 Okay so this is the continuation of the previous lecture so let me recall what our assumptions are we have a domain in C2 the product of C with itself it is an open connected subset of C2 and we have a function f a complex valued function of two complex variables the first variable being denoted by z second variable denoted by w and we assume that the function is continuous we assume that there is a point in the domain where the function vanishes and the first derivative with respect to the first variable does not vanish and of course we assume that the function for each fixed value of the second variable in the domain the function is an analytic function of the first variable okay see if this domain really confuses you for some reason you can simply assume that this domain is all of the complex the product of you know two copies of the complex plane okay you can assume D equal to C2 if you think that it makes life more easier to understand okay. So here is the implicit function theorem which says that if the function vanishes at a point where the first variable the partial derivative with respect to the first variable does not vanish okay which means by that I mean you consider it as function in the first variable freezing the second variable and then you substitute that point you take the derivative with respect to the first variable then substitute that point that is the partial derivative with respect to the first variable at that point and if that is not zero the implicit function theorem says that you can solve for the first variable in terms of the second variable So in this case you can solve for z in terms of w so you can write the explicit function g z equal to g g of w which will be an explicit equation for f of z comma w equal to 0 in other words in a neighbourhood of w0 what you will get is a function g of w such that if you write z equal to g of w and substitute it in this equation you will get f of g of w comma w equal to 0. So you have solved f of z comma w equal to 0 for z as g of w that is what the implicit function theorem says. And now so what we do is as you may recall the first hypothesis tells you that if with respect to the second variable free frozen as w0 this is an analytic function of z and at z equal to z0 this is a 0 and because the first derivative does not vanish it is a simple zero okay and therefore by we proceeded after that to show that there is an open annulus surrounding a finite circle of finite radius centered at z0 cross an open disc surrounding w0 where f does not vanish okay and for all values of w in this disc surrounding w0 we said that we could solve for z as a function of w okay with z lying inside a disc centered at z0 radius rho okay and of course this rho is chosen in such a way that it is the disc which isolates the 0 of the 0 z0 of f of z comma w w0 okay. So this is the in this disc the function f of z comma w0 which is a function of z has only 1 0 and that 0 is at the center that is how this rho is chosen and the rho the way the delta is chosen is a little involved but that is what I explained in the previous lecture and now for what I want to say is that for every w in this disc open disc centered w0 radius delta if I define g of w to be this function okay then my claim is that and if I call that g of w as z then f of z that z comma w is 0 okay in other words g of w solves the equation f of z comma w equal to 0 with z equal to g of w okay. So the explicit the implicit equation f of z comma w equal to 0 is solved for the first variable z in terms of the second variable w for all w in this disc alright that is what it says. Now I will have to prove I will have to prove this fact so the first thing I have to tell you is that I want to tell you something about this integral okay mind you for all points on this boundary circle f of the function does not vanish that is because of the that is because of this fact that the set of points where the function does not vanish contains this contains an annular open annular region containing this circle cross this disc centered open disc centered w0 okay. So I am dividing by this term and this term does not vanish mind you here the first variable lies on the on this disc centered at z0 radius rho and the second variable is constrained by this equation which says that it lies to within a distance of delta from w0 okay. So I am dividing by something that is not 0 alright and mind you if you look at what is happening here the variable of integration is I have called it as zeta because I do not want to use z alright I am calling the variable of integration as zeta and zeta is the first variable the second variable is fixed to w alright and if you take f of zeta, w you know that it is already an analytic function of zeta so I take its derivative with respect to zeta and then I substitute for the second variable to be equal to w okay you can also write this dou f by dou zeta of dou by you can also write this as d by d zeta of f of zeta comma w okay it is the same thing because you are freezing the second variable to w and you are taking derivative with respect to the first way right. Now the point I want to make is that since and you know of course the derivative of an analytic function is also an analytic function okay so since f of zeta comma w is an analytic function for fixed at w its derivative with respect to the first variable is also an analytic function for fixed values of the side for the fixed value of w and therefore this is an analytic function this is an analytic