 Okay so let us continue with our discussion, so you know basically we are studying zeroes of analytic functions and our aim is to begin with the argument principle which is essentially corollary of the residue theorem and then try to prove some of these important theorems the Roush's theorem, Hubert's theorem, open mapping theorem and the inverse function theorem. Begin with the argument principle, so let me start here, of course you can look at a proper proof of the argument principle in any standard book on complex analysis but I will try to tell you how you get it, so basically you are looking at, so you are basically having a you know a contour, a simple close contour here, so simple of course means that it is, it does not cross itself there are no self intersections and when I say contour it is piecewise smooth, okay. So when you write a parameterization for this contour then the function, consider it as a function of the parameter it is continuous and is differentiable with respect to the parameter and the derivative with respect to the parameter is also continuous, okay and this should happen piecewise, right. So that is what a simple close contour is and we are looking at a function f of z which is, so you know I will call this domain as the interior I will call it as D and I will call the contour as doh sub D, so the partial doh or del depending on what you used to this doh D is always the boundary of D and that is the bounding contour and f of z is assumed to be is analytic in on doh on the union D union doh D except for isolated poles in D, okay and of course f has no zeros on the boundary and f of z has no zeros on the boundary, okay. So this is the assumption, so which means that you see there are, so there are poles, they are isolated poles, they are isolated singularities and the fact that there are isolated poles already means that there are only finitely many of them, okay and so there are points z1, z2, etc, zn which are z1 through zn are finitely many poles, poles of orders well if you want n1 and so on oops, so let me use capital N1 N sub N, okay respectively, okay and of course you must understand that the fact that there are only finitely many poles is a consequence of topology because you see you already assumed that there are only isolated poles and you know function which has only isolated poles is called a meromorphic function that is the language that we use, a meromorphic function is a function which the only singularities that it has are isolated poles, okay in the region where it is defined. So of course it is not defined at the poles but places where it has singularities are supposed to be isolated poles, okay such a function is called a meromorphic function. So basically f is a meromorphic function but on the boundary it is analytic and it is never 0 on the boundary, okay and see there are that it has only isolated poles in this region in this domain D follows from the fact that it follows from a little bit of topology. Because if there are this region D along with the boundary becomes a compact set, okay D along with the boundary becomes a compact set and you know since you are in Euclidean space in a compact set if you have an infinite subset there will always be an accumulation point, so if the number of poles is infinite, okay then if you take this subset of poles that becomes an infinite subset and that being inside a compact set will certainly have a limit point, okay and that limit point certainly will also be a pole, okay it will certainly not be a point of analyticity and then you end up with I mean you will get a contradiction to the fact that all this that limit point will also be a singularity and you will get a contradiction to the fact that all the singularities are isolated because in every neighbourhood of that point you will have a singularity so it is not isolated you will get a non-isolated singularity, okay you assume that there are only isolated singularities and they are poles then because of compactness they are going to be only finitely many poles and as so these are poles and maybe it is a good idea to use let me use P1 to Pn to be the poles, okay so let me use P1 to Pn and then of course similarly there are only finitely many zeros, okay the reason is because I have already told you the zeros of an analytic function are also isolated, okay the zeros of an analytic function are isolated this is a I mean this is a result of this that you should have come across in a first course in complex analysis and what is the reason the reason is actually the identity theorem if you want, okay see if you have a if you have a if you have a zero that is not isolated, okay it means that in every neighbourhood of the zero you can find another zero, so you can find a sequence of zeros which go to zero, okay and therefore what happens is that your function is zero on a set which has an accumulation point and the identity theorem will then tell you that the function has to be identically zero, okay if a power and that again also is true if you look at a power series, if you have a power series and if it vanishes at a point and it vanishes at a point in every neighbourhood of that point that means you have a sequence of points where it vanishes and finally tends to a point the sequence tends to a point where also it vanishes then the power series has to be complete to the zero series all the coefficients will vanish, okay. So the only way in which you can have a non isolated zero is that it is zero everywhere, okay for an analytic function. In other words if you have zeros they have to be isolated, so the set of zeros will also form and will also be isolated and again the same compactness argument will tell you that there are only finitely many zeros inside, okay. So all those zeros by as say Z1, Z2 etc Z sub m, okay so Z1 through Zm are the finitely many zeros of orders L1 etc Lm, okay and of course there are no zeros on the boundary, on the boundary the function is analytic, okay there are