 Okay so let us continue with your discussion so you see what we have been doing is looking at zeros of analytic functions okay. So what we saw in the last two lectures was basically the so called Roush's theorem okay and that actually in principle tells you that if you you know perturbed an analytic function by here in a small way then the number of zeros that it enclose that is enclosed in a region is not going to change okay. Now of course this came out of the basically out of the argument principle and the argument principle in turn came out of the Roush's theorem okay. So now we are going to again continue with the study of zeros of analytic functions and of course I should remind you that the argument principle actually tells you gives you a method of counting the number of zeros and poles with multiplicity inside a closed contour in the region that is enclosed by a closed contour. Now what I am going to do this in today's lecture is try to actually look at the zero of a function an analytic function that is obtained as a seek as a limit of a sequence of analytic functions okay and what I am going to say is that essentially the zero of the limiting analytic function is gotten by you know it is gotten by zeros of the analytic functions that converge and taking limits of such zeros okay. So you have sequence of functions that converge to an analytic function if the analytic function has a zero then that zero can be gotten as a limit of zeros of the functions that originally we started with which converge to the given analytic function. So this and essentially this is what is called as Hurwitz's theorem okay but like so let me start of the discussion like this so you see suppose you have so you have the following situation so D is a domain okay which means it is an open connected site okay and of course a domain in C it is a subset of the complex numbers and you have suppose you have a sequence fk fk of functions analytic on analytic for every k greater than or equal to 1. So here is a sequence of analytic functions defined on the domain D and you all know from a first course in analysis what I mean when I say that fk converges to a function f point device. So f is limit k tends to infinity of fk exists okay so this sequence of functions converges to a function this is actually point wise convergence what does that mean it means you take any point z in D small z in D take the value fk of z for various k you will get a sequence of complex numbers and you take the limit of this sequence the limit of the sequence exists and whatever the limit is that is what you are calling as f of z and you do this for every z in D okay that means fk of z converges to f of z for each point of z in D this is point wise convergence and of course you know that point wise convergence by itself is not good enough because you know to begin with even if the fk's are continuous okay forget analytic suppose the fk's are just continuous and f is the limit then f need not be continuous. So what really helps is the notion of uniform convergence if you remember so you know if fk converges to f uniformly then and if fk each fk is continuous then f is continuous okay. So of course so you know if fk each fk is continuous and if I want f to be continuous I need uniform convergence but usually what happens is uniform convergence will not happen on a whole domain okay usually uniform convergence happens only on compact subsets of the domain in general this is the best condition to assume and this is what will happen. So let me write that down and what I am trying to say is that if I assume that fk converges to f uniformly on the whole domain then of course I will get that not only is f continuous if fk is continuous in fact f will become analytic if fk each fk is analytic okay but the fact is you cannot in general expect uniform convergence on a whole open set usually uniform convergence is to be expected on bounded closed and bounded subsets which are otherwise called compact sets because you know in the Euclidean space any subset is closed and bounded if and only if it is compact from basic topology okay. So you can replace the condition that fk converges to f uniformly on the whole domain by a slightly weaker condition which is fk converges to f uniformly on compact subsets of the domain okay and this technical condition is referred to in some of the literature as normal convergence okay. So if fk converges normally to f okay then the fact is that since each fk is analytic f becomes analytic okay. So it gives you the nice situation that a sequence of analytic functions does converge to an analytic function okay so that is the first piece of that is the first piece of information that we need. So let me write this down if fk converges to f uniformly on D then f is analytic on D okay this is a and of course you know what do I say what do I mean when I say fk converges to f uniformly so you know there is a difference between point-wise convergence and uniform convergence so let me remind you what does point-wise convergence means mean it means that if you take a point small z in capital D and you take the you evaluate the sequence of functions at that point you will get a sequence of complex numbers you will get the sequence fk of z and that fk of z will converge to f of z okay and fk of z converges to f of z means given an epsilon you can find index n such that fk of z-f of z can be made in modulus lesser than epsilon whenever k is greater than that index but if but all this is for a fixed point z in D okay but if you change the point the index that you will need if you change the point z then the index that you will need to make the distance between fk of z and f of z lesser than epsilon will also change in general that index will depend on z also but if it does not depend on z that is you are able to get an index such that fk of z the distance of fk of z can be made lesser than epsilon from f of z for all z irrespective of what z it is you choose the domain that is when you say that the convergence is uniform okay so the uniformness is and is that there is no dependence on which point of D you choose okay so uniform convergence is of course stronger than point of convergence and what I am making the statement that fk converges