function of zeta this is also an analytic function of zeta this is just simply the identity which is the identity function and what we have in the denominator is also an analytic function of zeta for this fixed at w so this integrand is an analytic function of zeta and it is and the denominator does not vanish so this integral is well defined okay and this integral is well defined and you are going to get you are going to get a function of w because you are going to integrate with respect to see for a path integral to be well defined all you need is that the integrand is a continuous function on the path you do not need anything else okay so for this function for this integral to be well defined I just need the integrand to be a continuous function but in fact it is a quotient of analytic functions with the denominator analytic function not vanishing so certainly it is a continuous function so this integral exists and after you integrated out the variable zeta goes away and what is left out is only a function of w I am calling that as g of w so this function is well defined alright. Now the claim is that this function solves f of z, w equal to 0 okay that is the claim now so and I have to also tell you that value of g of w if I call it as z then I have to tell you that the value z is taken by g at w only once okay that is what I want to say that is I am just saying that z is you know it is a simple 0 of f of zeta, w for w in this neighbourhood right alright so let me prove this so what I am going to do is let me go to this side of the board and again the method of proof is essentially the same as we did for the inverse function theorem if you look at the proof of the inverse function theorem you can see that I am more or less following the same ideas. So what I will do is I will do the following thing firstly we define 1 by 2 pi i integral over mod zeta minus z0 equal to rho and then I will write the same expression but I will forget I will not write the zeta so I will simply write do f zeta, beta by do zeta at zeta, w t zeta by f of zeta, w okay. Suppose we define this for mod w minus w0 strictly less than delta mind you this integral also makes sense this integral makes sense because again you know the integrand is the numerator is an analytic function it is just you take f and you freeze the second variable to be w then it is an analytic function of the first variable you take its derivative with respect to the first variable and you know the derivative of analytic function is again an analytic function therefore this is an analytic function and it is just the derivative of the function in the denominator so it is a logarithmic derivative actually alright and so in other words what this is is just you are just taking the integral of the logarithmic derivative of f but then if you take the integral of the logarithmic derivative of f and divide by 2 pi i what you would get is by the argument principle you are going to get the number of zeros minus number of poles counted with multiplicity. The first thing I need to say is the following first of all n of w is well defined and is an integer okay it is well defined because the denominator function does not vanish on the boundary on this path on the circle where the variable of integration lies the denominator function does not vanish because that is how we chose this that is how you have chosen this rho and delta okay. So first of all this integral is well defined and since it is the logarithmic derivative of the denominator it is going to be an integer it is a logarithmic derivative of the denominator divided by 2 pi i so it is going to be an integer which is going to be a number of zeros minus number of poles counted with multiplicity so it is an integer valued function and it is an exercise that we have done before in the proofs of several other theorems it is easy to check that n of w is actually an analytic function of w for w in this disk okay. So let me write that down also n of w is analytic in w for mod w minus w not less than delta okay we have we have proved a similar statement earlier okay. So this is so you see you have an analytic function which is integer valued and it is defined on this disk which is connected therefore it has to be constant okay so this is also an idea which we have used earlier so this implies n of w is equal to constant it is a constant integer and that constant will be equal to n of w not for all w with mod w minus w not less than delta this is what you will get okay. And you see but what is n of w not see n of w not if you calculate n of w not you are then looking at the number of zeros minus number of poles of the function f of z comma w not in inside this disk but inside this disk it is analytic and it has only one zero the zero is at the center at z not and it is a simple zero okay. So n of w not will be one okay so what this will tell you is that n of w is one okay so the moral of the story is that you know for every w with in this disk I can find I find that the function f of z comma w has only one single it has only one single zero at a point z which lies inside this disk that is what it says because what is n of w n of w is n of w is one okay we will tell you that you know it will have only one zero and that has to be a simple zero if it is a zero of higher order then this number will go up alright. So moral of the story is that for every w in this disk I can find a z which is lying inside the disk bounded by this circle such that f of z comma w is zero so what this