 Okay, so let us continue without the discussion about the inverse function theorem. So let me quickly recall you start with the analytic function analytic at a point z0 assume that the first derivative at z0 is non-zero that means you are saying that z0 is a 0 of order 1 of the function f of z-w0 which is also a non-constant analytic function and then you choose a disk centered at z0 radius rho such that on the closed disk z0 is a only 0 of f z-w0 and on the boundary of the disk fz-w0 does not vanish so if you take its modulus that is bounded below by a certain delta and for this delta you look at all w such that mod w-w0 is less than delta and the claim is that the that f is f is actually 1 to 1 on this disk and the inverse of f is given by this formula okay. So what we have proved so far is that we have proved that f is 1 to 1 on the disk we have proved that f assumes every value w such that mod w-w0 is less than delta every such value is assumed by f in this disk centered at z0 radius rho okay and which is a re-statement of the fact that f is 1 to 1 on this disk okay I mean given the fact that the number of times it assumes the value f assumes the value w is exactly the same as number of times f assumes the value w0 for any w and but the number of times it assumes the value w0 is 1 therefore the number of times it assumes the value w is also 1 okay. So you get both the you get the 1 to 1 nature of f okay now the only thing I have to show is that this is the this formula for f inverse w is correct okay so I let me let me try to explain at least the logic behind that formula okay so you know our so let us go back to this so let us go back to this diagram see you have on the so you have on the on the on the z plane we have this point z0 and we have this disk centered at z0 radius rho okay and then on the on the complex plane that is given by the omega w plane okay again we have this this other disk centered at w0 radius delta we have this disk okay and of course f of z0 is equal to w0 okay and f inverse is f inverse is kind of you know defined from here to here this is f inverse okay it is at least defined it is at least defined properly as a function because f is 1 to 1 restricted to disk to this disk alright and think for a moment as to why this formula why should you able be able to you know expect such a formula first of all you should get a feeling for why you should get a formula like that okay so what I am trying to say is see do the following thing choose I mean it is I just want to say that this formula is a it is just Cauchy's integral formula for f inverse if you look at it very carefully so you see what is so let me so let me you know go back and let me rub this off so that I can use this side of the board and let us for a moment you know recall the Cauchy integral formula the you know the first order Cauchy integral formula or the 0th order Cauchy integral formula okay the first of the integral formulas you see suppose you have so this is a Cauchy integral formula what is the Cauchy integral formula so you know you assume that f now let me not use f let me use h okay and let me use some other variable okay so let me use instead of using z let me use neta okay h neta analytic or holomorphic analytic function of eta on and inside a simple closed contour gamma in the in the in the neta plane so you see the diagram is something like this I have the complex plane and I have the variable is neta so I am calling it the neta plane okay I am purposely using neta because I do not want to use z okay and there is some well there is some simple closed contour gamma okay and mind you so it is a closed curve it is simple that means it has no self intersections and it is a contour means that it is piecewise smooth that is it is a continuous image of an interval that is what it is to say that it is a path for a continuous path further you want it to be piecewise smooth which means that if you parameterize that then as a function of the parameter it should be not only differentiable but the derivative with respect to the parameter should be continuous and that should be at least piecewise continuous on the interval okay that is what it means. So mind you all this is needed to be able to do the path integral on a contour okay so then you know if I take a point neta not here okay what is Cauchy integral formula it says that if you calculate 1 by 2 pi i integral over gamma of h of eta d eta by eta minus eta not what will you get you will simply get h of eta not this is the Cauchy integral formula this is the first Cauchy integral formula right of course if you differentiate both sides with respect to eta not and you allow the differentiation to go past the integral sign you will get the successive higher Cauchy integral formulas okay which will tell you that the higher derivatives at eta not are expressible in terms of this integral with a factorial appearing on top and then you will have higher powers of eta minus eta not appearing in the denominator okay alright so let us apply this to our situation in our situation we are on the w plane okay and on the w plane there is that point z not I am sorry there is a point w not and there is that disc centered at w not radius delta okay and we are having this function f inverse of w you are having this function f inverse of w you see finally we are going to our aim is to show that f inverse w is an analytic function okay of w that is our aim okay and then you have to show that