 So, this is the theorem which we want to prove today. So, one of the first thing that you asked last time was why does this product converge at all. And once we ensure that this converges then there is still one more thing that one needs to show which is that since f we know is of order 1. This function should also be of order 1 then their division will be an entire function of order 1 without 0's and therefore, it is equal to e to the a z plus b. So, these are two things we need to show that this product converges not only it converges it is actually an entire function of order 1. Now, to show this we will necessarily have to use some properties of these roots z i's. This will not be a convergent function for all possible z i's actually. And the property that we use of z i's is actually the fact that f the function we started with whose roots are these z i's is an function of order 1. So, this already says something about the roots. Why does it say something about the roots? See roots are very closely related or let us say number of roots are very closely related to the order of the function think of polynomials. A polynomial of degree k has k roots and that is what the number of roots is what determines the degree and degree is what determines the asymptotic growth of the function. The more the number of roots are the higher the growth is. This seems somewhat counterintuitive at first glance that you would expect that more roots are there the less will be the growth of the function. But actually eventually more roots allow a higher order of growth. And so since we know that the order of the growth of the function we can conclude something about the about the roots. Of course, the number of roots are infinite, but we can say that within a certain region how many roots are there and that count is what we will derive now. So, let us say we estimate the number of roots in some disc of radius r. And for this we prove the following lemma first that. So, let us say it has t roots z 1 to z t inside the disc. So, I am ruling out the boundaries. In fact, I will pick boundary r. So, that there are no roots that I can always do. Then this product of i going from 1 to t r divide by the absolute value of z i that is bounded by the order this the right hand side is essentially the order of the function at distance r the absolute the bound on the magnitude of the function. And so is bounded by the magnitude proof lemma is fairly straight forward. Let us define a function based on f which is the following. So, this g is simply f multiplied with a certain product. See the idea is to take away all the zeros of f inside the disc. So, what we are doing here is we are dividing f by the product of z minus z i which takes away all the zeros. But the rest of the multiplier is to ensure that the absolute value of g does not blow up or it stays bounded by absolute value of f around the boundary. And that is clear by the fact that if you look at firstly note that what should I say. Let us look at the absolute value of g z for when absolute value of z is r what is the absolute value of g z. And this is equal to this is the only trick I am going to play here. What is r square? When absolute value of z is r then z z bar is r square. So, I will replace this by z z now I take absolute value of z out in common. And now if you look at the product absolute value of z is r which cancels with the r here. And absolute value of z bar minus z i bar is same as absolute value of z minus z i because one is conjugate of the other. The absolute value does not change. So, the product is actually one. And so what we get is and now for g what we know is g has g is an analytic function on the disc. There is no poles in the disc inside the disc inside the disc it has to have finite number of zeros. Because every zero is isolated for any analytic function every zero is isolated that is something we showed long time ago. Therefore, inside a finite disc or any finite region there can be only finitely many zeros. If there are infinitely many zeros then there will be a yeah but then they will converge. See isolated means there is a delta. So, that delta disc around that point has no zeros for every zero there is a delta disc around it which has no zeros. But in this case if the zeros converge to one point then that point or around that point there are you can never find a delta with only with no zeros. If it does not matter whether that point is a zero or not just look at that point any for any delta disc around that point will actually have infinitely many zeros for any delta. Now, zero is isolated this cannot happen because if a zero is isolated then take any point on the plane there is a small enough delta you can find around which there is at most one zero or inside which there is a most one zero either that point is a zero then there is a small delta. So, that no there is no other zero if that point is not zero then there is a small delta which. So, that in that disc there is no zero not convinced let us look at this fine let us look at this let us go here. So, let us establish the following property on some domain D now let us do the lemma for any z in D do you believe this lemma f is analytic on domain D take any point in the domain if f z is zero if z is a zero of f then by definition of the fact that zeros are isolated there is a small delta. So, that delta disc around z has no zero no other zero there is only one zero on the other end if f z is not zero then it is some finite distance away from any zero of f inside the disc. So, take the now that is not precise argument. So, if f z is not zero then you take a small enough disc around f z and suppose it does contain a zero now around that point which is a zero there is another delta prime disc which is has no zero no other zero. So, let us so let us see this. So, this is the domain this is let us say this is z this is the disc around delta disc around z suppose it has a zero here this is a zero then there is a small disc around this which has no other zero and let us say if you look join this line from this point to this point then. So, I am trying to find say that argue that take the closest zero to z the question is can one precisely define the closest zero to z then it has to be zero that is simple enough. So, that is simple good. So, then there has