tells you is that I have a function from this disk to the z plane okay so this means we have a function mod w minus w not less than delta from this set to c with w going to let me call this as g of w okay if you want let us call that as z okay and in fact where it goes into is it goes into this disk the set of all mod z minus zeta not z not is less than rho it goes into this. So basically I have a function like this you know what this function is it sends to each w the point z such that f of z comma w is zero such that f of z comma w is equal to zero okay this is the it sends for every w to the unique simple pole of the unique simple pole of this integrand which is essentially the unique simple zero of f of zeta comma w namely it is unique value z such that f of zeta comma w such that f of z comma w is zero okay for every w I am getting a z right this is my function okay and in fact in fact there is also another way of looking at it and which is as follows namely I will have to tell you that I mean there is something here that I have to prove I will have to say that this see after all I am getting a function here w going to some function of w which I am calling as z but I am calling that function as g and I have written that the formula for g is this okay so I will have to tell you that this that g is given by this formula that g is given by this formula it gives a value z satisfying f of z comma w equal to zero okay so I will have to only show that this formula is correct alright. So how do I do that that is again very simple application of residue theorem so you see consider now let us consider the that integral 1 by 2 pi i integral over mod zeta minus z not equal to rho zeta dou zeta by dou sorry dou f by dou zeta at zeta comma w t zeta by f of zeta comma w for mod w minus w not less than delta consider this okay. Now what is the integrand the integrand is zeta actually t by sorry you know if you want I can okay so let me write it like this d by d zeta of f of zeta comma w divided by f of zeta comma w this is the integrand and consider the of course w is fixed okay this form this is defined for fixed w lying in this open disc so you consider this integrand as a function of zeta because the variable of integration is zeta okay. If you look at this integrand you will see that it is an analytic function okay and it is analytic not only it is analytic in the on this disc and it is interior alright and you will see that it will have only one simple pole namely that will correspond to a simple 0 of the denominator where that 0 is zeta equal to z which comes because of this okay. So this this is analytic in mod zeta minus z not less than or equal to rho with a simple 0 at zeta equal to z okay there is a value there is a value z where f of z comma w is 0 first that is all that comes from here okay that is because enough because it is enough for w is 1 so there is a z and there is only one z okay and the for that z f of z comma w is 0 for the given w and the 0 is simple of order 1 okay that is all that is all completely you know buried in this statement that n of w is 1 right now take that z that z will be the only pole for this function it will be a simple pole. So you know if you take 1 by 2 pi i integrate over a curve a function then by the residue theorem we will tell you that you will get simply the residue of the function at that point at that at its you will get actually generally you will get the sum of residues at various poles. So in this case I will simply get this residue at the simple pole okay so by the residue theorem by the residue theorem what will happen is 1 by 2 pi i integral over mod zeta minus z0 equal to rho of zeta dou f by dou zeta at zeta comma w d zeta by f of zeta comma w is actually residue of this function zeta if you want d by d zeta f of zeta comma w by f of zeta comma w at zeta equal to z this is the residue and this is the residue at a simple pole and you know there is a you know how to compute the residue at a simple pole you will just have to multiply it by the variable minus that pole and then take the limit as the variable tends to that pole that is how you find the residue at a simple pole. So that will be equal to limit zeta tends to z zeta minus z times the integrand which is I will get zeta d by d zeta f zeta comma w by f zeta comma w okay this is what I will get and well the and I claim that this is equal to z which will prove that this quantity is equal to z but but z is g of w if you want and so it tells you that this is the formula for g of w okay so it proves that formula that I have written here right. So so why is this true this is this pretty easy because this is limit zeta tends to z of zeta d by d zeta of f of zeta comma w by see the denominator I am writing this as f of zeta comma w minus f of z comma w mind you this is 0 f of z comma w is 0 by zeta minus z I am just pushing this zeta minus z to the denominator and you know now if I take the limit here I am going to get derivative I am going this what I am going to get here is just this these two going to cancel I am going to just get limit zeta instead of zeta and that is going to be z okay so that proves the formula the only thing that you will have to think about is how I can cancel this without ensuring that this derivative is non-zero okay that this derivative is non-zero at z okay and that again follows from the fact that n of w is 1 okay note that so let me write that down note that n of w equal to 1 ensures that d by d zeta f of zeta comma w at zeta equal