 Okay so let us continue with our discussion of the inverse function theorem okay, so I want to point out certain technicalities with respect to the inverse function theorem the proof that I gave in the previous lecture. So let me again give this look at the proof okay, so you have uhhh you have certain uhhh domain D uhhh the complex plane well it need not be bounded there is some domain D the complex plane which is the z plane and you have a function uhhh W equal to f of z defined on this domain. So this D D is an open D is a domain uhhh which is an open connected set okay and of course the open set is non-empty uhhh and you suppose we take a point z0 uhhh in the domain uhhh such that f dash uhhh the derivative of f at z0 does not vanish okay let z0 be a point in the domain with uhhh f dash of z0 not equal to 0 okay. Then of course if you look at uhhh the uhhh target complex plane this is the source complex plane with the source source variable is z the target variable is W this is W plane and uhhh W is f of z and of course the image of D will be f of D okay by the open mapping theorem uhhh uhhh uhhh f of D will also be uhhh an open connected set. See the first thing you should understand is if the derivative uhhh uhhh first of all the derivative is not 0, so it is not uhhh uhhh it is not uhhh it is a non-constant analytic function because if it is a constant analytic function the derivative will be identically 0, so it is a non-constant analytic function first. And uhhh the the the other important thing is that uhhh since this domain is actually connected uhhh it cannot be uhhh you know constant on some disc okay it cannot just be constant on a on a small disc right because identity theorem will tell you that if it is constant on a small disc then it is constant everywhere right that is because the domain is connected okay therefore this function is a uhhh is a non-constant analytic function this because of this condition alright. And uhhh the open mapping theorem says that a non-constant analytic function maps open sets to open sets and since D is an open set f of D will also be an open set. So, you see this f of D will be uhhh an open set in the uhhh uhhh in the complex plane by uhhh so by the open mapping theorem. And see the point I want to make is that so so let me write it here this implies f is non-constant, so f of D is an open set and you know under uhhh under continuous function the image of a connected set is connected. Since D is connected f of D is also connected so f of D is also connected and this implies that f of D is also a domain that is also a domain alright. So f of D uhhh I mean I have so D for me is uhhh uhhh some region with some which may be unbounded it may have partial boundary okay and then I will have similarly I will have some f of D it will be some region with boundary here and uhhh it may not be fully bounded in all directions okay. But it is an open uhhh connected set okay and you take uhhh let the value uhhh let f take the value w0 at f of z0 so uhhh put w0 equal to f of z0 okay. Now see what what f dash of z0 is not equal to 0 tells you is that uhhh z0 is a 0 of f of z-w0 uhhh of order 1 okay. f dash of z0 is non-zero implies z0 is a 0 of f of z-w0 of order 1 this is what it means alright. Because if you have 0 of higher order then all then all derivatives uhhh up to uhhh uhhh up to that uhhh uhhh uhhh all derivatives up to 1 less than that order will vanish okay. If it is a 0 of order 1 the first derivative will not vanish but the 0 derivative which is a function which will that will vanish. If it is a 0 of order 2 then the function will vanish at that point and the first derivative will vanish. It is a 0 of order 3 then the function it is first derivative and the second derivative will vanish the third derivative will not vanish okay. So, all derivatives up to 1 less than the order will vanish okay. So, the fact that f dash does not vanish means that z0 is a 0 of order 1 okay. This is essentially if you want uhhh this follows from uhhh looking at the Taylor expansion or by looking at the definition of what is meant by order of a 0 okay uhhh Taylor expansion of f uhhh at a in a disc surrounding uhhh in a disc centered at z0 okay. Now uhhh the point is that uhhh you look at the function. So, you know you choose uhhh uhhh now since f is a since f is a non-constant analytic function f of z-w0 is also a non-constant analytic function and you know the zeros of a non-constant analytic functions are uhhh uhhh the zeros are isolated okay. Therefore uhhh what does it mean give me a 0 then there is a disc I can find a disc okay a closed if you want I can even include the boundary of the disc and make sure that there is no other 0 of that function except the point at the center of the disc which is a chosen 0 okay. So, isolation