to be you cannot then there has to be a closest zero to z and if this was a closest zero then you take a disc centered here of radius. So, if this is delta prime then you take a disc of radius whatever delta one delta prime at here then it will not delta prime whatever some smaller number than this distance. So, it will not contain a zero. So, with this lemma in place now you can argue that in a finite domain there can be only finitely because if not then there will be a infinitely if there are infinitely many zeros then if you keep dividing this domain into smaller smaller pieces every piece there will always be one piece which are infinitely many zeros and keep shrinking it will you stay infinite. And that is not that is going to violate this lemma at some point I am saying compactness I am saying that take a this is a compact set of. So, if finite a disc is a compact set so then we go back. So, the finitely many zeros so this works out fine and so that is the was point of this trick that the absolute value of g is same as absolute value of f on the edge of this disc and further g has no zeros actually that is not very important, but g is analytic on the disc. The reason g is analytic is that all the zeros of this or the poles of g or let us say possible poles of g are also zeros of f so they get cancelled out. So, g is analytic and since g is analytic we can invoke the Cauchy's integral formula to conclude that g zero the absolute value of g zero is bounded by the maximum value of f z around the circumference. In fact, this happens at maximum value of g z around the circumference and this is same as we just saw maximum value g z around circumference in absolute value same as f z around circumference. And what is max f z around the circumference because f has an order 1 so at z equals r f grows like e to the order r to the 1 plus epsilon. And what is g zero g zero if you look at the definition of g when you plug the zero in here what do you get product of i going from 1 to t r square up there r down here. And minus z i so absolute value of g zero is therefore product i going from 1 to t r by absolute value of z i now f zero is some finite number is not it is a non-zero number by assumption some finite value whatever it is does not matter. So, we just put this together with the bound we just derived on the upper bound on absolute value of g zero what we get is that that is what dilemma is. And now you can see that this expression already says that you cannot have too many zeros here because absolute value of z i is always less than r. So, this ratio is always more than 1 and you are taking product of t of them and we know an upper bound on this product. So, you cannot have too many of these in fact with this already you can derive suppose we just take try to count how many z i's are there with absolute value less than r by 2 l. So, then we get i going to 1 to l r by absolute value of z i this is surely less than equal to i going from 1 to t r by absolute value of z i and this is e to the order r 1 plus epsilon. And what is this this is greater than equal to 2 to the l. So, what we get is l is order r to the 1 plus epsilon and l is remember the number of zeros of f of absolute value at most r by 2 r was arbitrary. So, this already gives a bound on the number of roots of f up to a certain whatever that certain value you give whatever bound you give you get the number. So, let us say if we denote by n r the number that is n r is order r to the 1 plus epsilon. And this is a very interesting conclusion of the fact that f is an entire function of order 1. Now, using this, but this is still not the full story of course, this is something very useful we will make use of this later on also one of the very interesting thing we can use this for is to bound sums of roots some expression of roots. So, here is another lemma for any delta greater than epsilon if you take the sum of 1 over absolute value of z i to the 1 plus delta this sum converges proof is very straight forward. So, I split this sum by or rather group this sum by absolute value of z i's and I do group them of absolute value between 0 and 1, 1 and 2, 2 and 4, 4 and 8. So, successive powers of 2 now this sum if you look at the thing inside absolute this is in the denominator. So, I can replace this by always or upper bound this sum by replacing this by 2 to the k minus 1 or 2 to the k which one gives me an upper bound smaller value of denominator will give me an upper bound. So, we take 2 to the k minus 1 for absolute value times 1 plus delta and now this sum becomes easy. So, this is how many roots are there between 2 to the k minus 1 and 2 to the k well that is we just derive that just use forget about the lower bound how many roots are there up to 2 to the k 2 to the k to the 1 plus epsilon and of course, some constant here. So, let us just take out the constant parts of this and this of course, converges this is geometric series converging to it converges to. So, and that is a very useful property for us and this is what we are going to use to prove the convergence we just saw back to the proof of theorem. So, we consider this product I would like to prove that this converges this is an entire function which means for any z this has takes finite value and let us say for some z which is this is a capital R right fix a z whose absolute value is capital R and I want to show that this is a finite value at that. So, I split this product as in 2 parts z i less than equal to 2 R and 1 is z i greater than 2 this part this is a finite product. So, this surely will converge to some finite value right this is all very nicely behave there are no poles here. So, it will converge. So, if I show that this converges to some finite value I am done. So, let us just focus on this part absolute value of z i I order this in product of z i then only consider now why is this finite we just decided right we just argued that in a disk of radius 2 R there are only finitely many 0. So, this is finite so that is that argument still holds. So, this is bounded and I want to show that this is bounded now because absolute value of z i is more than 2 R here if you look at z over z i the absolute value of this is less than half always for all I therefore, I can write this as e to the this part as e to the log 1 minus z over z i plus z over