to z is not equal to this is non-zero derivative is non-zero okay because of the derivative saying that the derivative is non-zero tells you that z is a 0 of order 1 of f of zeta comma w which is what n of w equal to 1 says okay if this derivative was 0 then n of w the value of n of w would have shot up it will not be 1 it will be some it will be an integer greater than 1 that is not the case okay that is why I can cancel these two guys I will simply get limit zeta tends to zeta which is z okay so the formula for and what is this z this z is the unique 0 of f of zeta comma w so it is a unique value lying in this disk the integer of this disk where f of zeta comma w is 0 okay so that finishes that finishes the proof of this claim okay now I come to the question more technical question as to when this function g that I have defined okay which gives the first variable explicitly in terms of the second variable when is that function g also analytic okay so the answer to that is it will be so when the function capital F is also analytic in the second variable so far what we have assumed is that the function capital F is continuous in both variables okay put together it is continuous as a function of two variables and we have assumed that it is continuous in the first variable for every fixed value of the second variable okay but what we have not assumed is that if you freeze the first variable then consider it as a function of the second variable that it is analytic with respect to second variable that we have not assumed so far we have assumed only continuity with respect to both the variables put together which will give you continuity with respect to each variable separately okay. So what I am saying is that if capital F is also analytic in the second variable for every fixed value of the first variable then this g that you have defined is actually an analytic function of w okay so let me write that down so facts number one if F of z, w is analytic in w for every fixed z for every fixed z then g of w is analytic in w this is the first fact okay proving this is again an exercise which involves techniques similar to ones that we have used so far so I will probably leave it as an exercise and in fact if a function is analytic then the natural see you have function which has which is given by a formula g is given by a formula and the moment you say it is analytic then you will then it is natural to ask what is the formula for the derivative okay so there is an answer to that and that is remark number two fact number two in the above case that is when F is a function analytic function in the second variable for every value fixed value of the first variable and g is analytic the derivative g dash of w is actually given by minus of dou F by dou second variable divided by evaluated at g w, w divided by dou F by dou first variable evaluated at g w, w where you know F is written as F of eta, this is partial derivative of the second variable and this is a partial derivative with respect to the first variable and then you evaluate at this point z, w but now z is g of w okay and this is everything so this is a formula for g dash of w in terms of F and g of w okay and so I leave it as exercises for you to check that these formulas are these formulas hold but the point I want to say is this finishes the statement of the proof of the implicit function theorem okay but what I want to next say is I want to say why this implies the inverse function theorem so you know as a corollary you have the inverse function theorem what you do is put you put F of z, w is equal to F of z minus w okay where you know F is analytic on a domain in C okay and F of z not is equal to w not or w not in that domain and F dash of z not is not equal to 0 okay. So the inverse function theorem says that you know you take a analytic function defined analytic function now of one complex variable assume that the function takes assume that the derivative of the function is not 0 at a point then you know what it says in sufficiently small neighborhood of the point the function is 1 to 1 and you can write out an inverse function okay and so you know if you put capital F of z, w to be small f of z minus w then you can see that capital F of z not, w not is 0 okay then number 1 you will see that capital F of z not, w not is 0 because it will be F of z not minus w not which is 0 and second thing is if you take the partial derivative of this with respect to the first variable okay then you will get F dash of z okay and if you substitute z not you will get F dash of z not and that which is not 0. So doh F by doh zeta of zeta F of zeta, zeta at z not, w not is not 0 which is you know which is the condition that we need to apply the implicit function theorem. So now if you apply the implicit function theorem what will tell you is that I can write I can solve for z as g of w such that F of g of w capital F of g of w, w is 0 which means you are saying that small f of small g of w minus small w is 0 which means you are saying small f of small g of w is equal to small w which means you are just saying g is inverse of f okay. So let me write that down, so you get the inverse function theorem and so we get for mod w minus w not less than delta, delta suitably small chosen suitably small you will get z equal to g of w such that f of g of w, w is 0 that is such that f of g of w is equal to w that is it means that g is the inverse g is f inverse, so you get the inverse function, so g becomes f inverse okay and in fact you can see