the analytic function is given by this formula this is what we want to prove but for a moment to get an idea as to how this formula came the first place let us assume that f inverse w is actually an analytic function of w okay then see take the w here take a w here okay and what you do is well you know see it is as I explained here the image of this disc contains the disc centered at w not and radius w okay in fact it will also contain more or less it will contain the boundary of the disc okay and what I want to say is for example you know if I take a small curve here gamma surrounding w okay and if you believe that f inverse w is an analytic function of w and what then what will the Cauchy integral formula give you for f inverse w it will tell you that if I take 1 by 2 pi i and integrate over gamma f inverse w f inverse of e eta d eta by eta minus w I am supposed to get f inverse of w this is just the Cauchy integral formula the Cauchy integral formula says that mind you whenever I write this eta the d eta at the moment I write d eta and eta is a dummy variable which lies on the on gamma okay so I did not want to write f inverse w d w because I want to put then I will have to put w minus w which does not make sense I wanted w here so I have changed dummy variable to eta okay and this is just the Cauchy integral formula which is valid provided f inverse is analytic function of w okay now you do the following you put zeta to be f inverse eta you put this if you put zeta equal to f inverse eta what you will get is that means that f of zeta is eta alright and it will mean that d eta will be f dash of zeta d zeta so f dash of zeta d zeta will be d eta okay if you write it like this symbolically the element of integration okay and you know this then this formula you know mind you f is of course f is analytic and I assumed f inverse also is analytic therefore f and f inverse are isomorphisms mind you okay they are isomorphisms and therefore what I am doing is I am simply in this integral I can make a change of variable if I make a change of variable what will happen to that integral that integral after change of variable will simply become I will get the following things so let me write it here you will get f inverse of w is equal to 1 by 2 pi i integral so now you see I am changing the variable from eta to zeta okay and now since eta lies in the w plane f inverse eta will lie in the z plane so zeta will lie in the z plane okay so what will happen is that I will have to take f inverse of this curve okay I am making a change of variable in my integration and if I plug it here I will get so let me plug it here f inverse eta so this will become zeta f dash of zeta d zeta divided by eta is f zeta minus w okay and if you look at it this is the same as this formula here that is the same formula that has appeared here only thing is that the boundary curve that I have taken which should be f inverse gamma is actually the is simply the circle centered at z0 radius rho with the positive orientation of course okay and that this can be replaced by that f inverse gamma can be replaced by this boundary curve is justifiable because in the region between the two curves there are going to be no singularities okay the version of Cauchy theorem says that between two curves one contain inside the other if there are no singularities then the integral over the smaller curve is the same as the integral over the bigger curve okay so in principle I can actually replace this f inverse gamma by mod z minus z0 equal to rho you must understand that since f inverse is analytic okay since f inverse is one to one f inverse of gamma will also be a simple closed curve because it is one to one continuous and it is conformal it is analytic so this will also be a curve simple closed curve it will also be a simple closed contour okay and the fact is that this can be replaced by that okay so the moral of the story is that if you assume that f inverse w is analytic as a function of w then it is very clear how this formula comes this formula is not mysterious okay you can derive it in this sense it is just another avatar of the Cauchy integral formula for f inverse w okay that removes the mystery out of this formula now what you do is that is just to you know psychologically satisfy you that this is not something mysterious this is something that you can guess very easily okay so there are two things in mathematics one thing is to guess the right formula then the second thing is of course to really prove it without any you know gaps without hand waving so now you have to prove this formula that you have to prove that this is the correct formula that can be done in a completely different way once you believe the formula you can always find so many ways to prove it so psychological difficulty is always in trying to first of all believe that certain formula is true so that is also very very important so that is what I demonstrated now so let us come back to proving this formula so maybe I will let me go somewhere for example okay so let me rub this off so that I can use this part of the board I do not need this anymore let me continue here okay so let us try to prove this to prove f inverse of w what w-w0 this is what you have to prove this is the this is the inversion formula this is what you want to prove okay the proof is pretty easy because you look at the function zeta f dash of zeta by f of zeta minus w look at this function look at this function for mod zeta