 Okay so what we will try to look at in this and the next lecture is the so called implicit function theorem okay. Now let me recall what the implicit function theorem is in the context of real analysis okay as far as real variables are concerned this is something that you should have seen in courses in real analysis. So the so let me put the title as the implicit function theorem so you know the so let us look at the could a very simple example so let me look at the case of the case of two real variables this is something that you should have come across in a first or second course in analysis real analysis. So we take the plane R2 and let us look at some equation of the form capital F of x, y equal to 0 so this is to be thought of as an implicit relationship between x and y because you are not able to write x as a function of y or y as a function of x this is called an implicit relation. If this can be equally equivalently written as x is equal to some function of y okay that means you have solved it explicitly for x and if you can write it of the form y is equal to a function of x then you have solved it explicitly for y okay but if in general you are given relationship like this where one variable is not written as a function of the other variable explicitly this is called an implicit function. So the standard example for example is x squared plus x squared plus y squared minus 1 equal to 0 which you know is the circle unit circle and so it is the unit circle centered at the origin alright and what you should notice is that you know if you calculate so if you calculate so if I take the example F of x, y is equal to x squared plus y squared minus 1 and if I calculate dou F by dou x then I will get if I partially differentiate this with respect to x I will get 2x okay and if I partially differentiate it with respect to y I get 2y okay and the set of points where the partial derivative with respect to x vanishes are the set of points where x equal to 0 okay and x equal to 0 corresponds to the y axis these are points on the y axis and of course I am looking at points on the curve also okay. So you know so if you look at if you look at F equal to 0 and dou F by dou x equal to 0 if you look at this set of equations okay so F equal to 0 means it is a point on the curve defined by F of x, y and it is also point where the first partial derivative with respect to x vanishes okay so this is this happens only so this happens only at the point x, y is equal to so you know if I want the first partial derivative to vanish then 2x must be 0 so x must be 0 and if x, y is lying on the curve then it should satisfy this equation and if x is 0 it means that y is plus or minus 1 so you get this is 0 plus or minus 1 okay so you will get so namely you will get these points so it is x equal to 0, y is equal to minus 1 and I will get this point which is x equal to 0, y equal to plus 1 okay and so if you take all other points on the curve these are points for which the first partial derivative with respect to x does not vanish okay. So if you take curve circle minus these 2 points 0 plus or minus 1 on these points of the curve what happens is that the first partial derivative with respect to x does not vanish okay and so what happens at these so at every other point of the curve what happens is that you can solve for x as a function of y so if you give me any other point on the curve okay then you know I can take the if you give me a point x0, y0 on the curve okay then I can so I get the point y0 here okay and I can find a small neighbourhood of y0 okay where x can be written as a function of y0 and for that function this is the graph of the function in that neighbourhood so you know if I take this piece of the curve in this piece of the curve at say at x0, y0 locally you can write x as a function of y okay locally you can solve so the point is so in this case you know I can solve for this is locally at y0 okay x is a function of y in a neighbourhood of y0 so you see x0, y0 is a point on the curve that is a point which is different from these 2 points okay so the first partial derivative with respect to x does not vanish and then locally at that point I can write the curve as a function of y okay and you know what this is in this you can write the function you can write it you know the function that I am talking about you can write x as positive square root of x can be written as positive square root of 1 minus y square okay that give me this branch of the curve and of course if my point is here if my x0, y0 is here then I will be able to write x as negative square root of 1 minus y square okay so here on this piece this is the graph of x equal to positive square root of 1 minus y square okay and this is the graph of x equal to negative square root of 1 minus y square so what you see is that if you choose a point where the first partial derivative with respect to x does not vanish you can write x explicitly as a function of y and mind you when I write x equal to g of y what it means is that f of g of y, y is 0 because it is a x equal to g of y is actually a piece of the curve so x equal to g of y will satisfy the equation of the curve okay and this x equal to g of y in this case if it is a point in the first quadrant then x equal to g of y is plus so it is positive square root of 1 minus y square so this is your g of y okay and in this case it is negative square root of 1 minus y square okay and that is what you will get whether you take a point here or here you will always get positive square root of 1 minus y square you take a point here or here you will get always negative square root of 1 minus y square and these