of zeros tells me that I can find a disc uhhh of radius rho centered at z0 such that on this disc including the boundary there are no other zeros of f f of z-w0 except this 0 at the center z0 which is a 0 of order 1 okay. So, uhhh by isolation of zeros by isolation of zeros there exists a rho positive such that z0 is the only 0 of f of z-w0 in uhhh mod z-z0 less than or equal to rho okay. In other words the function f of f of z assumes the value w0 only once in that closed disc and that is at uhhh the point z equal to z0 that is the center of the disc and the multiplicity is only 1 it assumes the value only once because if if it assumes the value twice then the order of the 0 uhhh of f of z-w0 at z0 will become 2 and so on but the order of 0 is the order of the 0 is 1. So, it assumes the uhhh the only 0 it has uhhh is at z equal to z0 and the multiplicity of the 0 is 1 that is what you must understand okay. Now now what you do is that you look at the function you look at the modulus of this function mod of f of z-w0. Now that modulus on this boundary is non-zero and it has a minimum okay and that is what we called as delta alright. See uhhh uhhh let delta be the minimum positive value of mod of uhhh f of z-w0 in uhhh on on the boundary disc on the boundary circuit okay. See if you look at mod mod of f of z-w0 that vanishes only at one point in the closed disc that is at the center okay at no other point it vanishes because if it vanishes at some other point that will become a 0 of f of z-w0 but we have assumed that there is only 1 0 because of isolation of zeros. So in particular on the boundary circle also it will not vanish okay and because it is modulus of a uhhh modulus it will be a uhhh it is a non-negative function okay. In fact it is a positive real valued function and it is a continuous function and it is a continuous function uhhh you uhhh when you restrict it to this set this is a compact connected set. Therefore the image of that continuous function on the real line will be a closed interval okay and the and this delta will be the left end point of that closed interval okay. A continuous the so I am just using topological facts uhhh that the continuous image of a connected set is connected the continuous image of a compact set is compact. A subset of the real line is connected if and only if it is an interval uhhh subset of the real line is compact if and only if it is both closed and bounded and an interval if it is a closed set then it has to be a closed interval okay. So well all that gives you that delta is the uhhh is the uhhh uhhh uhhh there is this value delta that we want and it is positive okay. Now what you do is uhhh uhhh now we define uhhh uhhh the function g of w to be equal to 1 by 2 pi i integral over uhhh mod uhhh zeta minus z not is equal to rho uhhh uhhh zeta f dash of zeta d zeta by f zeta minus w for mod w minus w not less than delta you define now you define this function okay. Now so look at what is happening see mod w minus w not is strictly less than delta is the disc centered at w not and radius delta in the target plane target complex plane the diagram I have already the way I have drawn the diagram it shows that the disc is already in the image okay the way I have written it it shows that the disc is already in the image okay but that that is actually the conclusion alright. See you define this function first okay you define this function and uhhh first of all to uhhh to say that this function is well defined you see what is the integrand the integrand of this function is this is this is this function okay and where is this function considered it is considered on the boundary circle alright mind you I am integrating with respect to zeta and w uhhh is not part of the variable of integration so after I integrated I will get something that depends on w and that is what I am calling as gfw okay. Now if you look at the integrand okay then you can see the integrand is continuous on this uhhh uhhh on this boundary circle because you see f uhhh f f is analytic therefore f dash is also analytic so f dash is continuous and uhhh this is just z f dash z by f of z-w this is what that integrand is as a function of z where z varies on this boundary circle okay I am only changing the variable of integration from z to zeta for an obvious reason which I which will soon become apparent okay but the fact is that uhhh this integrand is continuous mind you this denom the only problem with the integrand the continuity of the integrand is that the denominator can vanish okay but the denominator cannot vanish because if the denominator vanishes then I will get a value on the boundary circle for which f of uhhh is I will I will get a z if I get a zeta on the boundary circle such that f of zeta is equal to w okay then it means that uhhh the distance