that for fix z if I fix a value of z then f of z is a constant I will get constant minus w and constant minus w is certainly analytic function of w. When I write f of z minus w if I fix a w then this is f of z minus a constant and that is analytic because f of z is analytic, so it is analytic in the first variable if I fix if I freeze the second variable. Similarly if I freeze the first variable z then f of z becomes a constant, so I will get constant minus w and constant minus w is certainly analytic function of w with derivative minus 1 alright, so it is analytic in both variables alright. So according to the implicit function theorem this g that you get will also be analytic g that you will get will also be analytic it will be given by this formula and you will see that this is if you plug in capital F equal to small f minus w in this formula you will get back the formula that we got in the inverse function theorem, you will get back the same formula and you will also see that you will get more if you plug if you use this formula note that we get also of course we get as before f inverse of w is 1 by 2 pi i integral over mod zeta minus z not equal to rho zeta f dash of zeta t zeta by f zeta minus w ok which is the formula for the inverse function theorem that you get from here and also and further you get this from this you get a very nice formula you get f inverse derivative with respect to omega with respect to w is you take look at this you take partial derivative with respect to the second variable. So if I take partial derivative with respect to the second variable I will get minus 1 ok so I will get minus 1 divided by I take partial derivative with respect to the first variable if I take partial derivative with respect to the first variable I will get f dash of z and then if I substitute for z g of w I will get f dash of g of w so what I will get is I will get f dash of g of w so you get this formula ok and this formula is not very surprising because it is actually which is actually which is just product the product rule applied to so I should so here I have written g of w but I have to replace the g by f inverse so this is minus 1 by f dash of f inverse of w and of course you know f inverse w is z oh I am sorry ok yeah there is a minus here I forgot so I should get a plus yeah thanks for pointing that out ok there is already a minus here but then if I take partial derivative with respect to the second variable I will get minus 1 so it will become plus thanks for pointing that out but yeah so essentially what you will get is you know it is f dash of z times g dash of f of z ok is equal to 1 ok and that is just trying to it is just got by applying product rule to g circle f of z is identity so it is just g circle f is equal to identity on z see g is f inverse ok so what does this say this is g of f of z is equal to z this is g of f of z equal to z differentiate it you will get g dash of f of z into f dash of z equal to 1 so it will tell you g dash of f of z is 1 by f dash of z that is what you got so it is just product rule applied to this where g is f inverse ok that is not very surprising so ok so now let me let me end this lecture by saying something interesting about an open problem which is an open problem both in complex analysis and algebraic geometry a very important Jacobian conjecture ok so let me explain what this Jacobian conjecture is so what you do is you take c2 to c2 ok map given by z comma w going to f of z comma w g of z comma w ok so in general you know if you want a function that is going from c2 to c2 I will take a point with two complex coordinates and I have to produce two complex coordinates so I have to give you two functions each function will vary will depend on both coordinates so I call them as capital F and capital G ok in general the best kind of functions you will think of are that you know both f and g are for example analytic in both variables ok and in particular what I want to do is I want to look at the case when f and g are actually polynomials into variables ok suppose we have a map where capital F and g are polynomials so f and g are polynomials in the variables z and w ok and if we can find an inverse map of the same type namely c2 to c2 which is given by z comma w going to say let me use something else I will let me use some other notation so I will get I will use p and q ok so assume that you have a map like this which is called a polynomial map because it is given by two polynomials and assume that of course you know if you give me a polynomial map I do not know whether it is injective I do not know whether it is surjective but assume it is bijective and assume that there is an inverse map which is also polynomial map ok assume that there is an inverse map which is also polynomial map which is given like this ok. Check that Jacobian of this pair f comma g is equal to a non-zero constant number non-zero constant complex number and that will hold also for the Jacobian of p and q where of course Jacobian of f and g means you take determinant of you take partial derivative of f with respect to both variables dou f by dou z dou f by dou w then you do it for the other function dou g by dou z dou g by dou w this is the Jacobian determinant mind you f and g are polynomials in two variables if you take partial derivatives again you are going to get polynomials in two variables ok and the fact is that this determinant therefore will be a polynomial in two variables but the fact is that if you have an inverse like this