are the two branches of that curve actually okay depending on the point you have chosen but the idea is that what you see is that you get uniquely a function which solves your equation, you get a unique function of course you get this branch if you start at the point here and you get this piece of the curve you start at the point here, so the piece of the curve that you get depends on the point at which you are trying to solve the equation and that point has to have the property that the first partial derivative with respect to X does not vanish. So the moral of the story is that if the first partial derivative with respect to X I mean the partial derivative with respect to X does not vanish, then you can solve for X okay and similarly I can repeat the same thing with Y also if I look at the equation F equal to 0 is equal to dou F by dou Y, if I look at this then this is true if and only if the point X, Y is well you know I want dou F by dou Y to be 0, dou F by dou Y is 2 Y, so 2 Y is 0 which means Y is 0, so the Y coordinate is 0 and if Y coordinate is 0 and I am on the curve then I will get X squared minus 1 equal to 0 that will give me X equal to plus or minus 1, so I will get plus or minus 1, 0. So the situation is that I will get let me draw another diagram instead of cramming that diagram with too many more things, so here is my so I will get the points plus or minus, so I will get these two points 1, 0 and minus 1, 0, so well and these are the points where the first partial derivative with respect to the variable Y does not vanish and now you choose a point different from this on circle minus these two points plus or minus 1, 0 the first partial derivative with respect to Y does not vanish okay because these are the two points where the first partial derivative with respect to Y vanish and we removed them at say at you know if I take a point X naught, Y naught on which is different from these two points see for example I could have taken a point X naught, Y naught here you can solve for Y as a function of X okay, Y can be written as a function of X and that will be a solution of f of X, f of X, y equal to 0, so you will get f of X, f of X is 0 in a neighbourhood of X naught okay. So what will happen is that you see here you project down to X naught and there is a small neighbourhood of X naught you can find such that the graph of the curve in a neighbourhood of X naught will be given by an equation you know what it is going to be Y is equal to positive square root of 1 minus X square, so this is your f of X okay and if I had chosen the point here if I had chosen the point X naught, Y naught of course if I had chosen any point on the upper semi-circle I will get this function if I had chosen a point on the lower semi-circle if I had chosen a point X naught, Y naught here in the lower semi-circle then you know for the corresponding X coordinate I will get a neighbourhood such that the equation the implicit function can be solved for Y and it will be minus root of 1 minus X square, so here it is going to be Y equal to negative square root of 1 minus X square okay. So the moral of the story is that to sum up you have an implicit function in 2 real variables you can solve that implicit function for the variable X in a neighbourhood of a point where the first partial derivative with respect to X does not vanish okay and that is here that is what I have written here, if the first partial derivative with respect to X does not vanish I can solve for X as a function of Y, if the first partial derivative with respect to Y does not vanish I can solve for Y as a function of X and both will be explicit solutions of the implicit function the implicit equation I started with okay. So this is what the implicit function theorem in real analysis is and in fact this implicit function theorem can be generalized to several variables instead of just taking a function of 2 real variables okay you can have a function of several variables and then the implicit function theorem for several variables will tell you that you can solve for a variable as a function of the other variables provided the first partial derivative with respect to that variable does not vanish at a point on the on the on the locus where this function is defined okay this is what this is how it generalizes there is a and it is not very difficult to prove that theorem okay. So the case of 2 variables goes to extend the case of several variables and this implicit function theorem is of lot of importance in the study of manifolds and study of Riemann surfaces and things like that it is a it is a very important theorem now what I want to say is that there is an implicit function theorem which is also there in for 2 complex variables okay and it is again it also extends to several complex variables and what I want to tell you is that this implicit function theorem for 2 complex variables that I am going to talk about is also going to tell you the same thing it is the same statement except that instead of 2 real variables you are going to deal with 2 complex variables and it is going to say that whenever the first partial derivative the partial derivative with respect to certain variable is not 0 then you can solve for that variable in terms of the other variable that is what it is going to tell you but the point with complex analysis is that because of Cauchy's theory you get an explicit formula given by an integral which you do not get in the case of real function. So the advantage of complex analysis is that because of the existence of Cauchy theory you can write down the you can write down the local solution okay so that is what we are going to