no singularities on the boundary but I am further assuming that there are no zeros on the boundary, okay all the zeros are only inside and there are only finitely many, okay and of course when you take a zero or a pole you have to count it with multiplicity, okay you have to worry about the order of zero or the order of the pole, right. And well see the point is you know if you are looking at a let us assume that you are looking at a simple zero or a pole I mean you are looking at a single zero or a pole, okay. So suppose I have a point let me call just call it as Z0, okay and I surround it because it is isolated from the other zeros or poles suppose I surround it by a very small disc whose boundary is a circle, okay and again I can parameterize the circle as I mean if I take the radius to be rho small enough radius then there is no other zero or pole here, okay and then the function f of Z can be written as Z-Z0 to the power of n let me use something m if you want into g of Z, okay where g of Z is analytic at Z0, at Z0 and g of Z0 is not zero. Of course if m is either positive or negative, okay m is positive is Z0 is a zero, okay and m is negative if Z0 is a pole, alright. If m is positive then m is the order of the zero at Z0, if m is negative then minus m is the order of the pole at Z0, okay and now you know if you calculate d log f Z, okay what you will get is you will get I mean this is essentially by definition f dash of Z by f of Z this is the logarithmic derivative, alright and if you calculate it what you will get is well if you differentiate this with respect to Z you use product rule what you will get is m into Z-Z0 to the m-1 times g plus Z-Z0 to the m d dash of Z divided by f dash by f which is just Z-Z0 to the power of m g of Z this is what you will get, okay and if you expand it out then what you will get is as follows you will get well and that will turn out to be so d log f Z will be so you know when I say this I am worried about f only in the small neighbourhood of Z0 where I have chosen the radius rho very small and where there are no other zero supports, okay Z0 is the only zero okay and there is no zero or pole even on the boundary, okay so I am taking such a small neighbourhood such a small disc and that is possible because all the zeros and poles they are all isolated, okay they can be separated from each other by small discs, okay. So d log f Z will turn out to be well if I calculate this I will get well I will get m by Z-Z0 plus you know here I am going to get an analytic function see what you must understand is see g see the function g cannot see the function g cannot vanish anywhere here because the function g vanish at a certain point that will also be a 0 for f, okay but I have assumed f has no other zeros or poles. So in principle g does not vanish at the g does not vanish and at Z0 and it is analytic so this piece so the second term is g dash by g, okay and g dash by g g0 vanishing at Z0 okay is an analytic function, okay because you know if an analytic function does not vanish at a point then 1 by that analytic function is also analytic function at in a small neighbourhood at a point, okay. So if you want you can shrink further the neighbourhood if you really want, okay. So the fact is that the second term will be g dash by g dash by g that is a logarithmic derivative of g and that is analytic, okay and now you know if I integrate 1 by 2 pi i integral over gamma okay yeah integral over this gamma of d log fz if I do it you know I am going to integrate this and you know if I integrate this part that is the integral we have already calculated you will get m which is the residue. So you will get m plus if you integrate this part you will get 0 because that is Cauchy's theorem. Cauchy's theorem says that if you integrate analytic function you are going to get 0 for closed curve simply closed curve. So the net effect is that you get m. So what you see is that if you do if you take a small enough circle surrounding a 0 over a pole and you compute the integral 1 by 2 pi i of the logarithmic derivative you end up getting exactly the order of the 0 over the pole. And now all you have to do is simply surround each of the 0s and each of the poles by such small discs okay and then use the Cauchy's theorem in the region that is gotten by taking away from d these discs where the function f is completely analytic and the Cauchy's theorem will tell you that the integral over the boundary is sum of the integrals over each of the small discs but the integrals over each of the small discs gives you the number of gives you the order of the pole or the 0 and therefore you get the residue theorem namely you get you so implies by Cauchy's theorem that 1 by 2 pi i integral over the boundary d log fz actually will give you the number of you will get if there are 0s you will get all the l i's so you will get sigma l i minus and if it is a poles you will get the minus n i's okay. See mind you if what you understand is if this is a pole then m is negative okay and when I write n is an order of the pole p1 then this n1 will be this m will be minus n1 okay so what you will get is sigma l i minus sigma m nj which is precisely number of 0s minus number of poles n sub 0s number of 0s n sub infinity is number of poles okay. So this is this is basically the argument principle okay it is an application of residue theorem plus Cauchy's theorem so of course the question is why is it called argument principle okay so the answer to that is that this quantity here is actually the change in the argument of f of z as z varies over the boundary okay so why the reason why it is called the argument principle is the following you normally write log a plus i b where a and b are real numbers okay i is of course always square root of minus 1 a square root of minus 1 you always nively write it as mod of a plus i b plus i times argument of a plus i b okay this is how you define the logarithm plus but of course this is not the you know this is a