to f uniformly that is what when I make the statement that is what it means okay. So you must have come across this in first course in real analysis or complex analysis but anyway let me remind you now but of course is a very strong condition to expect uniform convergence on a whole domain is in general too much what you normally will get is you will get uniform convergence on closed and bounded or compact subsets of domain and that condition which is certainly weaker than this is called normal convergence and the fact is even if you weaken this condition to normal convergence still the limit function continues to be analytic okay. So let me write that down we say fk converges to f normally in D if the convergence is uniform on any compact which is equal to closed and bounded subset of D okay so this is called normal convergence. So normal convergence is actually uniform convergence on compact subsets okay and of course if you have uniform convergence on the whole domain then of course you have normal convergence because uniform convergence holds on the whole domain then it also holds on any subset of the domain okay and therefore this is a stronger condition than this this is a weaker condition but usually this is what will happen in practice and but the fact is even if you weaken this condition you still get that the limit function is analytic. So let me write that in this in such a situation f is again analytic on D okay so having looked at this of course what is our aim our aim is actually to show that in this situation that is if you have sequence of analytic functions converging to f normally in a domain then you give me a 0 of the limit function okay give me a 0 of the limit function then that 0 is gotten by 0s of the functions in the sequence by convergence okay that is a 0 of f is an accumulation point of 0s of fk that is essentially what Hurwitz theorem is okay. So but before that let me take a small diversion to explain why is it that if fk converges to f uniformly or even normally why is it that the limit function is again analytic okay. So I mean I am doing this purposely many of you must have come across this in a first course in complex analysis but I am just doing this so that you recall some basic things and it is to help you refresh your memory okay so you see you know so how does one prove this that is exactly what I want to say suppose fk converges to f normally in D suppose is the case okay. Now how do I show that and given of course that each fk is analytic how do I show f is analytic okay so just think about it you know if you want to show a function is analytic there are so many ways of showing one is of course you show that it is differentiable at every point it is differentiable one at every point which means you have to calculate the derivative at every point and show that the derivative exists okay. The other way is of course to write out the Cauchy Riemann equations and check that they are valid and also to check the first partial derivatives which satisfy the Cauchy Riemann equations are continuous okay but of course there is a deeper theorem which says that you do not have to check the continuity of the first partial derivatives it is enough to check that the Cauchy Riemann equations hold and of course it is very important that you assume f is you will have to use the fact that f is continuous okay. So first of all because you have normal converges it means that if I take a point and if I take a close disc surrounding that point a close disc surrounding that point in the in the domain D then of course it is a compact subset of D because it is closed and bounded and fk will converge to f there uniformly and since each fk is continuous analytic of course means it is continuous. So the limit function f will also become continuous okay so the continuity of the limit function will come automatically just because of uniform convergence okay and now I could choose such a disc at every point of the domain and therefore I get continuity at all points. So what I wanted to first understand is the moment I make such an assumption to begin with f is automatically continuous on the whole domain okay then of course our question is how do you show that f is analytic. So the of course as I told you one is to show it is differentiable at each point the other one is to show that the first power I mean the f satisfies the Cauchy Riemann equations and the third way is of course to show that f is locally represented by a convergent power series okay you see and but these things are not so easy to do in general they are not so easy to do what really you can use in principle to show that a function is analytic is to check the conditions of Morera's theorem which is a you know kind of converse to Cauchy's theorem. So you see if you remember so let me say the following thing how do we show so let me write this how do we show f is analytic so this is our question okay so I will draw a line here. So if you recall first is Cauchy's theorem okay suppose f so let me use g so let me use d I do not use the same d but anyway let it be let it be so let g from d to c be analytic okay then integral of g of z dz over gamma a simple closed curve is 0 for every simple closed for every simple closed contour gamma whose interior belongs to d okay. So of course whose interior and of course I should say gamma itself is a sub of d the contour itself should be in the domain and the interior of the contour should also be in the domain so that means there is a standard orientation on the contour the orientation is such that the interior of the contour lies to your left as you traverse the contour for example if you traverse if the contour is a circle if you traverse it in the anticlockwise sense then the interior of the circle will lie towards your left if you walk on the circle that is called orientation. And whenever you do a path integral you have to orient the path the path has to be oriented direction has to be given and if it is a closed if it is integral over a closed path or a loop then the orientation is always given is said to be positive if the region inside the contour is it lies to the left as you walk along the contour in the direction prescribed okay. So and of course when I say simple closed