minus z dot strictly less than or equal to rho look at this function look at this function on this closed disc okay that means I am taking the boundary curve and also the interior on this if you see this is an analytic function except for a simple pole at zeta equal to zeta see it is analytic in this closed disc except except for a simple I should not say with except for a simple pole at this point see that function is analytic on the closed disc the only place where it will lose analyticity is when the denominator vanishes and the denominator will vanish exactly at zeta equal to z okay and why that is because at zeta equal to zeta f of zeta is f of z which is equal to w and therefore the denominator is 0 that means z is a 0 of f of zeta minus w but 0 of what order 0 of order 1 because f is 1 to 1 it is only 0 of order 1 okay therefore it is 0 of order 1 of the denominator so it is a simple pole so it is a function that has a simple pole at only one point in the closed disc every other point is analytic now what is the residue theorem the residue theorem says that you take if you integrate a function over a boundary curve and divide by 2 pi i okay then you will simply get sum of residues of f at the poles so by the residue theorem by the residue theorem 1 by 2 pi i integral over mod zeta minus z not equal to rho of zeta f dash of zeta d zeta by f of zeta minus w is actually the residue of the function zeta f dash of zeta by f of zeta minus w at the point z because z is the only pole and it is a simple pole and what is the residue at the simple pole it is by taking a limit as z as the variable tends to the pole of the function multiplied by the variable minus the pole okay so if you calculate this this is actually limit zeta tends to z zeta minus z multiplied by this function zeta f dash of zeta by f of zeta minus w okay and if you compute this is actually equal to z okay that will show that the value of this is equal to z in other words the value of this is equal to z which is f inverse w and your formula is proved okay why is the value equal to z it is very simple you can write it out that is pretty easy to see this is actually limit zeta tends to z you see I will get zeta f dash of zeta by let me group this f zeta minus f of z because w is f of z mind you divided by zeta minus zeta tends to z but limit zeta tends to z of this is precisely f dash of zeta it is it is f dash of yeah limit zeta tends to this is f dash of z okay and limit zeta tends to z of the numerator is z f dash of z so what I will get is this will just be z f dash of z by f dash of z that is simply z and that calculation is valid because f dash of z is not 0 f dash is non-zero so that is that is something that you will have to convince yourself that f dash is non-zero it should be obvious okay so if you cancel f dash of z out you will get z and that proves this formula okay and so we get f inverse w is z is this 1 by 2 pi i integral over mod zeta minus z not equal to rho zeta f dash of zeta is zeta by f of zeta minus w you get this so you get this formula the inversion formula okay now the only thing that you will have to worry about is the statement that f inverse is an analytic function of w for w lying inside this disc and so in other words you have to show f inverse of w is actually a differentiable function of w for w lying in that disc centered at w not and radius delta okay and in fact the proof that it is differentiable in that disc is you know it is the same kind of technique that we have already seen it is the same kind of technique that we proved I mean which we used to prove that n of w that we defined in the proof of the open mapping theorem the function n of w which is just the number of times the function f assumes the value w we proved n of w is actually analytic function of w it is the same technique that you have you have to use to show that f inverse w as defined like this is an analytic function for w analytic function of w for of course mod w minus w not lesser than delta okay this is something that that can be shown right so let me outline that so that is the only thing that is left out to show that f inverse is an analytic function of w okay so probably I will write now I can now let me rub off this so let me write here it remains to show that f inverse is an analytic function this is what is this is the only thing that needs to be proved mind you that f inverse is already continuous you know why that is because you see f is already an open map okay f is a non-constant analytic function we have already seen in an earlier lecture that a non-constant analytic function is an open mapping so f is an open map but when f is an open map it means that if f has an inverse that f inverse is continuous so the continuity of f inverse is already there so you must remember that because of open mapping theorem f inverse is continuous already that is already there okay now we will now we can prove the analyticity of f inverse with respect to w okay that can be done as follows so let me just indicate what needs to be proved and then we can I mean you can fill out the details so you know let us instead of using instead of calling this function as f inverse w let us give it some other name let me call it as h of w so that is easier to write. So put h of w is equal to 1 by 2 pi i integral of mod z- integral of zeta-z0 equal to rho zeta f dash of zeta d zeta by f zeta-w for mod w-w0 strictly less than delta put this okay so let us call that