of f of zeta from w0 is less than delta but you know for every point in the boundary circle the distance of f of zeta minus w0 is greater than or equal to delta okay which is a contradiction. Therefore this the denominator is never going to vanish therefore this is a continuous function it is a continuous function on the boundary therefore I can integrate and if I integrate I will get a function so this function is well defined there is no problem about it okay to integrate a function I all I need is a continuous function on the curve the moment you have continuous function on a smooth curve or piecewise smooth curve okay you can always integrate it and if the resulting uhhh integral is well defined it is just the Riemann integral alright. So, this integral exists but it will depend on w because this w is uhhh is is is here on the right side okay after you integrate it I will get something that depends on w and that is what I am calling as g of w. So, this g of w is well defined there is no doubt about it alright and mind you g of w is a map uhhh defined on this disc and it is taking complex values the fact is that this g of w will actually go back into this disc and it will actually be equal to z where f of uhhh z is w. So the fact is g of w is well defined by the choice of delta uhhh delta okay but the fact is g is nothing but f inverse the fact is that this g is a formula for f inverse and f in and that uhhh g which is equal to f inverse is an analytic function for of w in this disc. So, the picture is that you are you are going back so your g is like this so g is equal to f inverse is defined like this okay and see the fact that the point is if you take the inverse image uhhh of if you if you take the uhhh inverse image of this disc under f okay if you take the inverse image of this disc under f. So, maybe I will be a little careful and say that uhhh uhhh that this disc I choose also uhhh delta small enough so that this disc is inside f of d okay. So, I perhaps it is not required but let me uhhh further further let me write this further let uhhh uhhh f of the let the image of uhhh the disc mod z-z0 uhhh uhhh strictly less than rho contain uhhh the disc mod w-w0 less than delta let me assume this okay. So, I mean w0 is in the image alright and I have chosen a delta but uhhh if necessary let me make the disc delta smaller so that I am I assure that this whole disc is in f of d okay let me assume that. So, the fact that f the disc is in f of d will tell you that every point in the disc is a value of f. So, there is always a z such that f of z is equal to w alright and the fact is that g what is this g of what is this g of w it is f inverse of w and that is the same as z. So, this this if you compute this you will get z okay where z is the uhhh uhhh is the point at which f takes the value w alright and more importantly you will also have that the uhhh the function will turn out to be injective on the inverse image of that disc okay okay which is uhhh which is the which is either you take the inverse image of that disc under f or you take the image of that disc under g they are one and the same okay. So, uhhh further so what is happening is further what you will have is f is one to one on uhhh f inverse of that disc okay which is the same as g of that disc and mind you see f inverse is uhhh uhhh f inverse is analytic f is f inverse is just g which is analytic. So, it is also an open map so the image of this open disc will again be an open but it will be inside this okay. So, what you must understand is uhhh that the the the the inverse is defined okay and so f will be one to one on this and uhhh f restricted to this disc f inverse of uhhh uhhh mod w-w0 less than delta is a holomorphic isomorphism okay. Now the fact there are the the the technical things that I want you to understand are two points okay the first point is that f takes uhhh f f is one to one on that disc uhhh on the on the inverse image of of this disc and uhhh the the reason for f being one to one is because of this formula okay and in fact you know uhhh why f is one to one on that disc is also because of this uhhh this function n of w that we define. See you can define the function uhhh n of w to be 1 by 2 pi i integral over mod uhhh zeta minus z0 equal to row uhhh uhhh f dash zeta d zeta by f zeta minus w okay if you if you look at this what is this this the the right side is 1 by 2 pi i integral over mod zeta minus z0 equal to row of d log f of zeta minus w this is what it is this is what the right side is and what is this uhhh what does this give you by the argument principle or the residue theorem what this tells you is you will get the number of zeros minus number of poles of f zeta minus w in inside this disc okay but inside this disc this will have no poles because this integrand