches like this you get all these branches zj of m zj of w okay and where are these branches defined well the branches are all defined on you know the point is that somewhere in the middle I am using the mth root function and the mth root function you know pretty well depends on the logarithm and the logarithm is analytic only on a slit plane so you know when I define this zeta to the 1 by m I am using a logarithm here okay see what is zeta to the 1 by m it is actually e to the 1 by m log zeta and when I do this I get m branches what are those m branches well I will get e to the 1 by m log I will take principal logarithm zeta and then I will get plus 2n pi i these are the various branches and all I have to do is now I have to read this n mod m so I will just have to put n equal to 1 etcetera up to m either I start I do from 1 to m or I start from 0 to n or r or n is equal to 0 to m minus 1 you will get exactly precisely m of them okay normally if this 1 by m is not there you will get many but because this 1 by m is that this n has to be read mod m okay. So these are you get so many branches and because of these m branches when you invert you get these m branches and where are these analytic see each of these branches analytic only on the slit disc I have to throw out this I have to throw out I have to throw out the portion of the negative real axis including the origin and when I translate it I will get I will have to throw out this region which is actually this line segment from omega not minus delta, omega not this line segment I have to throw it out. So the moral of the story is that the Zj of m the Zj of w are analytic on the slit disc mod w minus w not less than delta minus from this you throw out the you throw out this portion of the diameter the left of the centre which is w not minus delta comma w not you throw this out this is the slit disc they are analytic on that okay and why you have to do that is because these fellows in between these logarithms you want the you want these logarithms to be analytic so you have to throw this out and that that effects that effectively means you have to throw out here okay. So these are analytic but the point is where will they all glue to give a single valued function that will be on a on the Riemann surface which sits above this punctured disc so what happens is that the situation is the Zj of w glue together to give an analytic function Z of w on so you know let me use let me write Z of w tilde on the Riemann on the on a Riemann surface on an m sheeted Riemann surface over this punctured disc so you know situation is like this I have this punctured disc this is w not this is this is delta okay on this I have this Riemann surface you see yeah it is again it is little difficult to draw but then okay I have this m sheeted Riemann surface 1 2 etc up to m I have this m sheeted Riemann surface okay and on on sheet so I have this I have this function EZ of w tilde where you know w tilde is a point on this which goes down to a point w here okay each w has precisely m inverse images this is a cover this is a covering map okay this is a Riemann surface how do I get this Riemann surface is by taking m copies of this slit disc and then you know joining them together so that all the I am actually doing actually what I am doing is I am constructing the Riemann surface of this logarithm I mean I am actually constructing the Riemann surface of zeta to the 1 by m and simply translating it here I am not doing anything else okay so this is the this is that Riemann surface on this I have this function Z of w tilde okay now that function will go back to the thing that I started with namely it will go to the it will go to that disc centered at z0 radius rho I think here radius rho and what will happen is that you know if I start with w I will get so many points w1 tilde w2 tilde w3 tilde and so on the last sheet I will get wm tilde I will get all these points and if I take Z of wi tilde that is wj tilde I will simply get Zj of w I will get all those m those m functions they will all give me if I go to different pre images of this point there is this projection map which is a covering map every point has m pre images because after all I have taken m copies of this slit disc and you know I have I have cut and paste the negative real axis carefully okay. So every point will have m pre images and when I apply Z to each pre image I will get the corresponding Zj which will be the Z Z restricted to each sheet is the Zj okay Z restricted to the first sheet is Z1 Z restricted to second sheet is Z2 this way I get all the Z's but the beautiful thing is when I go up this all the Zj's they become they all glue together to give a single analytic function and that they give first of all give rise to a single function and the point is that function is analytic that is a beautiful thing each Zj if I started from here it is not analytic because I will have to throw out this I have to throw out this slit because along that slit it is it cannot be defined so that it becomes analytic but if I go to the Riemann surface I am able to cross this because the reason is because as I go from one sheet to the next sheet I am doing analytic continuation I am going from one branch to the next branch by analytic continuation therefore I am actually getting one function on the Riemann surface above so here is where I am using the fact that you know every Zj if I even though every Zj is not analytic on this on this slit I can analytically continue it to the next Zj across that slit and that is the reason why on above I get a continuously I get one single analytic function because I am there is no obstruction to analytic continuation along that slit that is why all of them glue to give you an analytic function. Now the point is that these the story is we have seen so much already the story is that all these Zj's they are all algebraic so you know so this is what I wanted to say so you know in fact I told you that if you write Z-Z1w you take the product up to Z-Zmw ok if I take this product then this part will be just if I write it out I will get the elementary symmetric functions in the Zi's I will get a polynomial monic polynomial in Z of degree m with coefficients elementary symmetric functions in the Zi's which is what I explained so I will simply get Z power m so I want to end by saying the following thing this A sub k's they are all the symmetric functions in the Zi's ok. The point I want to make is that this A sub k's you see the Zi each individual Zi is not analytic at that slit ok each individual Zi is not analytic at that slit. So if you take any elementary symmetric function of these Zi's which is one of these Ak's you do not expect it to be analytic on that slit but the fact is it is and what is the reason the reason is you see if you take a point here ok and go around once across you cross this slit if you go around once what will happen is the Zi's will get permitted ok because if you go around once effectively we have seen that if you go around once the origin the logarithm from one branch of the logarithm you move to another branch of the logarithm alright. So that will tell you that if you start with any point here and you go around once you are going to permute the Zi's but if you permute the Zi's the Ak's will not change because they are symmetric functions. So what is the moral of the story the moral of the story is each of these Ak's you go around once analytic continuation leads back to the same Ak ok therefore on the punctured disc each of these Ak's will define we will define a single valid analytic function. So all the Ak's will become analytic further the only problem is at the point W0 and that W0 also Ak's will be analytic why because when W is equal to W0 ok then each Zi of W0 will become Z0 so left side will become Z-Z0 power m and therefore what happens is that each Ak is an analytic function which is which has a limit at the point W0 therefore it is also by Riemann's removable singularity theorem it is also analytic there. So the moral of the story is all the Ak's are analytic on this whole disc so what you have done is you have shown that all the Zi's they are algebraic all the branches which live only on the slit disc they are any way solutions of an algebraic equation it is a polynomial in Z whose coefficients are analytic on the whole disc the coefficients are the Ak's so though the Zi's are analytic only on the slit disc their symmetric functions are analytic on the whole disc that is a beautiful thing ok that is the whole point that is here is where I am using the fact of analytic continuation alright. So it gives you a very beautiful picture of the behaviour of an analytic function at a critical point ok so I will stop with that.