see. So and there are I will try to see if I can give you applications of this one of the applications of the implicit theorem for example is the fact that if you give me an equation in 2 variables such that there is no there is no singular point a singular point is a point where is a point on the on the locus where all the partial derivative is vanished okay. See to be able to solve it as a function of one variable in terms of the other the partial derivative with respect to that variable should not vanish okay but if the partial derivative with every variable vanishes at a point then I am in bad shape I cannot conclude that I can solve the function for one variable in terms of the other variables okay because all that the implicit function theorem says is that if the partial derivative with respect to a certain variable is not 0 then you can solve in a neighbourhood of that point for that variable you can solve for that variable in terms of the other variables okay. So obviously if all the partial derivatives are going to vanish then you cannot apply the implicit function theorem and the points on the locus where all the partial derivatives vanish at the same time they are called singular points okay and the problem is that at singular points you cannot conclude whether you can solve of course both kinds of cases occur there are cases where you have singular points and yet you can solve locally for one variable even though the first partial derivative vanishes as you can look at the curve for example y square equal to x cube which is a cusp okay and there are also curves for which at singular points you cannot solve locally okay. So the main application to manifold theory is that whenever you have an equation which has no singular points for example the curve is this curve here is a circle and the circle has no singularities okay and one way of seeing it is that it has a unique tangent at every point okay and whenever you have a curve which has no singularities then the beautiful thing is that you can turn it into a manifold okay. So in this case the circle is a manifold one dimensional manifold if you had if I had two complex variables then I can turn the locus into what is called a Riemann surface okay. So one of the most important applications of implicit function theorem is that you can for a curve for an equation which is non-singular which has no singular points you can turn the locus of that equation the set of points satisfied by that equation into a manifold and you can do calculus on it okay. So that is one of the most important applications of the implicit function theorem that is why it is very very important for higher analysis okay. So let me go ahead with the complex version so what I am going to do complex version so what I am going to do is I am going to take a function f from c2 to c written as z,w going to capital F of z,w okay here is my function and of course you know when I when we talk about the implicit function theorem the function that we are dealing with has to have some obviously good properties for example I want at least that the function is continuous in both variables and you know it is partial derivative with respect to each variable exists and the partial derivatives are also continuous these are obvious assumptions that one makes okay. Maybe some of these assumptions can be relaxed a little but there is no harm in assuming that the function that you are dealing with is at least continuous in both variables in the first partial derivatives are continuous so what I am going to do is at least in the complex case let me be very strict and tell you what the exact conditions are I want f to f is continuous so I should at this point give you a word of caution when I say f is continuous I mean f is continuous as a map okay and that means continuity with respect to both variables okay and I want you to understand that continuity with respect to both variables implies continuity with respect to each variable separately but continuity with respect to each variable separately is not as strong enough as continuity with respect to both variables. So this is so this is stronger than saying that for every z fixed z f of z, w is a continuous function of w and for every w fixed f of z, w is a continuous function of z okay that means when I say that for example f of z, w is a continuous function of w for every fixed z that means I am saying in the variable w separately f is a continuous function it is called separately continuous with respect to second variable similarly when I say f of z, w is continuous with respect to z for a fixed w I am saying the function f is separately continuous with respect to the first variable okay these are very very weak conditions when compared to this condition which is continuity in both variables okay. So I am just this is what I am assume when I say f is continuous mind you it implies separate continuity it is implies continuity in each variable but it is far, far stronger than that that is that is one thing that you can easily find functions which are you know continuous in separately in each variable but put together in two variables they will not be continuous okay. So I am just trying to emphasize the point that this is not this is stronger than continuity in each variable alright now so I assume this okay and then I also assume the following thing. So here is what the so here is the implicit function theorem so let me just say simple form let me give you the simple form and then give you the more involved statement so what is the simple form the simple form is well if z0, w0 is a point of f of z, w equal to 0 okay that means is a solution