the integrand is is this guy the integrand is this guy and this the integrand is continuous the denominator does not vanish because of the way we have chosen delta okay. So this uhhh uhhh there the denominator does not vanish that means this has no poles alright and it will have uhhh only 1 0 and that see the uhhh I mean it will have zeros at the points uhhh uhhh zeta where f of zeta equal to w okay it will have zeros those will be the points z which will be mapped to w and the number of times uhhh if you count those points with multiplicity that is the number you are getting here this is the number of times the function f assumes the value w inside that disc okay this is an integer alright but what we checked was that this the varies for what values of uhhh w it is defined it is for all those w with mod w minus w not strictly less than delta okay. We check that this is actually an analytic function of w you write you know write uhhh going back to the uhhh proof of the open mapping theorem alright. So this is a is an analytic function of w of w integer value hence constant hence it is also equal to n of w not okay. You have an you have an analytic function which is integer value and it is defined on a connected set therefore that integer value has to be only one value that means analytic function is constant therefore it is the same value no matter what value for w I put in particular I can put w equal to w not. So n of w is same as n of w not but what does this but what is n of w not n of w not is one okay in that disc the number of times f takes the value w not is only one it takes the value w not at the centre of the disc z not and the multiplicity is one because the 0 is of order one for f of z minus w not at z equal to z not okay. Therefore what but so what you get you get for every w uhhh in this disc you get n of w is equal to one you get this fact okay. Now if you go back and if you think about this fact what this tells you it tells you two things it tells you that for every w such that mod w minus w not is less than delta n of w equal to one means there is always a z inside this disc such that f of z is equal to w that is the first point and the second point is there is only one z and the uhhh the multiplicity with which uhhh f assumes the value w at z is one because if z if it assumes the value w uhhh at z twice then this n of w will become 2 because this n of w is with is the number of times the function takes the value w the function uhhh f of z term f of z minus w takes the value w I mean takes the value 0 which is the same as number of times the function uhhh uhhh f of z takes the value w counted with multiplicity okay. So this is one so what this tells you if you know if you know recall this this is not necessary this is automatic this is automatic what it tells you is that uhhh the you need to you need to assume this so the originally the diagram that I drew was correct okay this whole disc will automatically be in the image this disc will be in the image in fact the disc will be not in f of not only in f of d this disc will be in the image of this disc itself that is uhhh uhhh which is what I assumed but it did not be assumed it follows from here okay that is one technical point and what what this will tell you this will tell you that f restricted to the inverse image of that disc is one to one because what does that tell you tells you every value w with mod w minus w not less than delta is assumed by f only once in the disc uhhh centred at z not radius uhhh uhhh row okay excluding the boundary. That means that is another way of saying that f restricted to the inverse image of this disc is one to one okay that is what I have written here okay f is one to one on the inverse image of that disc okay that follows from there okay and there there is another technicality that I wanted to tell you see if you go back to the proof last time uhhh of uhhh proving that uhhh g of w is an analytic function of w at some point we needed that uhhh uhhh for w equal to f of z we needed that f dash of z is not 0 okay we needed that the derivative does not vanish at z okay and I am saying now that is also a consequence of this okay see let me recall for your benefit uhhh how we got this g of w equal to z how did we get g of w equal to z. First of all uhhh for any w such that mod w minus w not is less than delta there is a z such that w equal to f of z that is because of the fact that n of w equal to 1. So, there is a certain z take that z okay then f of z that z will be w alright now my claim is that f dash of z is non-zero see if f dash of z is 0 then mean that means you are saying that f the derivative of f minus w is 0 that means the uhhh uhhh the 0 z of f minus w has order greater than 1 okay see f dash of z uhhh uhhh for for uhhh z in mod z minus z not less