is here mod of f zeta-w0 is always greater than or equal to delta for zeta-z0 equal to rho I mean this is how we chose the delta first and then I mean you first chose the rho and then the delta after that and this how we chose okay rho and delta. Now what do I have to show I will have to show that h is an analytic function of w okay so what I will do is I will have to calculate h of I will have to calculate I will have to show limit w delta w tends to 0 h of w plus delta w minus h of w by delta w exists I have to show that this limit exists for every w with mod w-w0 less than delta this is what I have to show if I show this I am showing that h is differentiable on the disc mod w-w0 less than delta and that the fact that the functions are differentiable at every point in a disc means it is analytic in the disc because analytic is basically differentiable once at every point in an neighborhood of a given point right but what is this quantity this is this quantity okay. So for those of you who are watching this lecture an exercise is to just stop at this point and then just recall how we prove the function n of w is analytic in the case of the proof of the open mapping theorem and literally follow the same ideas okay but anyway let me continue so what is this quantity this quantity is if you calculate it is in so it is 1 by 2 by I and then there is a delta w and I will get integral over mod zeta-z0 equal to rho and I will get zeta f dash of zeta so there is a zeta f dash zeta is a common fact it is a common term so what I will get is I will get simply get 1 by f of zeta-w-delta w-1 by f zeta-w this whole thing into zeta f dash of zeta dz this is what this difference will be okay and if you compute it it will be 1 by 2 pi I delta w well if I take LCM then I am going to end up with integral mod z-z0 is equal to rho I am going to get a delta w on top zeta f dash of zeta dzeta divided by f zeta-w-delta w into f zeta-w okay. And then since delta w has got nothing to do with the variable of integration zeta okay so I can cancel this delta w out and what I will get is I will get 1 by 2 pi I integral over mod z-z0 equal to rho of zeta f dash of zeta dzeta over a product of these two terms f zeta-w-delta w into f zeta-w okay and what I am supposed to show is that the limit of this quantity as delta w tends to 0 exists okay you if one is very naive one can one will pass the let the limit pass through the integral and if you apply the limit if you allow the limit to go inside the integral okay which is not always possible but if you naively do it what will happen is that the limit will be 1 by 2 pi I integral over mod z-z0 equal to rho zeta f dash of zeta dzeta over f zeta-w the whole square that is what you will get which is the expression that you will get if you take this and differentiate it with respect to w if you what is h dash of w is differentiation of h of w with respect to w so you have to differentiate this with respect to w okay that means you have to apply d by d w to this expression but then you know allow naively the d by w to pass through the integral if you allow that then you will be only differentiating the what is the the integrand and the variable with which you are differentiating is w which is which is different from the variable of integration okay so if you differentiate this this integrand with respect to w you will simply get zeta f dash of zeta by f of zeta-w the whole square that is what you are going to get and that is what you are getting here if you let the limit delta w tend to 0 pass into the integral okay so the moral of the story is from here how do you prove that the limit of this as delta w tends to 0 is this expression with f zeta-w the whole square the denominator okay. So let me write that out so let me get rid of this so that I can use that part of the board see we claim h dash of w is actually 1 by 2 pi i integral over mod zeta-z0 equal to rho into differentiate the integrand with respect to w which is zeta f dash of zeta d zeta over f zeta-w the whole square we claim this is this okay so in other words you are claiming the limit as delta w tends to 0 of this quantity is that that is limit delta w tends to 0 of 1 by 2 pi i integral over mod zeta-z0 equal to rho of this quantity which is zeta f dash of zeta d zeta over f zeta-w-zeta w times f zeta-w is equal to this quantity which is 1 by 2 pi i integral over mod zeta-z0 equal to rho of zeta f dash of zeta d zeta over f of zeta-w the whole square this is what you are claiming so you have to show the limit of this quantity is equal to this which is effectively differentiating under the integral sign and I tell you that that is always possible in Cauchy theory I mean that is the whole point about studying analytic functions okay in fact I told you that Cauchy integral formula itself is a example of differentiating under the integral sign the higher Cauchy integral formulas are gotten from the first Cauchy integral formula by the literally differentiating under the integral sign and all this is possible because you are working with analytic functions that is the whole philosophy so you see I have to show limit of this quantity equal to this okay but now how do you prove that the trick is you absorb this also to the left side and show that the limit of the difference is 0 okay and why I can absorb this into the limit is because this has got nothing to do with the limit the limit is with respect to delta w going to 0 and this has got no delta