z i all I have done is taken written this as e to the log. Now, again log has come here. So, you have to be careful but here again because of this property that z over z i only moves in a radius of half and it moves around 1. So, 1 minus z over z i it is minimum value is half maximum value is in the real line 3 by 2 and similarly, in the complex. So, it is a disk centered around 1 of radius half that is a range here. So, there log is completely analytic on that disk because it 0 is far away. So, I can take any and I can take any log of the infinitely many we just choose the principle log which is no addition to this term. So, I can write this not only that because again this is at most absolute value in half. So, I can replace this because it is analytic. So, I replace this by the power series in that small disk. What is the power series like? So, let us bring z over z i first log of 1 minus x what is the power series? There is a negative for with all the terms then j greater than equal to 1 and then you get z by z i to the power j upon j z to the j. Now, the first term of this is z over z i and this sum which cancels with this and that was the whole reason of sticking e to the z over z i. So, what is left out after the cancellation is e to the minus j greater than equal to 2 z to the j over z i to the j and this is the product. Now, let us take the absolute value of this. So, this product absolute value or less than equal to e to the real part of this real part I can always substitute by the absolute value of the upper the exponent and so I get j greater than equal to 2 z to the j divided by j absolute value of z i and absolute values. So, it is all gone minus is gone. So, I am saying that absolute value of this is e to the real of this. Now, real of this is at most real of a complex number real part of a complex number less than equal to absolute value of the complex and the absolute value of this whenever as soon as you take that the minus signs go away and then you take it inside. So, it is sum of absolute values just take less than equal to that. Now, let us take out the first term and inside this we have let us forget about this j sitting in the denominator this is only reducing the exponent. So, forget about the j. So, we just take this as. So, this again becomes a geometric series and in this geometric series z absolute value of z over absolute value of z i is at most half. So, it converges and remember that z i this always converges to at most what 2 no matter what z i am choosing here. So, I can write this as e to the 2 z square by and this of course, now I can write as and this sum we just showed the previous lemma is bounded sum over i 1 over absolute z i square. In fact, we can do slightly better here. So, let me take this out and do the slightly better analysis which we will use later on see z over z i is always less than 1 in this sum that is by choice. So, z over z i squared is less than equal to z over z i to the power 1 plus epsilon or 1 plus delta. So, this is I can write this as less than equal to e to the sum over z i greater than 2 r to z to the 1 plus delta by z i to the 1 plus delta. And what we conclude is that the absolute value of that product is bounded by e to the order absolute value of z to the 1 plus delta. So, we not only show that the product is bounded we all here we are also showing that the that product is of order 1 it is a function is an analytic function of order 1 and that completes the proof of the theorem because all we needed to show was that this product is bounded now or not does not complete the I should take that back it completes first part of the proof of the theorem that is showing that this function is an analytic function or an entire function of order 1 with precisely the same set of 0s. So, that we have concluded. So, now consider f z divided by this this function is an entire function this function has no 0s also because all the 0s cancel out of f and this. So, it is an entire function without 0s if we can get that the order of this function is 1 we have proved the theorem we know that the order of f is 1. So, if we can prove that order of 1 over this product is 1 then we are done. So, now we will show that 1 over absolute value of this is at most e to the order of f. Now, when we do this using the same thing as we just did actually just little bit of an extension of what we just did. So, 1 over the product I greater than equal to 1 of this we write it as 1 over product absolute value of z i is less than absolute value of z by 2 times 1 over this 3 product is given any z split the roots in 3 groups absolute value less than absolute value of z by 2 between z by 2 and 2 z and more than 2 z this is the only infinite product the other 2 are finite products. For this infinite product we have just shown that the product is bounded by this which is the function of order 1. So, this we are already done this is already 1. So, what is left is these 2 finite products if we consider this this is equal to 1 over product z i less than z by 2 1 minus z over z i in absolute value or let us say consider this is equal to we just consider this absolute value is equal to let us take the product. Now, this I can ignore why because the gap between z and z i is substantial it is at least absolute value of z by 2. So, I do not want to ignore this I am all I am saying is that this is less than equal to product z i absolute value of z i is bounded by of course, z by 2 absolute value of z minus z i is at least absolute value of z by 2. So, I can replace it by z by 2 and here I can I need to bring e guy e up there when I take up this is certainly true now of course, this cancels each other out. So, all that you are left with is e to the sum z i less than z by 2 z over z and how do we handle this we cannot handle the sum can we handle this sum now we need a cheap trick we cannot sum 1 over z i we do not know if that is bounded. But remember this is a finite sum not only is this finite sum it is only up to z i is less than z by 2 so this is at most. So, that means z is more than z i always in this sum. So, I do the reverse of what I did earlier and replace this by z to the 1 plus delta over z i to the 1 plus delta as always true I am increasing the exponent. So, this becomes higher and now we can bound this. So, that is it for today.