than rho and with w equal to f of z has to be non-zero why because if it is 0 if f dash of z is 0 then the derivative of f of uhhh the uhhh you it will contradict n of w equal to 1 you will get n of w greater than 1 for otherwise okay which is not correct n of w strictly equal to 1 what is n of w n of w is the number of times f assumes w and how should you think of it you should think of it as the number of zeros of f f of zeta minus w n of w is number of zeros of f of zeta minus w and z is a 0 of f of zeta minus w because z is f of w uhhh sorry w is f of z. So, z is a 0 of f of zeta minus w and it is 0 of order 1 because if it is 0 of order greater than 1 which is what uhhh will happen if the derivative of f of zeta minus w with respect to zeta is 0 which is the same as saying at at uhhh which is same as saying f dash of z is 0 then it will then this n of w will become 2 at least 2 but n of w is 1. So, that fact forces that you know uhhh f dash is non-zero okay. So, what what I want to tell you is that uhhh uhhh that is you also get that f dash of z is non-zero for all uhhh z uhhh in f inverse of that disk. So, when you take the inverse image of this disk here you will get some you will get a uhhh open uhhh neighbourhood of z0 okay and the fact is that open neighbourhood uhhh will be a neighbourhood restricted to which f is not only 1 to 1 the derivative of f will also not vanish and it is the non-vanishing or derivative that which actually makes this into an analytic function because what what is because uhhh how do you get the g of w is actually z that unique z for which f of z equal to w how do you get that you get it by applying residue theorem here. See 1 by 1 by 2 pi i integral over uhhh mod zeta minus z0 equal to rho of this function zeta f dash of zeta uhhh by f of zeta minus w d zeta this is actually residue of zeta f dash of zeta by f zeta minus w at uhhh z uhhh at uhhh uhhh at z this is a residue why because see the for this function this function has a simple pole at zeta equal to z which lies inside this inside the region bounded by this contour that is the only simple pole. So the so the fact is that when you take the residue of this function at z okay because this is a residue at a simple pole so here uhhh uhhh this is I am just applying the residue theorem. So when you take residue of the simple pole this is just limit zeta tends to z of zeta minus z times that function zeta f dash of zeta by f zeta minus w by f zeta minus w but what is w? w is f z okay then if I put push this zeta minus z to the denominator I will get limit zeta tends to z f zeta minus f z by zeta minus z which is f dash of z. So I will get f dash of z in the denominator and the numerator will become e z f dash of z and I will have to cancel off this f dash of z in the numerator and denominator and that will give me z this is just by the residue theorem but to show that so what does it show it shows that g of w is that z is equal to this z. So in other words it tells you that uhhh uhhh w is g inverse z but w is f of z so it tells you that g inverse is f okay. So this is just residue theorem and to apply this I have to cancel off this f dash of z when I take the limit here and when can I cancel f dash of z I can cancel f dash of z only if f dash of z is non-zero the derivative does not vanish and why is that guaranteed that is again guaranteed because of n of w equal to 1 that is what I want you to understand this is the crucial point okay because n of w is 1 f dash of z is non-zero alright for the unique z such that f of z equal to w therefore I can cancel this f dash of z here and and get the fact that g of w is actually z which is f inverse w okay so that is the technical point. So now if you put all these things together uhhh what you actually get is the following so so let me write this as a uhhh theorem or let me write this as a corollary that is to you know uhhh this is to make you uhhh this to sum up the relationship between when a function is uhhh 1 to 1 uhhh how that is related to the fact when a function has uhhh non-zero derivative so what are the two situations? So uhhh so the first the first statement is if f is uhhh analytic and 1 to 1 on a domain then f uhhh inverse is holomorphic or analytic that is f is a holomorphic isomorphism or analytic isomorphism. So you see the first corollary is that a 1 to 1 analytic function is a isomorphism a 1 to 1 holomorphic function is a holomorphic isomorphism it is a very very strong statement it is a corresponding statement for real valued function is not true a 1 to 1 real valued differentiable function okay uhhh is not uhhh differentiable isomorphism you can find a 1 to 1 real valued differentiable function you can find a 1 to 1 uhhh maybe even infinitely differentiable function okay uhhh which is 1 to 1 but for which