multiple value thing in fact you can add to this argument you normally add you know plus 2 n pi okay and various values of n are supposed to give you the various logarithms the only requirement is that a plus i b should not be the 0 complex number should not be 0 okay when the problem is that when the 0 the argument of 0 is not defined okay so mod of 0 is of course defined argument is not defined so it should not be 0 so you of course you get a logarithm for every non-zero complex number and all these logarithms they all differ by integral multiples of 2 n pi I mean integral multiples of 2 pi i okay and in the same way you can write well you can write log fz as you know log mod fz I think I have forgotten there is a there is a ln here which I have forgotten okay yeah I have forgotten this yeah I have forgotten that so this is ln which is log natural log to the base e and this of course is also log to the base e for that matter only thing is that this is a complex logarithm that is a real logarithm to both to the base e okay. So if you write log fz you will get again naively you will get ln mod fz plus i times argument of fz plus 2 n pi you can write something like this okay but then there is a this is okay if you fix a particular value of z okay but then the only problem is that you know if you try to write this uniformly for all values of z in a domain then you might end up in trouble in the sense that you see the problem is that this log fz need not define analytic function okay you what you will get is you will have to you will have to do some kind of slitting of the domain to you have to throw away some points from the domain to get what is called a branch of this logarithm okay and the only case when you will be able to get a I mean one case where you can always get a logarithm will be if your domain is simply connected and the function never vanishes on your domain you can always find a logarithm okay. But the point is that you know so you know you can naively write log fz is d ln mod fz plus d if you want argument of fz you can write it like this where d is supposed to be the difference as you change z along an arc okay you can change z along an arc and you can make sense of the difference in these values and of course here you choose a particular branch of the logarithm. So what I will do that I would rather put a capital A and you should be able to choose a particular branch of the logarithm. So in particular what I am saying is that you can do this on a nice arc for example you can do this on this boundary arc you can always do this on the boundary arc you can write that and why you can do that is because on the boundary z can be written as a smooth function piecewise smooth function of a real parameter okay. So then you can so you know on doh d if it is which is parameterised which is parameterised by gamma of t alright where t is a t way is in an interval on the real line okay gamma piecewise smooth which means that you know gamma is continuous it is differentiable with respect to t, continuous with respect to t, differentiable with respect to t and the derivative with respect to t is also continuous piecewise that is what piecewise smooth means okay. So gamma piecewise smooth with respect to t okay what you can do is you can write you can write integral over doh d, d log fz as integral from a to b if you want integral over gamma of d log f of gamma of t and that will be integral over gamma of d ln mod f of gamma of t plus integral over gamma d argument of f of gamma of t okay where of course here you can choose a particular single valued branch of the of the logarithm and actually what will happen is so you know what I am just trying to compute what this what this integral is what is the logarithmic integral over a simple closed contour like that. If you compute it what will happen is see this part will be 0 okay the this part of the integral will vanish okay and this part of the integral will give you the difference in the argument of the function from the starting point to the ending point of course you know when we do this integration over the over the contour when you parametrize it then you know you choose some point which you take as gamma of a and that is also equal to gamma of b. Gamma is the parametrization of the path and the parametrization is a map so you think of parametrization as a map ab is the is the interval on the open line I mean on r1 and gamma is a function and of course gamma of t has a real part has an imaginary part and for different values of t you are going to get different points you are going to get different complex numbers and as t changes from a to b gamma traces this path and the starting point is equal to the ending point so gamma of a is the same as gamma of b okay and the fact is that if you if you go around once like this the first integral will vanish okay and the second integral will essentially give you the change in the argument of f of z as you move across the across gamma so what you will get is see this will turn out to be just f of argument change in the argument of f, change in the argument of f of z as you go around once along the boundary dow d which is now parametrized by gamma okay this is change in the argument so the fact is that if you calculate this you know this logarithmic integral you are actually getting the change in the argument okay you are getting the change in the argument of the function alright and mind you in all these calculations I had taken arg to be a fixed set determination a fixed branch of the logarithm. If I take a different branch of the logarithm you know any two different branches of the logarithm will differ by a constant multiple of 2 pi i but since you are taking the difference this value will not depend on which branch of logarithm you took okay instead of taking arg which is one branch suppose I had taken arg prime which is some other branch okay then arg and arg prime will differ by say some 2m pi but when you take the difference the 