contour by contour I mean a curve which is piecewise smooth so it is a continuous image of an interval a closed interval with starting point equal to the ending point and the interval can be divided into close sub intervals such that the parameterizations are given by smooth functions they are given by functions which are not only differentiable but the derivative is continuous okay so contour is a piecewise smooth curve okay. So this is Cauchy's theorem right so what it essentially tells you is that you integrate an analytic function over a closed curve like a loop you are going to get 0 okay. Now what is Morera's theorem it is like a well partial converse to Cauchy's theorem so here is Morera's theorem if and what does it say? It says if g from D to C is continuous and if integral over gamma g of z dz is 0 for every gamma as above okay so I should say something here for every not gamma as above slightly weaker for every gamma in D for every closed curve simple closed contour gamma in D, gamma in D then g is analytic this is Morera's theorem. Morera's theorem is like a converse to Cauchy's theorem but it is a partial converse because you see in Morera's theorem you have to assume already that the function g is continuous okay and then so that is a you need continuity of g that is one extra thing that you need but what you leave out is you are just saying that over any loop sitting inside D the integral of the function is 0 which is the condition of Cauchy's theorem but it is weaker because here the loop had to be such that the interior of the loop had also to be in D whereas here I am not putting the condition there I am just saying that the loop the interior of the loop need not be in D it is not required okay that means this will work for a region with a hole this will work for a region with a hole and to give you an idea why this is true you see you see the so you see suppose you have idea of proof of Morera's theorem see suppose see suppose I have a suppose my domain D is like this it might have some holes suppose this is my domain okay the shaded region it might have holes okay so suppose but the domain is of course an open connected set okay if it has no holes then it is called simply connected and that is the condition that any closed loop can be continuously shrunk to a point without going out of the domain for example if there is a hole then a loop surrounding that point cannot be shrunk continuously to a point without going outside the domain. So this is not simply connected this I mean domains like this are called multiply connected they are not simply connected and see what look at the condition see the condition is see suppose I fix suppose I fix how do I use that condition that the integral over any closed loop is 0 see you take a point you take a point Z0 you fix a point Z0 okay and take any other point Z okay what you do is you just join Z0 to Z by a path gamma okay and then what you do is you look at the you look at the function H of Z given by integral over the gamma of f of Z dz look at this function okay now first what I want you to understand is that f is continuous f is continuous therefore f is continuous on the whole domain okay therefore f is continuous also on the arc and you need continuity on the arc to be able to compute the arc integral or the path integral because the continuity is basically done using the notion of Riemann integral you just form Riemann sums over the arc you parameterize the arc okay which means that you think of this as an image of an interval on the real line and then you are actually integrating over that integral the composition of f with gamma where gamma is a parameterization of the arc okay so the point is that and of course that will involve that will need the fact that the that the parameterization of this arc is piecewise smooth okay so what I want to tell you is that this is well defined this is well defined because f is continuous okay so you know so if you want I can in fact write it as t equal to a to t equal to b f of gamma of t into gamma dash of t dt where gamma is a map from the closed interval ab into d such that gamma traverses capital gamma from z0 to z and gamma and gamma dash t is piecewise continuous on ab so this is how you write the so let me see whether I have written it correctly I think probably I should not put a gamma yeah z is so yeah so you formally write it as z on the on the on the curve on the on this path capital gamma you write a point omega as omega is equal to gamma of t then d omega will become gamma dash of t dt and that is how you get this gamma dash of t dt okay so I think there is a little bit of probably I should I should avoid z here because the same z also is used here so let me use omega let me use omega because that is better because you know what will happen is omega if you take a point omega on the on the arc then omega is just gamma of t where t varies small t varies from a to b and then you will see that d omega will be gamma dash of t dt and that is how I replace formally this d omega by gamma dash of t dt we do this formally but then you can actually make it rigorous by the definition of the Riemann integral okay this is actually limit of Riemann swaps okay but mind you what you have used in the process is mind you to do define a Riemann integral now you see this is an integral on the real line on the on this close interval on the real line and you know to integrate a function you know it should be at least piecewise continuous f is already gamma is continuous and f is continuous so the composition this is just the composition f circle gamma so that is continuous so it is a continuous function of t and gamma dash of t is piecewise continuous because that is what I told you what a contour means the contour of path is always assumed to be piecewise continuous that means the parameterization small gamma of the contour capital gamma must have a derivative which is piecewise continuous so you see unless I do not unless I have unless I have these continuity of these two things I cannot define this integral okay that is where the technicality comes in okay you have to notice that okay so the point is in any case I can define this but the more important thing is I defined it based on gamma and I am