 Hi, I'm Zor. Welcome to Unizor Education. We will talk about something which is very interestingly called the fundamental theorem of algebra. It sounds really very impressive, if I can say so. Well, the name is historical, and some time ago algebra was primarily concerned with solving polynomial equations. And the higher degree of the polynomial equation, the better. How to solve the polynomial equation of third degree, fourth degree, etc. Well, there is actually some history behind it. I think it was Cardano who came up with the formula of solutions for the polynomial of third degree. I'm not sure about the fourth, but starting from fifth degree, I think it was French mathematician Galois who basically proved that it's not possible to express solutions of the polynomial equation of the fifth and higher degrees in radicals. Just as some kind of a formula, basically, which combines the coefficients of this polynomial. Anyway, the fundamental theorem of algebra is very interesting, and I think it is really quite amazing property of the polynomial equations. First of all, it considers the polynomials only in the field of complex numbers. So, we're talking about polynomial of nth degree, which means the old, let me use the argument z as traditionally used for complex variables. So, the polynomial of the nth degree of the argument z, its general form is this, where a zero should not equal to zero because we are claiming that this polynomial is of nth degree, which means the z to the power of n must have a non-zero coefficient. So, this is a general form of the polynomial of nth degree, and it is assumed that z argument is a complex number as well as coefficients from a zero to, oops, that's my mistake, this is a, n, z, n. So, all coefficients from a zero to a n are also complex numbers. So, the fundamental theorem of algebra says that this polynomial, if used in the equation, must have at least one complex solution. So, the result is some kind of a complex number z, which if substituted, it will result in zero. It's not an easy theorem, and this is one of those theorems which I'm just telling without proving that. Obviously, there is a ton of material about how to prove this particular theorem. It's just beyond the scope of this particular course. However, well, first of all, it's very easy to remember it. Like, any polynomial has at least one complex, mind you, solution. Now, why am I saying complex? Well, for obvious reason, this has no solution among real numbers. But there is a solution among complex number, x equals i, and x equals minus i. So, again, any polynomial has a root solution of this particular equation is called the root. It has at least one root. Now, there are a couple of corollaries, well, something like simple consequences from this particular theorem, which I would like actually to present to you with the proof. And here it is. So, let's assume that we have this. This is a polynomial of the nth degree. Now, I know it has at least one complex root. Now, the very simple sequence of this statement is as follows. Let's say c is a root of this particular polynomial, which means if I will substitute it here, I will have a0 c to the nth plus a1 z to the n minus 1, et cetera, plus a... Yes, I'm substituting c, which is a root a minus 1 c plus a n equals to 0. So, this is given. I know that this c exists because of the fundamental theorem of algebra. And the statement of my theorem is the following. That this particular polynomial can be represented as z minus c times some other polynomial, but the degree should be n minus 1. So, this is b0 z to the n minus 1 plus b1 z to the n minus 2 plus, et cetera, plus bn minus 2 z plus bn minus 1. So, this is a general polynomial of n minus 1st degree. The statement is that if z equals s is a root of this polynomial, then it can be represented as a product of z minus c and some other polynomial of degree n minus 1. And I'm going to prove it. Now, just to make my life easier, I might actually use a sigma symbol. Now, sigma means a summation. Well, just in case I'm sure many of you have seen it, I'll just repeat what I mean. So, this particular polynomial I can always represent as sigma of ai z to the n minus i where i from 0 to n. It means I'm summing these elements, varying the index i from 0 to n. If it's 0, it's a0 z to the n, which is this one. If i is equal to 1, it's a1 z to the n minus 1, which is this one, et cetera. But up to the very last one, when i is equal to n, I will have an and z to the power of 0, which is 1. So, I can use this notation and now I can rewrite it as this. So, this is corresponding with bi z. In this case, it's n minus 1 minus i, right? Because it's an n minus 1's degree polynomial, and i from 0 to n minus 1. So, this is basically another formulation of the same theorem, which I might actually use it just as some kind of a shorthand. So, I have to prove this, knowing that c is a root of this polynomial. Now, the fact that this is a root, I will write again in this format, c to the n minus i is equal to 0, where i is from 0 to n. So, this is given, and this is what I have to prove. Well, let's do it in a very, very simple manner. We will just multiply this by this. I'll open the parenthesis. If I will multiply z to all these numbers, I