2m pi will cancel out so this quantity which is a change in the argument that is not going to change okay. So I mean roughly what you must think is that what you must try to understand is that when you calculate this logarithmic integral over the boundary you are actually getting the change in the argument of the function okay and the change in the argument of the function could be 0 or it could be something it all depends on what the function is and it depends very much on the zeros and poles of the function inside that is what the residue theorem says what the residue theorem actually says is that it says that you see the change in argument of the function is 2 pi times the difference between the number of zeros and number of points counted with multiplicity. So this is another way of looking at the argument principle and this is what lengths the argument principle it is name okay so when you calculate the when you calculate the logarithmic integral I mean you calculate the integral of the logarithmic derivative over a closed curve what you actually get is a change in the argument and the argument principle says that this change in the argument is 2 pi times the number of zeros minus the number of poles inside okay that is why it is called the argument principle. Now the advantage of that is that we can now prove Roush's theorem so here is Roush's theorem so let f of z and so let me use l of z and b of z the analytic functions on d union dou d where you think of this kind of a diagram okay so d is simple closed contour which is piecewise and I mean dou d is a simple closed contour which is piecewise smooth and d is the interior and you take this whole region along with the boundary and it is analytic there okay there are 2 both functions are analytic there and with mod l of z strictly less than mod b of z in on the boundary okay. So you know the small and the small l and small b are supposed to you must think of small l as little and small b is big and I am just saying that the bigger one is really bigger than the little one okay in modulars on the boundary okay then what Roush's theorem says is that then the number of zeros of b of z and b of z plus l of z inside d are the same okay this is Roush's theorem so you just see what it says it says see I have a function b of z which you think of as a bigger function it is bigger than l of z which is smaller function and what do you mean by bigger in modulars it is strictly greater than the modulus of l of z on the boundary okay then by adding to bz this little function lz you are not going to change the number of zeros inside you are not going to change the number of zeros inside and so this addition of this little this lz to that bz is thought of a small analytic perturbation okay you can think of it as a small analytic perturbation and Roush's theorem is says that if you take the function analytic function bz and analytically perturb it the resulting analytic function is not going to have different number of zeros than the original analytic function okay so this number of zeros the count of the number of zeros is not going to change. So you know the proof of this is basically an application of the argument principle so let me explain let me first explain what is the idea of the proof the idea of the proof is here are two functions okay you want to show that they have the same zeroes inside your domain alright it is bounded by this simple closed curve. Now of course mind you there are no poles okay there are no poles here the functions are completely analytic and of course mind you the function b has no zero on the boundary b has no zero on the boundary because you see on the boundary mod bz is strictly greater than mod lz and mod lz is certainly greater than or equal to zero so that will tell you that mod bz strictly greater than zero on the boundary so b has no zeros on the boundary mind you b has no zeros on the boundary alright. Now I want to show that b of z and b of z plus l of z have the same number of zeros that is what I want to show how do I show it how do I use the argument principle the argument principle says is the number of zeros times two pi is the integral of the logarithmic derivative and that is also equal to the change in the argument of the function. So if I want to show that these two have the same number of zeros inside all I have to show is that the change in the argument for both is the same because it is the change in the argument that counts the number of zeros so all I have to show is that change in the argument of this and the change in the argument of that over this boundary is the same okay. So in other words I have to show that the dr of bz and the dr of change in the argument of bz plus lz they are the same as you traverse as z traverses along the boundary curve okay so that leads you to look at this function if you look at bz plus lz by bz okay you will see that of course you know this is 1 plus l of z by bz okay and so you know what this will tell you is that argument of you know the argument of quotient is the difference of the arguments so you will get argument of bz plus lz minus argument of bz is argument of 1 plus lz by bz okay this is what you will get and therefore what this will tell you is that you know so now I want you to look at this quantity here so this equation will tell you that the change in the argument of bz plus lz minus the change in the argument of bz is equal to the change in the argument of this quantity and mind you I can divide by b because b does not have any zeros okay and it does not have any zeros on the boundary so mind you all this is I am writing this down only on the boundary because I have to compute the argument change as z varies on the boundary b may have zeros inside that is not the point the point is this is being done on the boundary so I should write this on to d it does not make sense if you take z inside because z could be a 0 of b and then I cannot divide by b of z okay so this happening on the boundary