the inverse function is not differentiable. For example simply take the function from the real line to the real line given by x going to x cube just going just take the function x going to x cube the inverse function is well defined by the cube root okay which which exists the real cube root alright. So it is 1 to 1 function and it is infinitely differentiable x cube is just a polynomial so it is infinitely differentiable but if you take the inverse function which is x to the 1 by 3 it is not a differentiable function at the origin okay. So the moral of the story is a 1 to 1 differentiable function of 1 real variable is not uhhh the inverse function uhhh it need not be differentiable but that does not happen in functions of one complex variable as you know whatever happens in functions of one complex variable is completely different from what is what happens with functions of one real variable and this is one remarkable thing that happens that any 1 to 1 analytic or holomorphic function is automatically an isomorphism the inverse function also becomes holomorphic it becomes analytic that is a very deep fact okay and uhhh how do you how do you prove this statement how do you prove a statement like this I am saying this proof follows from here from our argument see because you see the movement you say f is 1 to 1 okay the movement you say f is 1 to 1 uhhh what it means is that if you fix a point z not in the domain where f is 1 to 1 okay the fact that f is 1 to 1 will tell you n of w not will be 1 the fact the say saying that the function is 1 to 1 means that uhhh if you fix a point z not in the source domain and you take its value f of w f of z not as w not then it tells you that z not is a 0 of or simple 0 it is a 0 of order 1 of f of z minus w not. So, n number of times f assumes the value w not is 1 but then by this argument the number of times the function assumes the value any nearby values will also be 1 and I told you that this argument actually tells you that the uhhh the derivatives cannot vanish because the movement the derivative vanishes then the number of times that the the the function takes the value at that point goes up okay. So, the moral of the story is that if f is 1 to 1 then you are in this situation this n of w this will become 1 and this n of w becoming 1 will force that f dash is non-zero. So, in particular if if if f is analytic and 1 to 1 on a domain then the then the derivative of f cannot vanish the derivative of f cannot vanish and f inverse will be defined f inverse will be uhhh defined and it will be holomorphic by this argument and mind you to check a function is holomorphic it is enough to check at every point. So, the same argument that we have here I can go through the same argument I can fix z not then I can take this w not then I will get that f inverse will be holomorphic when I restrict this disk but then I can cover the target domain by all by such disks by changing z not in the source domain. So, I will get that the inverse function f inverse is holomorphic at every point I mean after all holomorphic at one at a point means it has to be holomorphic in a or analytic in a disk surrounding that point and I can cover the whole whole region. So, I can check that f inverse is holomorphic locally but that is good enough to check the def inverse is holomorphic because checking holomorphic holomorphicity or analyticity is a local check okay it is like checking continuity if you want to check a function is continuous on a set it is enough to check on an open cover okay similarly if you want to check a function is holomorphic or analytic it is enough to check on an open cover. So, I can take the open cover to be given by such disks okay so the moral of the story is if f is analytic and one to one on a domain then f inverse is analytic and f is an analytic isomorphism in particular in particular f dash is not 0 on the domain f dash can never vanish okay. So, if it is one to one one to one one to one analytic functions very strong condition it is a terribly strong condition it tells you that it is an isomorphism okay and in particular it also tells you that the derivative will never vanish. So, this is the first corollary this what is the second corollary the second corollary is the other condition instead of assuming f is one not one to one suppose I assume derivative of f does not vanish then what do I get okay the answer to that is the second corollary which says in that case you will get a local holomorphism okay. So, let me write that if f is analytic on a domain analytic or holomorphic on a domain and f dash is not equal to 0 on the domain by which I mean that f dash never vanishes the derivative first derivative of f never vanishes on a domain then f is a local isomorphism it is a local homo it is not a in particular f is