but I am writing it only as h of z where z is the end point of gamma so the question is what is the dependence on gamma am I being careless about the dependence on gamma but the answer is that is exactly where I am using this condition that the integral of okay I think I have messed up something so it should be g here sorry it should have been g because it is this g I am trying to show is analytic okay so I am thinking of this g okay which is of course somehow I change notation from f to g okay so it is g that I am worried about so my situation is like this I have domain d I have the function g defined on d with values in complex numbers and it satisfies the conclusion of Cauchy's theorem that integral over every loop is same okay and the extra condition is g is already assumed to be continuous so I look at this integral okay and the fact is that this does not depend on gamma because you know if I took some other path gamma prime okay it will also this will also be equal to integral over gamma prime of g of g omega d omega this will also be true that is because integral over gamma followed by the inverse of the path gamma prime g is minus gamma prime will be 0 that is here is where I am using the condition where the integral over a closed path is 0 so this therefore hz is well defined it really does not depend on what path I am choosing and it also does not worry if there are holes in the region mind you okay now you see now it is a matter of now it is very easy it is something like the fundamental theorem of integral calculus see what you are actually saying is you are saying that h is the integral of g so you can immediately say that the derivative of h must be g okay so that is a fundamental theorem of calculus kind of statement and what this will tell you is that you will get from this that g dash of sorry h dash of z is actually g of z you will get this this is just a version of the fundamental theorem for integral calculus so all you are saying is in other words what you are saying is that you know if g is continuous and g satisfies this condition that integral over every loop is 0 then g has an g has an anti-derivative okay in other words there is a function so I should say then g is the derivative of function okay g is the derivative of a function but you see but what does this tell you see it tells you h is differentiable it tells you h is differentiable on d that means h is analytic on d okay but if but you know one thing if a function is analytic then all derivatives exist and the derivatives are also analytic therefore since so this equation in one go tells you that not only is h analytic because it is differentiable once everywhere but because it is analytic its derivative is also analytic and that is how g is analytic and that is how you prove Morera's theorem okay so this implies h is analytic and that implies h dash is analytic h dash is g is analytic okay so that is the that is a sketch of the proof that is how Morera's theorem is proof okay. Now it is this Morera condition that is very useful to check a function is analytic for example in this situation so you know so let us go back to that situation suppose fk converges to f normally in d suppose it is true so you know so you take a so here is your d maybe it may have some holes one does not bother and I take a point z0 in d so here is I take a point z0 in d what I have to show is that I have to show I want to show that if each fk is analytic I want to show that f is analytic that is what I want if each fk is analytic then so is f this is what I want to show how very simple to show a function is analytic on a domain show it is analytic at every point okay so what I will and mind is showing analytic at every point means it is analytic saying that it is analytic at every point is same as saying analytic in a neighbourhood of every point okay so I have to concentrate I should take an arbitrary point and I am concentrated on a neighbourhood of that arbitrary point which I will take to be a disc so you know what I will do I will choose a small disc I will choose a small disc here which is given by let me call this as d sub z0 and what is this choose d sub z0 is open disc centered at z0 small enough radius set of all z in d set of all z such that mod z-z0 is let us say lesser than some rho subset of d choose for z0 in d such a small disc you can because you know d is basically a domain therefore d is an open set therefore given any point in d there is a small disc surrounding that point which is in the which is in your domain so choose such a disc but what I will do is I will also assume that the boundary if you want I cannot the way I have drawn it even the boundary of the domain is inside d but probably I do not need it okay now you see now what you do is you look at you try to check f is analytic inside this d z0 okay. If I want to check is check that f is analytic inside this d z0 I will have to show that the mind f is already continuous why is it continuous because fk converges to f normally means fk converges to f in every compact subset so it will converge to f in the closure of this disc which will which is this disc along with the boundary circle okay so if you want let me let me also so let me write that assume assume that assume that d z0 closure is contained in d which means it is just saying that mod the set of all points such that mod z-z0 equal to rho is also in d okay that is the boundary of the disc that boundary circle is also in d okay then you see then f then d z0 bar is compact and so fk converges to f uniformly on on on d z there is uniform convergence because that is what normal convergence means it means uniform convergence on compact subsets okay and by let me remind you a compact set is something that is closed and bounded the when you take a closure of this disc it is of course closed and bounded okay so it is compact now so so what happens is since each fk is continuous so is f because uniform limit of continuous function is continuous okay there is no problem about that and so I have a continuous so if I restrict my attention to this disc I have a function f it is continuous I want to show it is analytic I can check the condition of Morera's theorem all I have to check is that give me a loop inside that