where b does not vanish okay so this will tell you that dr the change in the argument of bz plus lz minus the change in the argument of bz is equal to change in the argument of 1 plus lz by bz this is what it says but the fact is that this is 0 so the reason is you see what you should understand is mod lz is strictly less than mod bz on the boundary okay so mod lz by bz is strictly less than 1 okay and therefore if you look at this quantity this quantity will lie only on the right half plane 1 plus 1 plus lz by bz lies in the right half plane right I think that is I mean so you see this is a complex number lz by bz is a complex number that lies in the unit disc alright and to that you are adding 1 that means you are translating it to the right by 1 unit okay and therefore it has to go to the right of the y axis the imaginary axis right so this lies in the right half plane so you know basically so you know so this means see for now the point is that if you look at the argument of this you see there you see if you have you know if you have a variable point let me call it as omega which is which is which is say moving okay. Suppose it moves from omega 0 to let us say omega 1 okay then you see the if omega 1 and omega 0 are the same and you are looking at the right half plane okay if it is on the same half plane then the change in the argument will be 0 okay so essentially what it means is see if you go from here to here the change in argument will be literally you know this angle minus this angle it will be this total angle alright but if you no matter how haphazardly you go okay but if you come back to the same point okay then your change in argument will be 0 alright. So what this tells you is that this is 0 the change in the argument is 0 and that will tell you that the change in the argument of B plus L is the same as the change in the argument of B as you go along the boundary but then the argument principle will therefore tell you that the number of zeros of B plus L is the same as the number of zeros of B inside the region bounded by the boundary that is the argument principle because argument principle actually tells you that the number of zeros is controlled by the change in the argument. So to show the two functions have the same number of zeros all you have to show is the change in the arguments are the same for both functions okay so this implies that by argument principle see the whole point is you know why I am saying one half plane it is because you know it should not happen that the curve should not go around and come back if it goes around and come back it will pick up a change in argument for example you know if it went around across the origin and came back then it picks up see the moment it because the argument is measured with respect to the origin right by joining the point to the origin. So if you go around the origin certainly you are going to the argument is going to change by some quantity but if you are on the same side of the origin on the same side of a plane of a half plane passing through the origin the argument is going to be independent is only going to depend on the initial point and the final point no matter how you move so long as you are in the same half plane okay so that is the whole point that is why I need that it lies in the right half plane okay I mean I know for sure that it is not going to wander somewhere here and then you know come out come back all the way to this because if you did that I will pick up a 2 pi and if it does that twice I will pick up 4 pi and if it does that in a different direction I will get minus 2 pi minus 4 pi and things like that such things do not happen because it never wanders outside this right half plane and that is the that is this condition that is because of that condition okay. So the argument principle BZ and BZ plus LZ have the same number of zeros in D okay so this is an application of argument principle that is that is Roushi's theorem so there is a you know there is a another avatar of the Roushi's theorem which you can maybe try to prove as an exercise so let me rub this part of the board but then I mean this what you must understand is that this theorem is pretty powerful the proof is the proof is pretty easy okay because you have used basically you have used Roushi's theorem and you have used Cauchy's theorem and so you have used literally all the all the all the results that you have done in a first course in complex analysis alright so it is a very powerful theorem and to illustrate how powerful it is it is very easy to deduce fundamental theorem of algebra from this okay so but but but before that let me give you a stronger version of Roushi's theorem let F of Z and G of Z be analytic on D union to D and let mod of F of Z plus G Z be strictly less than mod F of Z plus mod G Z on the boundary then F and G have the same number of zeros okay this is another version of course you know mod of F plus G always be less than or equal to mod F plus mod G but on the boundary it is strictly less than that is the condition that tell you that will tell you that F and G will have the same number of zeros inside so this this other version of Roushi's theorem is I have written it is a stronger version but probably it is actually an equivalent version and you can easily deduce it is an exercise to prove that again essentially using argument principle and this can be deduce from that that is you can as an exercise also deduce this from that and I think you can also do it the other way because essentially it is going to depend only on argument principle okay so what this tells you is that if you want to say that the number of zeros of two analytic functions is same all you need to check is that the triangle inequality becomes a strict inequality on the on the boundary that is what it says okay so yeah so you can try that then of course let me give an example as to how powerful Roushi's theorem is so you can get fundamental theorem of algebra and the fundamental