 Okay so let us continue with our discussion of studying the behaviour of a function in a neighbourhood of the critical point. So we have discussed in the previous lectures you know the how to look at the logarithm the various branches of the logarithm and various branches of the power function okay and for a fractional power alright. So what we are going to do now you start with start with an analytic function f of z having a critical point a critical point at E z equal to z0 so that is f dash of z0 is 0 the critical points are the points where the derivative vanish assume of course we assume that f is non-constant on the domain on which it is defined which includes the points z0 and this will tell you that f dash is also non-constant and well on the other hand f dash is an analytic function also and z0 is a 0 of that and you know the zeros of a non-constant analytic function are isolated. So I can write that so there exists an epsilon greater than 0 or a rho greater than 0 such that well f dash of z is not 0 if mod z minus z0 is epsilon equal to rho okay. So z0 is a 0 it is isolated it is an isolated 0 of the derivative so there is a deleted neighbourhood and of course I should not include the value z0 so I should put this greater than 0. So for every point in this deleted disk close disk centre at z0 including the boundary except for the centre at every other point the derivative does not vanish okay. So any other 0 f dash will lie outside this disk apart from z0 okay z0 is only 0 there in this closed disk and that is just due to the fact that the zeros of a non-constant analytic function are isolated right. Now suppose the order of the 0 of z0 of f dash is m minus 1 and of course I am assuming m is greater than 1 right suppose this the order of 0 is m minus 1 the reason for m minus 1 I could have taken it as m but the reason I am taking it as m minus 1 is because you will see that it helps in our argument otherwise throughout my argument I will have to keep saying m plus 1 which is not very interesting. So well so suppose the order of the 0 of f dash is m minus 1 at z0 okay we say that z0 is a critical point of f of order m minus 1 okay. So the order of the critical point is the order of the 0 of the derivative okay and of course the critical value is the value of the function at that point the critical value is f of z0 this is the critical value corresponding to z0 alright and now the reason why I took m minus 1 is because we can consider the function f of z minus f of z0 okay consider the function f of z minus f of z0 of course this is also an analytic function because it is the analytic function f minus a constant alright. So this is also an analytic function but the point about this function is that it has a 0 at z0 okay and all its derivatives the first m minus 1 derivatives at z0 will also vanish because the derivatives of these are the same as the derivatives of f okay because this differs from f only by a constant. Therefore what happens is that z0 becomes a 0 of order m for this function okay which has a 0 of order m at z0 okay that is the reason why I took m minus 1 here because I get m here if I had taken m I would have got m plus 1 alright. Now this is an argument that we have seen many times so what we are saying is that f takes the value see f takes the value z0 f of z0 at z0 with multiplicity m okay that is the same as saying that fz minus fz0 takes the value 0 at z0 with multiplicity m okay and you know we have already seen this argument that in a sufficiently small neighbourhoods surrounding z0 the multiplicity is constant okay consider delta to be the minimum over mod zeta minus z0 equal to rho of mod of f of z minus f of z0 look at the minimum value of this mod okay and so you are on the boundary of that circle centered at z0 radius rho alright I want to say that f the modulus of fz minus fz0 is always positive on the boundary circle okay and if not I want to modify it so that becomes true perhaps it is already true. Let me do the following thing let me assume let me also assume that I not only assume that f dash is not 0 on this but I also assume that I will also assume that mod of f of z minus f of z0 that is also non-zero okay because that is also non-zero on this deleted neighbourhood okay the reason I can do that is because z0 is 0 of the analytic function f of z minus f of z0 and that 0 is also isolated because f of z minus f of z0 is a non-constant analytic function because f of z is also a non-constant analytic function you take a non-constant analytic function and add a constant to it the resulting analytic function is also not constant so let me write that down assume probably it is not necessary probably it will follow but anyway let me assume it for safety assume that f of z is not equal to f of z0 in 0 less than mod z minus z0 less than rho okay if which means that you know I have already chosen rows is that the derivative does not vanish but maybe if that row does not work I can make the rows smaller I can choose a smaller rows so that this is true and that will happen because z0 is an isolated 0 of f of z minus f of z0 because f of z minus f of z0 is a non-constant analytic function okay. So that means that on the boundary also this is not 0 I mean this is not equal to this which means that the modulus of the difference is non-zero okay so it means that this delta is positive so this will tell you that delta is positive delta is a positive number and then you know once I have this I can define for w with mod w minus w0 less than delta that is the reason I need the delta we have been through this argument many times but anyway define it to be the number of times f takes the value w okay. If you put w equal to w0 which is f of z0 I have not mentioned what w0 is probably I should do that now the critical value is f of