locally one to one but not necessarily but need not be globally one to one okay. So, you know so the whole purpose of this lecture was to try to tell you what is the difference between assuming one to one on the function f which is analytic and assuming the derivative does not vanish if you assume it is one to one then it is an isomorphism it is a very strong condition if you assume it is the derivative and if you assume one to one then the derivative cannot vanish but if you assume you do not assume it is one to one but if you assume the derivative does not vanish then is need not be one to 1, it could be many to 1 but the point is locally it will be 1 to 1, so locally it will be an isomorphism by the first part, it will be a local isomorphism okay. So and what you must understand is that is what it says that is what this argument also says what we have said is that if f dash of z0 does not vanish okay then we have found this disk where f dash in the inverse image of this disk you know that f dash does not vanish and there it is an isomorphism we have already proved and this happens for every point z0 locally okay globally it need not be an isomorphism and what is the illustration for this, the illustration for this is the if I will give you a simple illustration for this that will help you to think take the function z going to z cube, simply take the function z going to z cube. So you know here is a complex plane z cube is f of z, f of z which is my w okay, so this is the z plane which is the complex plane and then here is a target plane which is the complex plane which is the w plane and you know what will happen is 0 will of course go to 0, origin will go to the origin but if you take a point different from the origin then funny things will happen in fact as you will know you know I take this ray which for which this angle is 2 pi by 3 okay this is the ray on which the and then I take this other ray which is a mirror image of this which is for which the angle is 4 pi by 3 which is 240 degrees this is 120 degrees okay this is 4 pi by 3 and you know these the cube roots of unity will lie on these 3, so one will lie here and then this is omega complex cube root of unity and then there will be another the other one is of course as you know omega squared they will lie on this ray and the fact is what this function will do is it will map a whole sector this whole sector of 120 degrees will be mapped on to the whole plane because you know any point here can be written as Re power i theta z equal to Re power i theta if I apply z cube I will get r cube e power i 3 theta, so as theta varies from 0 to 120 degrees 3 theta will go from 0 to 360, so this whole sector will be mapped on to the whole disk I mean on to the whole plane similarly this sector from 120 degrees to 240 degrees that will again be mapped once more to the plane and from 240 degrees to 360 degrees that will again be mapped once more, so this so as you go around once if you take the unit circle okay which passes and you go around once from 1 to omega to omega squared and come back the unit circle there will go 3 times the unit circle will be traverse 3 times and what is going to happen is you see if you omit the point 0 here then you can omit the point 0 there and the what about the derivative the derivative is what derivative is 3 z squared the derivative will not vanish if you omit 0 but so the derivative is non 0 but still the function is not 1 to 1 it is 1 to 1 only on this any single sector if you take there it is 1 to 1 on the whole it is 3 to 1 okay on the whole it is 3 to 1, so what you get is it is many to 1 but if you restrict it to each of these sectors it becomes 1 to 1, so it becomes locally 1 to 1 if you omit the origin and what is this what is each sector it is the inverse image of the deleted plane the punctured plane. So you take the target complex plane remove 0 and take the inverse image then each of these will be each sector here with the origin removed there will be 3 inverse images and on restricted to each of these this will be a holomorphic isomorphism okay and the problem is at the point 0 because the derivative vanishes and at that point there is no neighbourhood of that point where the function will be 1 to 1 and that is the reason why the points at which the derivative of an analytic function vanishes they are called critical points because you cannot find any neighbourhood where the function will be 1 to 1 it will however small a neighbourhood of that point you take the function will become many to 1 okay whereas the derivative does not vanish you can always find a neighbourhood where the function is 1 to 1 and in fact even a holomorphic isomorphism that is what the second corollary says okay. So this is an illustration of both of these things okay. So I will stop here.