theorem of algebra is the theorem that you take a polynomial in one complex variable with complex coefficients then all the zeros of the polynomial you can find all the zeros and they are going to be complex numbers okay that is that is precisely the fundamental theorem of algebra so I mean the reason why is called the fundamental theorem of algebra is because see in algebra what you do is that you try to extend number systems because you are trying to solve equations so you know for people who have done courses in algebra you know that you know you can you start with the natural numbers and then you extend them to integers okay and then you extend them to rational numbers and then to real numbers and then to complex numbers and the point is that every time you extend it because you are not able to solve equations okay so yeah from natural numbers the counting numbers 1, 2, 3, 4, 2 the integers you extend because for example if you take the equation x plus 1 equal to 1 the solution is x equal to 0 it is not here so you have to include 0 and if you take an equation like x plus 2 equal to 1 the solution is minus 1 so you need to have negatives that is how you come to the integers and then here again you do not get a solution for an equation such as 2x equal to 3 the solution is x is 3 by 2 so you go to rational numbers you invert integers, non-zero integers and you get rational numbers and then these are all fields okay and then somehow you get the reason for moving from rational numbers to real numbers is actually to fill all the gaps which are the irrational numbers so it is more topological it is real numbers is a kind of topological completion of the rational numbers and then which for example in a first course in analysis real analysis you would have seen is constructed by the method of Dedekind cuts where you define real numbers to be equivalence classes of Cauchy sequences of rational numbers and equivalence being 2 Cauchy sequences of rational numbers are considered equivalent if you put them together as a even and odd subsequence of a bigger sequence and that sequence continues to be Cauchy okay and then the point is that going from Q to R does not help because an equation such as x squared plus 1 equal to 0 which is which will be the roots of minus 1 cannot be solved here so you have to go to complex numbers adding I to the real number system gives you the complex number system okay and then the question is now if you have an equation if you have polynomial equation over complex numbers the question is are there going to be polynomial equations for which you do not have solutions I mean the question is do I have to further extend it to something bigger and the fundamental algebra says you do not have to do it what it says is that you know now it says that the complex numbers are algebraically closed which is the fact that you take any polynomial in one variable in one complex variable with complex coefficients then all the zeros are complex numbers okay so you do not have to go you do not have to extend the number system further so that is the fundamental theorem of algebra. So every polynomial P of z is equal to you know a not plus a 1 z plus etc a n z power n a n not equal to 0 ai complex numbers has exactly n zeros in C counted with multiplicities so this is the fundamental theorem of algebra and the way one proves it is you know it is just using Rochier's theorem I will tell you first in words very it can be very elegantly expressed in words so what you can do is you can get rid of this you know or do not get rid of it I mean you look at this polynomial you know as you make mod z bigger and bigger then the modulus of this polynomial will be will depend on the leading term okay so in other words if you make mod z bigger and bigger that means if you make mod z greater than say a large positive real number r okay that means you are looking at the exterior of a circle centered at the origin radius capital R for R large you are looking at the exterior and this is called a neighbourhood of infinity if you want okay if you think of the complex numbers are sitting inside the Riemann sphere with the point at infinity on the north pole okay this exterior of a circle is a neighbourhood of infinity okay and what happens is that for when you go in for mod z greater than capital R are sufficiently large then you see except for the leading term all the other terms they become very small okay in modulus the leading term will dominate all the other terms in modulus okay so you know so what it tells you you look at only this function you take a such a large R okay and look at only this function okay and think of the rest of the terms as a perturbation it is a perturbation because the modulus of the rest of the terms is going to be very small when compared to the modulus of this because this is a leading term and you have taken mod z equal to R for R very large. So what is Roush's theorem going to tell you it is going to tell you that the number of zeros inside mod z equal to R okay that is in the disc mod z less than R is going to be for this whole function is going to be the same as the number of zeros of this big function which is a n z power n but a n z power n has n zeros a n z power n has automatically n zeros at the origin it is z equal to zero to zero of order n. So Roush's theorem will immediately tell you that this polynomial will have n zeros and they all can be found in inside a disc of sufficiently large radius. So it is a beautiful I mean you get fundamental theorem of algebra just like that why by thinking of the leading term as you know the big function and rest of the terms as a little function and choosing a disc centered at the origin with very large radius okay how large as large so that the modulus of the leading term dominates the modulus of the other terms some of the other terms okay. So this tells you how powerful Roush's theorem is okay. So probably I will stop here and we will continue in the next lecture.