 Alright so we were discussing the so called implicit function theorem and basically the implicit function theorem for example in two variables says that if the derivative with respect to a particular variable is non-zero then you can solve you can solve the equation you can solve the equation for that variable okay. So I stated the implicit function theorem and I will try to prove it and I also told you that the implicit function theorem is stronger than the inverse function theorem okay. So let me write it as the proof of the implicit function theorem this is the proof of the implicit function theorem right. So let us recall what we started with so we have say D is a domain C2 okay D domain which means it is an open connected set it is an open connected set but mind you it is a subset of C2 okay and maybe the arrow is not from C2 but it is from D so let me D is sitting inside C2 okay and I am having a function so I call the two variables the two complex variables as Z and W okay and I have a function f defined on this domain so it takes the point Z, W to f of Z, W okay so it is a complex valued function of two complex variables alright and assume first that for every W in the in D for every W in D fixed for every fixed W in D so let me write that again carefully for every fixed W in for every fixed W okay f of Z, W is analytic in Z okay. So you see if you fix a W fixed value for W then there will be values of EZ for which this function if you see if you fix a W then this becomes a function of Z okay and you can make sense of this as a function of Z and what will happen is of course for every Z such that the point Z, W is in this domain alright of course this is not defined for all values of Z it is only defined for all those values of Z such that the point Z, W is in this domain okay. Now for all those values of Z this is a function of Z okay and the fact is it is an analytic function of Z that is a holomorphic function of Z this is what we assume alright assume also also that for a point Z0, W0 in D the following things happen number 1 f of Z0, W0 is 0 okay that is this is a point this point with coordinates Z0 and W0 is a 0 for the function f number 2 you assume that dou f by dou zeta okay why I am writing dou is because there are 2 variables involved right but actually I am fixing the second variable I am finding the derivative with respect to the first variable which I can do because it is with respect to the first variable it is an analytic function it is a holomorphic function which means it is a differentiable function so I can derive I can take the derivative and I take this derivative and evaluate it at this point Z0, W0 and that I assume that the derivative is non-zero okay I assume that the derivative so all I am saying is the function has a 0 at the point Z0, W0 and the first and the function and the derivative of function with respect to the first variable Z at Z0, W0 is non-zero okay then the implicit function tells you that in a neighbourhood of W0 you can solve for EZ as a function of W okay this is what so let me write that then the implicit function theorem when the implicit function theorem says that we can find G of W for W in a neighbourhood of W0 such that F of G of W, W is 0 okay which means what you are doing is which means you are saying that you can solve for F of EZ, W with you can solve for the EZ such that F of EZ, W as a function of W in other words let us so let me write this in other words we can solve for EZ is equal to G of W for W in a for W in a neighbourhood of W0 and Z in a neighbourhood of Z0 basically so the whole idea is a function of two variables as a 0 at a point with respect to the first variable if the derivative is non-zero at that point then you can solve for the first variable as a function of the second variable in a neighbourhood in a suitable neighbourhood of the point corresponding to the first variable and a suitable neighbourhood of a point of the point corresponding to the second variable this is what it says alright. So in fact of course I have stated with this for the first variable but you can state it for the second variable also okay so if the so the general idea of the implicit function theorem is you have function of several variables and assume that the partial derivative with respect to one of the variables does not vanish then that variable can be solved for as a function of the remaining variables in a suitable neighbourhood that is what it says. So in this case the first variable Z and the partial derivative with respect to the first variable does not vanish so you can solve for the first variable as a function of the second variable which is the second variable is a remaining variable and in fact the truth is that this holds for several variables not for just for a function of two variables it actually holds for several variables and of course there is also a real you know there is a there is a version of this for real differentiable functions also which is the usual implicit function theorem for real differentiable functions okay. So here I am considering complex function of two variables but you can also consider complex function of n variables similarly I can also consider real differentiable function of n variables and the statement of the implicit function theorem is always the same it is just that you can solve for that variable with respect to which the partial derivative does not vanish that is all okay. Now let us see how this is achieved so you know the first thing I wanted to start with is see now you see f of z, w0 is an analytic function of z okay it has a 0 of course analytic is the same as holomorphic okay complex differentiable in a neighbourhood it has a 0 at z equal to z0 because f of z0, w0 is 0 that is our assumption so z0 is a 0 of this analytic function and mind you this is not a constant analytic function because it is first derivative with respect to the first variable is non-zero at that point okay. So since the first variable is non-zero what it tells you is that this 0 is a simple 0 and it also tells you that the analytic function is non-constant considered as a function of the first variable okay since so this is by assumption 1 and by assumption 2 since dou f by dou z okay dou f by dou zeta zeta, eta at z0 if I put w not here at z0, w0 r just let me put it as z equal to z0 and zeta equal to z0 is not 0 okay. So you take this function as a you freeze the second variable you put eta equal to w0 then it becomes only a