z0 equal to w0 I should tell you what w0 is alright. So when I put w equal to w0 I will get n of w0 okay and but then I can change w inside this alright because in this disk this is not going to vanish okay and except yes f of z minus w is certainly going to be greater than or equal to delta alright on the boundary on this boundary so this integral is well defined alright. So the moral story is that this number which we have seen many times before is going to give you the number of times f of z takes the value w okay and we have seen before that n of w is analytic in w and this integer value hence constant okay this is an argument we have seen many times before so n of w is equal to n of w0 and n of w0 is number of times f of z takes the value w0 which is f of z0 and that is m so this is equal to m. So what this tells you is it tells you the following thing for every w with mod w minus w0 less than delta there are n points in mod z minus z0 strictly less than rho at which f takes the value w okay. So let me write that down that is for every w with mod w minus w0 strictly less than delta there exists n, n there exists m points counted with multiplicity counted with multiplicity means some points may be repeated okay, z1 of w etc, zm of w in mod z minus z0 strictly less than rho such that f takes at each of these points f takes the value w okay this is what n of w equal to m means right. So the diagram is something like this okay so the diagram is something like this so here is my complex plane this is the w plane I mean this is the z plane the source plane and then there is the target plane which is in the complex plane and this is the w plane and I have this well I have this disc here centered at z0 radius rho and I have this function f of z w equal to f z which is a mapping and it maps into and I am looking at the at a disc centered at w0 with radius delta okay this is my mapping and what I am saying is that if you give me a point w here if you give me a point w here then you know you get all you get m of these points here okay and they may be there may be repetitions it may it means that all the m may not be distinct point it some point may be repeated with multiplicities okay they could all be one point with m times multiplicity okay which is what happens if w equal to w0 if w equal to w0 then the only z for which f of z is w0 is z0 okay but it is not to be thought of as one point it has to be thought of as m point m points because it is multiplicity m the 0 of f of z minus f of z0 at z equal to z0 has multiplicity m so you should think of z0 as being repeated m times even though it is the same point so when I draw these points that many of them could have be one and the same but these are the zi of w okay. So you know if you look at it it looks a little like it looks a little like the implicit function theorem see I what I am saying is if you take the equation w equal to f of z okay and you take this critical point z0 then I am getting a disc such that if I try to solve for z from w equal to f of z okay then I am getting z equal to m values z1 of w etc zm of w and these are solutions of w equal to f of z namely if I plug them in f of z I get w okay so I am solving for z from w equal to f of z okay so what it tells you is that you are getting m solutions you are getting m solutions for the equation w equal to f of z in the neighbourhood of a critical point z0 that is what it says okay and you know the way you should think of these so you know there are these so we have these functions z1 etc up to zm there are these functions and the way you should think of them as well you should think of them as you know inverses of f okay. So I am putting a I am putting a I am not writing f inverse the convention is to write f inverse only when it is said theoretically an inverse at least in which case f has to be injective alright but here f is certainly not injective okay because derivative has vanished at a point you cannot expect it to be injective at all and therefore I should not write f inverse so I am putting f and putting the inverse in a bracket to tell you that you know this inverse function okay the inverse function has m solutions z1 through zm and these zis are functions of w okay okay and they all solve the equation w equal to f of right now we need we just want to understand what these functions zis what these functions are are they continuous or the analytic etc etc okay and we also want to know how this mapping looks like how can you draw this mapping or how you can visualise this mapping okay. So what we are going to do is we are going to we are going to break this down by using what we have something that we have already seen namely we are going to use the fractional power function, branches of fractional power function which have which come from branches of the logarithm there are m branches for the 1 by mth power of a variable okay and they come from the logarithm by choosing the various branches of the logarithm we will use that to break this map down okay and see how it looks how this map looks and the point is that up to conform conformal equivalence that means up to a holomorphic isomorphism this map really looks like z going to z power m this map from here to here looks like z going to z power m that is the whole point alright that is the reason why we studied the mapping z going to z power m and its branches in the previous lecture right. So let me explain that so what we are going to do is we are going to do the following thing so you know so let us study in a neighbourhood of in this neighbourhood of z0 what is happening see we have we have z0 as 0 of order m of f of z-f of z0 I mean which is f of z-w0 so what we are going to get is f of z-f of z0 is z-z0 to the power of m times some g of z okay where well g of z0 is not 0 and g is