w term. So that is you have to show limit delta w tends to 0 of if I absorb it onto the left side I will get 1 by 2 pi i integral over mod zeta-z0 equal to rho of if I take this minus this okay if I take LCM what I will end up with is I think I will get a delta w into zeta into f dash of zeta into d zeta divided by f zeta-w-delta w times f zeta-w the whole squared is 0 this is what it will be it will turn out to be okay because if you push this whole term inside the limit on the left side okay and then take the integral out and if you take zeta f dash zeta d zeta out and you take LCM this is what you will get okay. So and which is again correct because if you again naively let limit delta w tend to 0 inside the integral okay there is a delta w on top so the integrand will vanish therefore the integral is 0 integral of a 0 integrand is 0 if you want to think of it like that. So this is again still you know it is still about trying to pass the limit into the integral that is what you must appreciate but we have simplified it to a very nice level but so but now what you do is to show that the limit of something is 0 you show you use now you go back to basic analysis and say when is the limit of one a certain quantity equal to 0 as delta w tends to 0 you use the epsilon delta condition you say that given an epsilon greater than 0 you can make the modulus of this quantity smaller than epsilon provided you choose delta sufficiently small such that mod delta w can be made less than delta provided you can find such a delta okay. So you go to the epsilon delta definition of a limit and that leads you to estimate this integral okay and to estimate the integral you use the ML formula which is the which is the formula which says that modulus of the integral is less than integral of the modulus. So if you if you apply that well let me let me quickly write that out it is pretty easy so what you do is you do the following thing you just you just estimate let us estimate this modulus of 1 by 2 pi i integral over mod zeta minus z0 equal to rho of this guy delta w zeta f dash of zeta t zeta by f zeta minus w minus delta w e to f zeta minus w the whole squared modulus of the integral is less than or equal to integral of the modulus so it is 1 by 2 pi integral over mod zeta minus z0 equal to rho okay mod delta w and I will get mod zeta mod f dash of zeta mod d zeta by mod f of zeta minus w minus delta w e to mod f zeta minus w the whole squared okay I will get this and the point you have to understand is that finally the if you if you look at see mod of fz mod of f zeta minus w is bounded below by a certain delta w okay which is positive because you are on this boundary circle alright and if you recall this is how we proceeded in the proof of the open mapping theorem and for mod delta w sufficiently small since mod of f zeta minus w minus delta w tends to mod of f zeta minus w as delta w tends to 0 you can you can ensure that modulus of f zeta minus w minus delta w is also greater than or equal to delta w by 2 okay because this is approaching this and this is greater than or equal to delta w so you can certainly make delta w small enough I mean this is small little delta w okay as this this delta w approach is 0 okay that this is greater than or equal to small delta w and that this approach is that will tell you that this also can be made lesser greater than or equal to half of that okay which is positive okay and then of course mod z mod zeta mod f dash of zeta they are both bounded mod zeta f dash of zeta is is less than or equal to some m double prime on mod zeta minus z not equal to rho because z f dash of z is an analytic function f is analytic so f dash is analytic and z into f dash of z is also analytic and you have looked so it is continuous so mod of that is continuous and you are looking at a continuous function on a compact set compact connected set so it has a minimum it has a maximum and this is the maximum value so it is bounded okay. So what you will get is that so all this will tell you and if you integrate over mod zeta minus z not equal to rho mod d zeta you will simply get the length of the arc namely the perimeter of the circle okay which is 2 pi rho so what you will get is you will get that this quantity is less than or equal to 1 by 2 pi into I will get this is greater than or equal to delta w so reciprocal of this is less than or equal to 1 by delta w and square of that will be 1 by delta w squared so this will and this will contribute to a 2 by delta w so I will get a 2 by delta w then I will get a 1 by delta w squared okay and then I will get of course I will get a mod delta w on the numerator that will come out as this and this mod zeta f dash of zeta will give rise to an m double prime and what is left out is integral of mod d zeta over this curve which is 2 pi rho okay so which is actually constant times mod delta w this is what it is and that will go to 0 as delta w tends to 0 that will go to 0 as delta w tends to 0 because this constant is independent of the triangular delta w okay. So moral of the story is that the limit of this quantity as delta w so as triangular delta w goes to 0 is certainly 0 and that is what you wanted to prove and that completes the proof of the fact that this function is an analytic function of w and that function is precisely f inverse w if I recall therefore this tells you that f inverse is an analytic function of w okay and that completes the proof okay so I will stop here.