function of the first variable you take the derivative with respect to that variable and then substitute the point z0, w0 I mean which essentially means you have to substitute z0 because you have already plugged in w0. Then the first partial derivative is non-zero implies what it implies is that it tells you that the function f of z, w0 is a non-constant analytic function of z because had it been a constant analytic function then its derivative would have been 0 which is not the case. So it is a non-constant analytic function of z and with a simple 0 at z equal to z0 okay and the simple the simpleness of the 0 is precisely the fact that the derivative does not vanish, if the derivative vanishes then you know the order of the 0 is more than 1 okay. So that the derivative does not vanish with respect to the first variable tells you that in the first variable the 0 is a simple 0 okay. Now you know that the 0s of a non-constant analytic function are isolated so I can find a disc centered at z0 and sufficiently small radius so that for all points in the disc including the boundary circle the function does not vanish okay except at the center of the disc where it vanishes that is I can isolate that 0 by a closed disc okay. So since the 0s of an analytic function are isolated there exists row positive and such that f of z, w0 is not 0 for all z with 0 less than mod z-z0 strictly less than or equal to row. So I can find this row okay yeah but there is something that I should have told you earlier the function that I start with okay so this is an assumption that I should have told you but anyway it is not late to say that. So the function that I start with this to begin with I at least assume that it is a continuous function in both variables okay I assume f is continuous that is something that you assume it is a of course you do not want to work with functions of several variables which are not continuous so f is a continuous function of both these variables okay I mind you that is in general that is very strong when compared to assuming that f is continuous separately in each variable okay so I am saying f is continuous on D okay. Now you see to give you a picture of the domains of z and w I want you to just recall the so called product topology you see recall product topology on C2 which is C cross C you see what you so you know the complex plane which is R2 it has the same topology as R2 okay and the topology is given by taking for a base the open disks centered at point and having positive radius alright and you know that this is a base because you take any open set in the complex plane it can be written as a union of such disks okay because any open set if you give me a point of any open set there is a sufficiently small disk surrounding that point which lies in that open set and now if you take all such disks corresponding to various points and take the union you will get the open set okay so the topology on the complex plane is given by disks as a base for the topology open disks okay. Now what happens in the product topology in the product topology what you do is that you take when you take a product of two topological spaces and make it into a topological space what you do is you do the following thing to prescribe open sets there so what you do is you take you take a open set in from each topological space okay you pull it back by the projection maps okay and you take products alright and then you and a union if you take unions of all such products that will give you all the open sets okay so all I want to say is that any if you give me any point in the plane I mean in C2 then you know you can you can you can and if it is inside an open set then you can find that point sitting inside a product of disks which is in that open set okay this is a fact from topology so if so you know if z0,w0 is a point of is contained in D which is contained in C cross C which contained in C2 C2 which is so you know I will even put z,w C cross C and there are these projections P1 the first projection projection onto the first coordinate P2 is projection onto the second coordinate okay then you know there exists well I should say rho there exists rho and lambda positive such that such that you know ball centred at the open ball centred at z prime radius rho cross the open ball centred at w prime radius lambda is contained in D okay this is this is a fact from the from the definition of product of policy where of course where you know the disk centred at rho this centred at z prime radius rho is set of all z such that mod z-z prime is less than rho and the disk centred at w prime radius lambda is set of all w such that mod w-w prime is less than lambda so it will contain a product of disks okay. So this is so this kind of so what you must understand is that whenever you are working with a point here you can always think of a disk you can think of this point as being as sitting into in a product of disks a disk centred at the first coordinate cross a disk centred at the second coordinate okay for sufficiently small values of the radii right this is because of product of product of policy and in fact product of policy tells you that this these projection maps are actually you know the projection maps are continuous and the projection maps are even open maps they take open sets to open sets so the fact is that you know if I take the image of the open set D under P1 the result is an open set in C which will contain z prime therefore it has to contain a disk alright and if I take the image of under D of P2 I will get an open set again in C which will contain w prime and so it will contain a disk containing w prime alright and then you can choose these radii are small enough so that this product is inside D okay. Now so all I am trying to say is that you know when I say that you can find a row such that for all z in a disk centred at z0 radius row f of z, w0 is defined it makes sense because z0, w0 is a point of D and therefore you know I can find such a disk okay the image of D under the first projection will contain z0 so it will contain a disk centred z0 right. So let us go ahead with this so the point is now I have this disk in which this disk centred z0 radius row in which f as a function of z has only 1, 0 at the centre which is simple 0 okay so in particular mod so f of z, w0 you see this is not 0 for z on mod z-z0 equal to 0. See on the boundary of the disk the function does not have any 0s okay see this f of z, w0 has only a 0 at z