analytic in mod z-z0 less than rho okay this is very simple this is just by the Taylor expansion of f okay you have this follows from the Taylor expansion of f about what is the Taylor expansion the Taylor expansion is you know it is f of z is f of z0 plus z-z0 into f dash of z0 plus z-z0 the whole square by factorial 2 f double dash of z0 and so on you go on up to z-z0 to the power of m-1 m-1 by factorial m-1. f to the m-1 derivative z0 and then I will get z-z0 to the power of m times something I will call that something as g of z this is what I will get alright and well the you know from the power m onwards whatever is there I am taking z-z0 power m outside the bracket outside the brackets as common and whatever is inside is anyway a power series it is an analytic function so I call that as gz okay this is the Taylor expansion and mind you all these terms will vanish because z0 is a critical point of order m-1 okay so that means f dash has 0 of order m-1 at z0 so you know f dash f double dash they all will vanish alright so the so f of z-f of z0 will become z-z0 power m into g of z that is exactly what I have written here so this is this is simply from Taylor expansion alright now and mind you g of z0 is not going to be 0 alright that is because f mth derivative of f with respect to z0 is norm 0 right because it is a the order of 0 of f dash at z0 is only m-1 it is not m alright so well now what we are going to do is we are going to do the following thing choose probably it already is true but maybe what I will do is I will choose row smaller if you want to make sure that g does not vanish in mod z-z0 less than row okay choose row smaller if needed to make to ensure g of z is not equal to 0 in mod z-z0 strictly less than row okay I want g not equal to 0 reason why I want g not equal to 0 is because you know I want to write g I want to write branches of g of z to the power of 1 by m I want to do that alright and you will see why I want to do that alright that is why I want g not equal to 0 alright and I am saying g is not equal to 0 it is already true in fact I do not have to choose row smaller alright that is because I have already assumed that f of z is not equal to f of z0 in this okay so this this sentence is unnecessary anyway let it be there so well now so we have m branches of g of z to the 1 by m in mod z-z0 strictly less than row okay. So you see this is again I proved a lemma the previous lecture saying that you know if you have a function which is non vanishing on a simply connected domain then you can find branch of the analytic branch of the logarithm of that function okay and what is what is g of what is g of z to the 1 by m see g of z to the 1 by m is e to the 1 by m log g z this is what is by definition so you know and you know if I have an analytic branch of log g z then e to the 1 by m log g z will give me an analytic branch of g z to the 1 by m alright. So well of course there are there are going to be m branches but nevertheless let me choose one branch so let me write that down we may take so if you we may take integral from z0 to z of d log g of z well this is supposed to give me log g z-log g z0 and if I want log g z I have to add log g z0 log g z0 to this as an analytic branch of log g z in mod z-z0 strictly less than row. So this is the this is an analytic branch of the log okay and of course where when I do this integral from z0 to z I can choose any path okay the integral is independent of the path this is what we saw last time. So this is a this is a branch of logs analytic branch of log g log g z and once you have this branch I can write a branch of g z to the 1 by m as e to the 1 by m into this branch okay. So that is how I get an analytic branch of g z to the 1 by m alright. Now put h of z to be well you know look at look at this expression z-z0 power m alright now this g z can be written as g z to the 1 by m whole power m and I can take m and m common okay and so I can write I can take the mth root and write that as h of z. So I am writing h of z is equal to z-z0 into g z to the 1 by m where g z to 1 by m is analytic branch of the logarithm as I it is an analytic branch as I have defined it here okay. You put h of z is this this is already this g z to the 1 by m is an analytic function and z-z0 is analytic function those is so that this is therefore an analytic function in the disk and the point about this function is that if you calculate it is derivative at z0 it will not vanish okay. So then h is analytic in mod z-z0 is less than rho and h dash of h dash of z0 is not equal to 0 because you know if I differentiate this I will differentiate it using the product rule alright and when I differentiate that is I keep z-z0 constant and differentiate that g z to the 1 by m and then if I substitute z0 it is going to vanish because there is z-z0 outside and for the other I have to add it to the this term kept constant and the derivative of this which is 1 okay and if I calculate it I will get g of z0 to the 1 by m. h dash of z0 is actually it is actually equal to g of z0 to the power of 1 by m this is what it is and mind you g of z0 is not 0 g doesn't vanish where have I written that g of g doesn't vanish at z0 g doesn't vanish at z0 so g g of z0 is a non-zero complex number so it has a logarithm and this is one of and one of the logarithms is one of the logarithmic values will be picked by this branch that I have chosen okay and that is the value here okay and therefore h dash is a non-zero number alright. So what this will now tell you it will tell you the following it will tell you that you know if you it will tell you that h dash is going to be 1 to 1 I mean the mapping h is going to be 1 to 1 in a smaller neighbourhood of z0 probably in this neighbourhood itself by the statement h dash of z is at let us differentiate that