of this that is z0 f of z0, w0 is 0 okay and if and number one if f is analytic in z for each fixed w okay so if you fix a w then this function becomes a function of z and as a function of z for a fixed w you assume it is analytic okay that means I can partially differentiate this with respect to the variable z because it is analytic with respect to z okay and the partial derivative of f with respect to the first variable okay so you see I am let me write it very carefully f I write f as a function of zeta eta I differentiate partially with respect to the first variable okay then I substitute z0, w0 suppose the so I am just saying that the partial derivative with respect to the first variable is nonzero at the point z0, w0 okay then you see there exists a delta positive so let me not say that and I will say then the first variable f of z, w equal to 0 can be solved to give e z is equal to g of w in a neighbourhood of z0 in a neighbourhood of w0 okay so you see this is the so of course this implies that you know f of g w, w, w0 okay so it is the same statement literally what is a general statement the general idea of the implicit function theorem is whenever you take a point on the locus satisfied by the equation if whenever you the first partial derivative with respect to certain variable is nonzero then you can solve explicitly for that variable in a neighbourhood of the point corresponding to the other variables okay so here z0, w0 is a point on this locus it is a point that satisfies f of z, w is 0 okay and the first partial derivative of f with respect to the first variable does not vanish at that point then I can solve for the first variable as a function of the second variable in a neighbourhood of the second variable point corresponding to the given point so the second variable point is w0 so in a neighbourhood of w0 I can solve okay I can solve for z as a function of w in a neighbourhood of w0 provide the first partial derivative with respect to z is not vanish and to make sure that I can really differentiate this with respect to the first variable z with respect to the first variable I need and I want to do it in a and I want to do it not just at that point I want to do it in a neighbourhood of that point so I will have to assume that the function is analytic in the first variable at least in a neighbourhood of the point z0 that is what I have assumed here okay I am assuming that if you freeze the second variable then as a function of the first variable it is analytic that allows me to differentiate it with respect to the first variable okay so this is the first part of the statement then what is the second part of the statement the second part of the statement is the following the second part of the statement is well if it tells you more it tells you that so you can ask see this function z equal to g of w is a function of w in a neighbourhood of w0 okay you can ask when that is analytic after all that is also a function of a complex variable you can ask the question when that is analytic okay and of course then you can ask the question as to what is the derivative of that so this gives the remaining parts of the theorem gives you answers to these questions okay so you know if so if f of z,w is an also analytic in w for every fixed z for every fixed z okay which is condition similar to this one okay in here we have assumed that the function f is analytic in the first variable for every fixed value of the second variable okay and now I am assuming the other way round I am assuming the function is also analytic with respect to the second variable for every fixed value of the first variable then g of w is analytic the g of w that I wrote that is the explicit solution z equal to g of w which is the explicit solution of f of z,w in a neighbourhood of w0 that g of w becomes analytic and what is the so if it is analytic you can ask me what is the derivative is there simple formula for the derivative and the answer is yes and the derivative is given by the following formula and moreover g dash of w okay which is by this I mean d by dz of g of w sorry g by dw of g of w is given by the following formula it is minus of you differentiate f partially with respect to the second variable and you divide by the derivative of f with respect to the first variable and then evaluate this whole crazy thing at the point g of w,w okay this is the derivative of the analytic function g with respect to w right mind you the numerator I have the derivative negative of the derivative with respect to the with respect to the second variable and the denominator I have the derivative with respect to the first variable and I am taking the I am taking the quotient and then I am evaluating it at g w,w that gives me a function of w and what is that function of w that function of w is the derivative of g of w with respect to w that is what the formula says okay and mind you we have already assume that the derivative with respect to the first variable is not 0 in a neighbourhood okay so what will happen is in that neighbourhood this will never you will get a suitable neighbourhood where this never vanishes and that is what justifies dividing by this mind you dou f by dou zeta at g w not which is z not,w not is nonzero okay so it will it will it will work in a neighbourhood of w not okay so the denominator will not vanish so this division is this division makes sense okay and so this is the formula for the derivative alright and fine so what if I put these two together it tells me that if you give me a complex function of two variables then if you want to look at the zero locus of that function of two variables then at a