equal to z0 on the boundary circle it does not have okay. So this is the boundary circle centred at z0 radius row on this boundary circle it does not have any 0s so it is a and mind you therefore if you take the mod of f you will see that of course I have forgotten to write not equal to 0. So f is not 0 so mod f is not 0 and mind you mod f is a continuous function okay and this continuous function is being taken on this compact set this is a circle which is closed and bounded so it is compact okay in Euclidean spaces closed and bounded subsets are precisely the compact sets. So you have a continuous function namely mod of f of z, w0 on this compact set so it is uniformly continuous and it has it attains its bounds so it has a certain minimum value okay. So it has a certain minimum value but let me not worry about that what I want to say is so I want to say that f inverse of c star which is c-0 contains the product mod z-z0 is equal to row cross w0 okay. See for each this is the same as saying that the value of f at any z, w0 is not 0 so long as z lies on the boundary circle this is what it says. Now I want you to you know really look at this diagram so here is so here is my here is my d which is sitting inside c2 you know I have z0, w0 here and basically I have in the so I have these two projections p1 and p2 and under this projection into c I have this I have this I have this disc centered at z0 and radius row I have this disc and in p2 of course I have a I have disc centered at w0 and radius lambda okay which is inside this copy of c okay I have this picture. And the function is defined for every z here and every z there right now you see but you see this f inverse c-0 is an open subset of c okay and therefore f inverse of this is an open subset of d after all it is a set of points where f does not vanish f inverse of this is f inverse this is the complement of f inverse of 0 this is just the complement of f inverse of 0 so it is a set of points where f does not vanish and therefore this is an open set this is an open set. So you have this open set which contains this product okay now the point about this product is this is a product this product is actually on one side this is the circle the unit circle on the other side it is a point it is a point cross the circle cross a point it is a circle cross a point and that is homeomorphic to the circle if you want but the fact is that in any case it is I want to point out that this is a compact set this is a compact set because if you want product of compact sets is compact if you want to use Tickenhof's theorem but that is too much I mean it is very clear that this cross circle cross a point is just the same as the circle topological okay and because the circle cross a point 2 if you project it onto the circle that will be topological isomorphism if you take this first projection circle cross a point to the circle that will be a topological isomorphism a homeomorphism and under a homeomorphism the image of a compact set is a compact set therefore in any case this is a compact set okay that is the fact I want you to remember. Now what you do is well you take any point zeta on the circle okay then at that point if I take f of zeta, w0 you will see that f of zeta, w0 is non-zero okay for any zeta with mod zeta minus so I am I seem to be okay let me not change notation let me keep still keep it as z for any point z let me continue using z with mod z-z0 less than I mean sorry equal to rho okay f of z, w0 is non-zero this is something that we already this is what we noted first okay and you see so what happens is that you see I have this point z, w0 in D I have this z, w0 which is inside D okay and that is a point at which the function does not vanish so it is in this open set so it is an open subset so it is an open set it is a point not only in D but it is in this open subset of D set of points where f does not vanish and therefore there is a product neighbourhood of this there is a product neighbourhood consisting of a product of a disc surrounding this point and a disc surrounding w0 which is contained in this by whatever I told you about the product of function okay. So there exists delta z so let me keep using let me use rho z greater than 0 and lambda z greater than 0 such that well so let me write it here D contains f inverse of c-0 which contains well this disc centred at z this centred at z radius rho z and this centred at w0 radius delta z so this is delta z I am sorry this is lambda z and here it is rho z. So this disc centred at z this is radius rho z okay that is a disc like this I mean the point is that I can make this I have assumed that the function does not vanish on the boundary of the disc okay and that means that there is an open set which contains this disc okay where the function so you know I can always choose a rho z sufficiently small even z where the function does not vanish so what will happen is D contains this open set D on which f is defined contains this open subset where f does not vanish that is an open set because f is continuous and the inverse image of an open set is open under a continuous function and this contains this product neighbourhood which is mod z- so I now I will have to use another some other notation so it is mod zeta-z lesser than rho z cross on the other side it will be mod w-w0 lesser than delta z lambda z so it will contain this which I mean which of course contains the point which of course contains the point well z,w0 where f does not vanish right so all I am doing is that the set of points where f does not vanish is an open set because f is continuous and you take a point where f does not vanish this point lying with the first coordinate lying on the on this boundary circle the second coordinate fixed to w0 then I am just saying that you can find this point inside a product neighbourhood okay inside this okay and what you must understand is that the fact is that the main fact I am using is that this is an open set and a point inside an open set is always contained in a product neighbourhood in that open set that is the fact about the product topology that I am using okay and now what you do is now you do this for all you do this for all z varying on this on the circle you do this for all z varying on the circle then what you will get is you will get an open covering of this compact set because you have covered every point on the circle cross w0 so what you will get is if I now vary z then all these product neighbourhoods