if this polynomial map has an inverse map is also a polynomial map then I want you to check that this determinant which is a polynomial in z then w is not actually a polynomial it is a constant and it is a non-zero constant ok this is probably just follows by using essentially product rule kind of argument ok. So you should be able to show that this is very this can be shown trivially ok with some basic knowledge of calculus now the Jacobian conjecture is actually the converse ok the Jacobian conjecture is the converse which says that you know conversely if you give me a pair of polynomials f and g with the property that the Jacobian is a non-zero constant then the result is that there is an inverse map which is again given by a pair of polynomials ok there is an inverse map which is again given by a pair of polynomials that is the Jacobian conjecture and the fact is that the Jacobian conjecture I have said it for two variables but you can set it for any n number of variables of course you can check that if I try to write this for one variable it is trivial to check that it is true ok. So the statement is really serious from the two variable case onwards so this is a Jacobian conjecture for two variables and for you can write that for n variables ok so but the fact is that it is still unsolved even for two variables and it is a statement only involving polynomials which are so easily understood ok. So it is a it is a it is a very deep question and it answers to answer this question people have gone into algebraic geometry they have gone into complex algebraic geometry they have gone into lot of complex analysis but the question is still open. So it is a it is a kind of question that if you solve some young person who is who is taking this course of lectures solves he will surely he or she will surely get a prize equivalent to the Nobel prize in mathematics so you will certainly get a fields medal if you solve this ok. So it is such a deep problem but it is so easily stated. So let me state it the Jacobian conjecture states that conversely if the Jacobian determinant of f, g is a non-zero constant complex number of course non-zero constant complex number then there exist an inverse map given by a pair of polynomials P and Q ok and I should say that this is not only for two variables it is also for n variables for n greater than 2 as well that is also that is also the general case of Jacobian conjecture but the fact is even for two variables it is not solved ok and the reason why I am talking about this when I am talking about the implicit function theorem and the inverse function theorem is that you see what is the inverse function theorem that we have seen says it says that see whenever you know the derivative is non-zero then locally you can invert ok whenever the derivative is non-zero at a point then you can find sufficiently small neighborhood of the point where the function is 1 to 1 so you can write the inverse ok. And this is the inverse function theorem for one variable but you can make the same statement for you can make a similar statement and inverse function theorem for several variables. So you know for example if you want for two variables the inverse function theorem for two variables will be give me a function like this and assume that f and g are if you want analytic do not even assume they are polynomials of course if they are polynomials they are analytic ok but assume f and g are more general analytic functions in both variables ok then calculate the Jacobian ok. The inverse function theorem tells in the two variable case that at every point where the Jacobian does not vanish ok I can invert the function in any boat ok so what I want to tell you is that from the view point of the inverse function theorem the Jacobian non-vanishing will tell you that locally this map can be inverted locally I can invert this map and the inverse by the implicit if you use the inverse function theorem or the implicit function theorem locally the inverse functions that you get they will again be analytic functions ok. So what analysis tells you is that this Jacobian non-vanishing will allow you to invert the map locally and the inversion is achieved by analytic functions ok that is what analysis tells you but the conjecture is very strong the conjecture says that you can find a global inverse and that global inverse is given by just two polynomials in the two variable case and similarly if it is a n variables case it tells you that the glow there is a global inverse which is given by n polynomials ok so you know what you will have to prove is that somehow the local functions that you get which invert you have to show that all the local functions you know they all can be chosen so that they all glue together to give a global function and you have to show that that global function is given by two polynomials the two variable case otherwise it is given by n polynomials in the n variable case. So that is the gap between what we know in analysis and what the Jacobian conjecture demands and of course I must tell you that you know if you prove that if you prove that there is a situation where you know the Jacobian is a non-zero constant but the map is not bijective then also you have given a counter example to the I mean if you are able to find you know if you can find a contradiction to this statement namely you find a counter example then also it will be a great result but no such counter example has been found ok so you are invited to try this or think about this.