disc if I if I show that for any closed loop inside the disc a simple closed contour inside the disc the integral of f vanishes then Morera's theorem will tell me that f is analytic inside the disc okay and in this way I can cover the whole region by small discs and cover all the points that will tell me f is analytic everywhere okay on d that is what I want so so you know suppose gamma inside dz not is a simple closed contour suppose it is simple closed contour then integral over gamma of f of z dz or let me keep using f omega f omega d omega is by definition integral over gamma limit k tends to infinity of fk of w omega d omega okay and now I am again using another important fact you see when a sequence of functions converges uniformly to a function this uniform convergence is so powerful that not only does it make the limit continuous if the original functions are continuous it also allows you to interchange integral and limit okay so long as on the region you are integrating in this case is a loop on that the convergence is uniform okay and of course this loop is contained this simple closed contour this loop is contained inside this and there of course uniform convergence is going on so uniform convergence gives you the authority to interchange limit and integral so what will happen is this can be written as limit k tends to infinity integral over gamma fk omega d omega I can do this the reason why I can go from here to here is because of uniform convergence mind you but then each fk of omega is analytic by Cauchy's theorem this is 0 therefore this is 0 therefore by Morera's theorem f becomes analytic in dz not but then since z not is arbitrary f becomes analytic in d and that is the proof okay Morera's theorem implies f is analytic in dz not since z not belong into d was arbitrary we get f is analytic in d okay so this is something that you need to know if you take a normal limit of analytic functions on a domain then the limiting function is certainly analytic okay fine so alright so now we come back to this question which is the question that is answered by Horowitz's theorem and that is about what is going to happen if you take 0 of f okay and Horowitz's theorem basically says that is 0 of f is going to come from 0s of fk beyond a certain stage okay so let me write that down suppose z not belonging to d is a 0 of f f is the uniform f is the normal limit of these analytic functions fk okay suppose z not is a 0 of f Horowitz's theorem roughly tells us that z not is obtained as a limit of 0s of fk for k large this is what this is essentially what Horowitz's theorem says okay so 0 of the limit 0 of the limit function the limit analytic function of a normal family of functions that normally converge to a analytic function then the 0 of the limit function comes as a limit of 0s of the original functions beyond a certain stage okay now so let me so let me write down let me write down the let me write down Horowitz's theorem properly let me write down the exact statement then we will try to see how to prove it so here is Horowitz's theorem so let fk normally in d suppose each fk is analytic in d then by what I have told f is also analytic on d okay I should say on d if you want let z not in d be a 0 of f of z of order m or order m not okay then there exists row greater than 0 with mod z minus z not so I should say then there exists row greater than 0 and n greater than 0 such that each fk for k greater than or equal to n has precisely m not 0s in mod z minus z not less than row and further all these 0s converge as k tends to infinity to z not okay so you see the statement is something like this let me again explain so here is your domain d because I have been drawing a bounded domain the domain need not even be bounded okay I am just drawing a bounded domain mind your domain is just an open connected set so it could be unbounded okay but we are really not worried about boundedness because the normal convergence because normal convergence ensures that so along as you restrict your attention to compact subsets the convergence uniform that is what you always need okay so you see the point is if you give me z not which is 0 of the limit function f then I can find a row a disk of radius row centered at z not such that beyond a certain stage all the functions in the sequence they also have the same number of 0s as the order of the 0 of the limit function and all these 0s as you as you decrease row okay all these 0s so in fact I should say I should say all the 0s converge in fact I must say as row tends to 0 to z not okay so here if I take a particular row then there is a row first of all such that f sub capital N f sub capital N plus 1 and so on all the functions beyond the index N all those functions have exactly the same number of 0s as f has as the limiting function f has but of course when I say number of 0s of course a 0s should be counted with multiplicity okay and the fact is that if you make row smaller then you will get 0s which are you know closer and closer to z not so obviously z not will be an accumulation point of the set of all 0s and in other words as you make as you choose any one of these 0s for each row and make row smaller and smaller you will get a sequence of 0s of the corresponding f's and they will all go and converge to z not. So if you think of it diagrammatically it is like it looks so you know it looks something like this if you look at fk probably you will have few 0s then if you look at fk plus 1 suppose m not is 5 I have 1, 2, 3, 4, 5 then maybe you know and this is z not okay then if you take if you so maybe I should draw a bigger diagram here expand this so the situation is like this here is z not and these are the 0s of fk k sufficiently large and if you count all these 0s with multiplicity the total will add up to m not which is the multiplicity of the 0 of f at z not and the point is as you make this row smaller all these 0s will converge to z not which means you know they will just coalesce they will just blend into one another and become one point it will become a point of order m okay this is what happens this is what Hurwitz theorem says okay. So I will stop with that and in the next part we will in the next lecture I will explain how to prove this theorem okay.