will give me an open covering of this product set which is compact and therefore it is it is possible by the definition of compactness to extract a finite sub cover okay so I should remind you that the standard definition of compactness is that given any open cover you can always extract a finite sub cover okay that is a set is called compact if it is contained in a union of open sets then it is enough to take only finitely many of those open sets and take the union and it will be contained in that sub union okay which is called a sub cover cover simply means that the set is contained in the union of your collection okay and the fact that every cover has an every cover open cover has a finite sub cover is compactness and we prove in a first course in topology on euclidean spaces that this compactness of a set is equivalent to it is being both closed as well as bounded. The boundedness is of course that it is contained in a finite large enough ball okay that means all of its points if you measure the distance from the origin the distances are bounded okay and the other the closeness is the condition that it contains its boundary and there is no if you take a sequence of points in the set then the limit will also be in the set there is no boundary point limit point which is not in that set that is the closeness and we prove in a first course in topology that subset of euclidean space is compact defined only if it is closed at boundary okay. So that is to tell you that for any open cover for this admits a finite sub cover so you know that means that I can just take finitely many z i's okay and if I take finitely many z i's that means I will get finitely many rho z i's okay the union of which will cover this disc on this side and here I will get finitely many lambda z i's and among them I can take the least and then you will see that what you will get is that an annular an annular open annular region containing this circle cross a disc is where the function will not be non vanishing okay you will get that okay. So let me write that down the collection mod zeta minus z is then rho z cross mod w minus w not less than lambda z where the collection or let me write the collection of this the collection of sets where z lies at z varies on mod z minus z not equal to is equal to rho is an open cover covering of the compact set open cover of the compact set given by this product mod z minus z not is equal to rho cross w not and hence and hence admits a finite cover a finite sub cover okay thus f inverse of c minus 0 contains an open annulus contains a an open set of the form open annulus centered open annulus centered at z not containing this circle mod z minus z not equal to rho cross a disc centered open disc at open disc centered at w not so which is that is I can write you know I can write mod z minus z not greater than some I can write rho plus epsilon I can rho I can put rho minus epsilon cross mod w minus w not less than delta where of course this delta is actually the delta is a minimum of the lambda the minimum of lambda zi over I with zi coming from the finite cover corresponding to the finite cover okay. So basically all this is to say that if the function does not vanish on the circle cross this point then it cannot vanish on open annulus containing the circle cross small disc surrounding that point okay that is what I am saying right. So in particular what it means is that for all w such that mod w minus w not is strictly less than delta okay for all w in this disc alright the function f of z comma w does not vanish with z here that is what it means okay thus f of z comma w is not 0 for all z in for all z comma w with this condition rho minus epsilon less than mod z minus z not less than rho plus epsilon and mod w minus w not strictly less than delta this is what happens right. Now so this is a this is an annulus containing the circle cross disc centred at w not okay the function does not vanish there we already know that you see f of z not comma w not is 0 we know that where is that there it is we have f of z not comma w not is 0 okay that is f of z not f of z comma w not has a 0 at z not and that 0 is a simple 0 we know this okay what I am now going to say is that I am going to say for every w with w lying in that disc mod w minus w not less than delta I am going to say that there is exactly 1 z lying inside mod z minus z not less than rho where f of z comma w is 0 and I am going to say that that 0 is simple 0 namely 0 of order 1 okay. So the claim is the following claim for every w with mod w minus w not less than delta there exists a z with mod z minus z not less than rho and with f of z comma w equal to 0 and z is a simple 0 so I should say and in fact I should say the following thing and in fact z being a simple 0 of f of zeta comma w and moreover z is given by g of w which is 1 by 2 pi i integral over mod zeta minus z not equal to rho zeta dou f zeta comma eta over dou zeta evaluated at zeta comma w not w d zeta by f of zeta comma w okay. So here is the claim so all I am saying is that you take this w in this disc okay take w in this disc then I can find a z inside this disc such that z equal to g of w okay and this how this z depends on w is that z will be a function of w and what is that function of w it is given by this formula okay. So saying that f of z comma w is 0 and saying that z is g of w the z is given by g of w which is this function is the same as saying that f of g of w comma w is 0 so what I have done is I have solved z for w as z equal to g of w and this is the so what I have done is the explicit equation f of z comma w equal to 0 has been solved as z equal to g of w for w in a neighbourhood of w not okay and this is being done provided the you are solving for the first variable and remember that the partial derivative with respect to the first variable is not 0 okay so this is the point so this is what the implicit function theorem is. So now we will have to prove this you have to prove this statement okay and mind you in all these before I end this lecture let me tell you that it is not necessary that this g is analytic okay you cannot you cannot ensure that this g is analytic but then if you what we will see later is that if this function capital F is also analytic in the second variable for every fixed value of the first variable then you can show that this g also is analytic as a function of w and there is a nice formula for the derivative of g okay which we will see in the next lecture.