it is d by dz of z-z0 g of z to the 1 by m and this is going to be I keep z-z0 constant and then differentiate this I am going to get 1 by mg of z to the power of 1 by m-1 into g dash of z plus I keep gz to the 1 by m constant and I differentiate this and I just have to say that this does not vanish probably does not but maybe let me do the following thing if necessary let me shrink let me make rho smaller alright so that I ensure that h dash does not vanish because after all h is an analytic function therefore its derivative h dash is also an analytic function and so it is a continuous function if a continuous function does not vanish at a point then there is a disc surrounding the point where it does not vanish just by continuity okay. So you know well if needed make rho smaller to ensure h dash of z is not equal to 0 mod z-z0 strictly less than rho okay and well and you know and in fact you can even ensure that but probably I do not even need this what I really want is I want a neighbourhood where h is 1 to 1 okay so that I can invert h alright or to even have h1 to 1 okay this I can do right and I can do this because of the that is because of the inverse function theorem I am using the inverse function theorem here see I h dash is non-zero at a point so it is non-zero in a neighbourhood and then inverse function theorem says that wherever the derivative is non-zero you can invert in a smaller neighbourhood okay. So you make rho smaller if you want and h will become a 1 to 1 analytic function which you know is an isomorphism on to its image because that is what the inverse function theorem says a 1 to 1 analytic function is a isomorphism on to its image the inverse function is also h inverse will also be analytic alright. So now what is the advantage of this the advantage of this is that this diagram can now be split up okay how can it be split up can be split up like this can be split up in the following way then we can understand the nature of these functions z1 of w through zm of w alright. So what we do is we do the following thing we have this so I have this disk here in the z plane so this is disk centred at z0 radius rho and well what do I do I apply h so I put I put a new variable I call zeta is equal to h of z alright I apply h and when I apply h what I am going to get is well mind you h is 1 to 1 okay h is a 1 to 1 function h is analytic alright and I have chosen rho small enough so that h is 1 to 1 okay and you analytic function is conformal okay it is a it maps an analytic function with nonzero derivative okay is a conformal map alright in fact I have assumed that of course h dash is also non vanishing there alright and in fact that will follow if h is 1 to 1 alright and therefore the mapping h is conformal which means that it will preserve angles between curves alright. So what I will get is I will get a if I take the image of this disk I will get something like a disk alright a slight distortion if you want but then essentially it is going to look probably like this I am just drawing it like this but then the point is that z0 is going to go to 0 okay z0 is going to go to 0 because h of z0 is 0 because h of z is z-z0 gz to the 1 by m so if I put z equal to z0 h will go to 0 so it will map the point z0 to the origin alright and well I can choose you know a small enough neighbourhood here of the origin alright so this is how h is going to map and then you know now what I am going to do is that I am going to put another mapping here this is eta equal to zeta to the m okay when I do this the combined map will be z going to h of z to the power of m and h of z to the power of m is z-z0 power m gz which is f of z-f of z0 see I am trying to come to f okay so if I you know how this mapping behaves okay so you know this is what we saw last time well you have this disk and you know if I draw it for m equal to 3 you know how it is going to look like if I start with ray like this alright then the inverse image is going to be 3 of these things I am going to get 3 separated by angles of 2 pi by 3 okay and for any general m you are going to get the inverse image of our ray like this is going to be m rays alright separated by angles of 2 pi by m alright and you know that each sector here of sectorial angle 2 pi by m is mapped by zeta going to zeta power m onto the whole disk alright so this is what we have seen and you know well and we know that there are branches of this here alright we know that there are branches what are the branches the branches are zeta to the 1 by m these are the branches okay and you know what those branches are we have written down those branches zeta to the 1 by m branches are there are n branches and the branches are given like this they are given as e to the 1 by m log zeta this is the first branch zeta to the 1 by m second branch is e to the 1 by m log zeta plus 2 pi i by m this is the second branch and it goes on like this until the mth branch which is e to the 1 by m log zeta plus 2 into m minus 1 pi i by m okay these are the branches of there are m branches okay in this case m equal to 3 you will have 3 branches alright for example if m equal to 3 and this is the point 1 under the 3 branches you will get the cube roots of unity alright in general you will get the mth roots of unity if this point was the point on the real axis with coordinate 1 okay so these are the branches we have seen this okay and all these branches live they are all you know analytic on the slit plane when namely on the slit disk you have to cut out this portion of the negative real axis along with the origin then you know that these all become analytic functions and where they will all live together as a single analytic function it will be above in the in the in the Raymond surface for zeta to the power 1 by m which is the m sheet at covering over