point on the zero locus of the function of two variables if the first partial derivative does not vanish for which I need the function with respect to the first variable to be analytic for every fixed value of the second variable then I can solve for the first variable in terms of the second variable as a function in a neighbourhood of the point corresponding to the second variable and further this solution which is the explicit solution this function is itself analytic and when does that happen that happens provided the function is also analytic as a function of the second variable for every fixed value of the first variable and in that case you have a formula for the derivative of this explicit function the explicit solution of the implicit equation and that is given by this ratio of derivatives negative ratio of derivatives that is what it says okay and okay so this is the kind of simple form okay now what is more what is the more involved form more involved statement the more involved statement is the following it will tell you it will tell you it will tell you what these neighbourhoods are it will give you more information choose row greater than 0 such that for 0 less than mod z-z0 less than or equal to row f of z, w0 is not 0 okay such see such a row exist see f of z, w is a analytic function for in z for every fixed w so you put w equal to w0 so f of z, w0 is an analytic function of z and it has a 0 at z0 because f of z0, w0 is 0 so z0 is a 0 of this analytic function of z and you know the 0s of an analytic function are isolated and mind you this analytic function is not identically 0 why it is not identically 0 is because it is first partial derivative is non-zero in a neighbourhood of z0 okay I mean it is not 0 at z0 and therefore it is not it cannot be 0 in a neighbourhood of z0 alright so it is certainly not the it is not the it is not a constant function it is a non-constant analytic function of z and you know a non-constant analytic function has 0s which are isolated so the 0 z0 of this function is isolated so there is a small neighbourhood there is a deleted closed disk you know centre at z0 of positive radius on which there are no other 0s so that is how you get this row okay we have already seen this several times in earlier lectures okay and then there exists a delta greater than 0 such that in fact g of w is given by 1 by 2 pi i integral over mod zeta minus z0 equal to rho the integral being taken with respect to the anticlockwise sense of zeta d dou f of zeta comma eta zeta comma w by dou zeta d zeta by f of zeta comma w okay. So this is the formula for g whenever w is at a distance of is a distance less than delta from w0 okay so this is the more involved statement gives you a formula for the explicit function g which is the function of w okay so what this tells you is that that is for every w with mod w minus w0 strictly less than delta there exists a unique z with mod z minus z0 strictly less than rho such that z equal to g of w is a solution of f of z comma w equal to 0 that is f of g of w comma w is 0. So this also tells you note then note that g which is defined from set of all w such that mod w minus w0 is less than delta to the set of all z such that mod z minus mod z minus z0 less than rho is a is a 1 1 function it is an injective function it takes w to gw which is z that is exist there is a unique z okay so you get a unique solution z equal to gw for the implicit for the implicit function f of z comma w equal to 0 and I think I am more or less done that is right so this is the extra information that you get this is the extra information that you get if you want the more involved statement okay this is the extra information that you get now it is very easy to tell to give you a corollary a very simple corollary a simple corollary of this is the inverse function theorem okay so the corollary is the inverse function theorem okay so what you do is put f of z comma w is equal to f of z minus w for f analytic okay you put f of z comma w to be f of z minus w where f is an analytic function of z okay then f of z minus w for every fixed w f of z minus w is an analytic function of z and for every z fixed z f of z will become a constant and a constant minus w is certainly an analytic function of w with derivative minus 1 so what you will get is you will get you will get that you will get the inverse function theorem so in particular you will get that if w not equal to f of z not there exist delta greater than 0 such that for mod w minus w not less than delta you will have g of w z equal to g of w will be given by 1 by 2 pi i integral over mod zeta minus z not equal to rho and you know if I plug in here I will get zeta f dash of zeta t zeta by f zeta minus w which if you recall was the formula that we got for g is actually f inverse okay. So this is the formula that we derived for the inverse function in the inverse function theorem and we get this formula and we get this if you get a formula for the inverse function theorem also as a corollary of the implicit function theorem so the implicit function theorem is always a more stronger statement than the inverse function theorem okay and therefore the philosophy is that the fact that our injective analytic function is a bi-holomorphic map okay if a function is injective and analytic that it is a holomorphic isomorphism is actually a corollary of the you know the implicit function theorem in some sense okay so the implicit function theorem is stronger than the inverse function theorem right. So what I will do is in the next lecture I will try to explain how to get the proof of the implicit function theorem okay I will stop here.