this over the punctured disk alright so we have seen this last time and then now what you do is you now you know already when you go from here to here to here you already got f of z minus f of z0 so to get f of z f of z you have to add f of z0 so what you do is you take this mapping this which sends eta to eta plus f of z0 this is just translation if you get this then finally end up with this original diagram you will get this disk well you will get an image the image will be something that contains this disk alright will be will be bigger but this point will now be w0 okay. So this is the whole map w equal to f of z pictured in a neighbourhood of small disk surrounding z0 and in its image containing a small disk surrounding w0 this is how the picture looks like alright and what one needs to do is to look at the solutions alright the solutions in the in this direction okay there are m solutions for every w here there are m points which I call z1w z2w etc zmw okay these are m solutions and I want to tell you that these solutions are also it is obvious that these solutions will be you know analytic on a slit disk okay and you can write down now what these z1w through zmwr alright in fact they are going to be so let me write them down maybe I have space here to do that so you know so this is what I should get if I take the first one I will get this then if I take the second one I will get h inverse e to the 1 by m log w-w0 plus well 2 pi i by m which you can take out and write as e to the 2 pi i by m times this and so on I will get and I will get zm of w to be h inverse e to the 1 by m log w-w0 plus 2 into m-1 pi i by m so these are the functions okay in this case for example you can take this to be principal branch if you want you can take it to be the principal branch so this log I you can take it to be the principal branch then you will get the other branches but the truth is that you need not take it to be the principal branch you can take it to be any one branch then you will get the remaining branches you can of course if you want take the principal branch there is no problem okay and so you see you get these functions the point with these functions is that these zi are analytic on you see they will be analytic on they will not be analytic on the whole slit on the whole disk but I will have to slit out the if you take the principal branch of the logarithm you will have to slit out the portion of the negative real axis from I mean the portion I mean you have to slit out this piece okay you have to cut this out alright and so I will write it on I will write this as mod w-w0 equal less than delta minus this is the line segment from w0-delta to w0 I throw out w0 and this line segment okay on this is a slit disk the slit disk okay these functions are all analytic on the slit disk and what happens is that they are all functional inverses for this so you know see you have this u f is of course not 1 to 1 it is it is m to 1 it is m to 1 alright so there are these functional inverses so that is the reason I am putting f to the minus 1 inside a bracket because it is not 1 there are so many of them and these are the zj's j equal to 1 to m they are the functional inverses okay because this followed by this will give you identity if you plug in zj equal to if you plug in for zj and here I should have written w equal to fz not fz0 should have been w equal to fz see if you plug in for zj in f okay f of zj will give you back w I mean it will give you back correct it will give you back w okay all these zj's are functions of w mind you because this is the w plane this is the complex plane this is the w plane these functions are all functions of w so you take this zj of w you plug it in f of z you will get back w so this followed by this is an identity map that means you have you have m inverses you have m solutions for this equation w equal to f of z in a neighbourhood of the critical point that is what is happening okay and the mapping looks the mapping can be described like this okay so in fact you know if I if so you know if I take the inverse image if I take the inverse image of the disc here this is after all translation again I will get this disc centred at the origin okay and then if I you know when I go like this I am taking the fractional mth power okay so you know it is going to look like this and then from here to here when I go by if I go like this it is h inverse and h inverse is conformal therefore you know this will this will result in something like this okay you will get something like this this is for m equal to 3 so more generally if you write m if you take any m you will get m you know you will get m curves centred at z0 and going out radially okay and that is how the mapping looks like okay so you know if you forget this essentially this is just a distortion of this map this map is just a distortion of this map and what is this map this is z going to z power m it is just a distortion of the power map so what you are saying is if a function f you look at a function the behaviour of the mapping at neighbourhood of a critical point of order m-1 you know up to a conformal twist okay it will look like z going to z power m that is what we have proved okay and therefore it will have m branches the inverse function will have m branches and these are the branches alright and where can you make sense of them all as a single analytic function what you have to do is that you will have to put a Riemann's surface over this which is an m sheet at cover on that all these will become analytic okay and they will become a single valued function so the moral of the story is that you get a single valued inverse for a function even at a critical point but the inverse leaves on a m sheet at covering that is the point okay that is how it looks right so I will stop here.