will have, now let me put it explicitly without the sigma sign. It will be b 0 z n minus 1 plus 1 from z, it will be z n plus. Then I will multiply z to the next one, so it will be d 1 z n minus 1 plus, et cetera, plus. The very last one would be when i is equal to n minus 1. So, it will be b n minus 1 z to the power n minus 1 minus n minus 1 and 1 from here to the first power. It will be just this. So, this is the result of multiplication of z times all of these numbers. Now, I have another one. I have minus c here. If I will multiply minus c by all of these guys, I will have the following. I will have minus c b 0 z to the n minus 1. Now, it's not an accident why I put this member under this member because they have exactly the same power of z. So, the first member would be b 0 times z to the n minus 1 and times minus c would be this. And I put it under the z to the n minus 1. Now, to get the power 1 of z, I have to use i equals to n minus 2. So, it will be minus. These are all minuses. Again, c b n minus 2 z and then for i equals to n minus 1, I will have minus c times b n minus 1. So, this is the result of opening the parenthesis and this is expression which I will have. And my question is, is or is not possible for this to be equal to this? Let's just think about how can we make it. Actually, it's very simple. Can I choose b 0 equals to a 0? Yes. In which case my z to the nth power would be exactly equal to my original, where is my original? My original polynomial is this. So, for z to the power of n, this polynomial has a 0 coefficient. This has b 0. So, if I will assign b 0 equal to a 0, now this polynomial and this will have exactly the same coefficient as z to the power of n. Now, how about the power z to the n minus 1? Well, I have b minus c, b 1 minus c b 0. It should be equal to a first, right? Because the a first, a 1 is the coefficient at z to the n minus 1. Can I do this? Well, actually it's very easy. b 0 I have already established. It's this, which means b 1 should be equal to a 1 plus c a 0, right? So, I can establish b 1 using this equation. Now, how about the next one? Well, I didn't put it here, but do you understand that this would be this, right? From which b 2 is equal to a 2 plus c b 1. c multiplied by b 1, which is a 1 times c plus a 0 times c square, right? If I multiply c by this, I will have a 1 times c and a 0 times c square. Now, b 3 minus c b 2 equals a 3. From which b 3 is equal to a 3 plus a 2 c plus a 3 c square a 1 c square plus a 0 c cube. So, this process actually can be repeated for all these members. Now, the very last one, I will write it down and I will just write it down by analogy with this, but obviously it can be proven by induction. It's a very simple thing. The very last one here would be of b n minus 1 minus c b n minus 2. It should be equal to a a minus 1, n minus 1. From which b n minus 1 should be equal to a n minus 1 plus a n minus 2 c plus a n minus 3 c square plus etc. plus a 0 c cube a n minus 1. Again, it can be proven by induction from all these absolutely equivalent equations. And the only thing which is left here is this. And it should be equal to the coefficient which is free of z or z to the 0 degree which is a nth, right? So, in addition to all this, I have this. Now, if I didn't have this, I have already found all my coefficients from b1 to bn minus 1. Now, obviously, I can't really just use these without checking that this particular equation is also true. Is it true or not? Well, let's just find out what it is. If this will be a true equality between these two things, then these solutions are actually everything which we need. These are exactly the coefficients of this polynomial of n minus 1 degree which is needed to be multiplied by z minus c to get the original polynomial. So, all I need right now is to check if this expression for bn minus 1 substituted to this would give me an identity. So, let me wipe out this. You don't need it anymore. It's already done its deed. Now, well, let's find out. I get this one and I should multiply it by minus c, right? Or, by the way, instead of this, I can do it this way. a n plus c b n minus 1, it should be equal to 0, right? So, that's the equivalent equality which I should check. All right. So, let's check what bn minus 1 is and we find out what it is, multiplied by c. Now, what happens? a minus a n minus 1 times c then plus a n minus 2 times c square plus, etc. Plus, the very last one would be, well, the one before last would be a1 c to the n minus 1 and the very last one would be a0 c to the n's, right? Each member is multiplied by c, so this exponent is increased by 1. This is what I have from this. Now, a n plus this one and it should be equal to 0. Is it equal to 0? Uh-huh. Absolutely. Why? Here is why. Because c is the root of the original polynomial. So, original polynomial, which is this one, has a root c which means if you substitute c instead of z, you will get 0 and this is exactly what we have to check. So, this last equation really is always satisfied if I choose b's like I did before. And that actually finishes, this is the end of the proof that any polynomial of the n's degree can be represented as a product of z minus c where c is a root and some other polynomial of degree which is one less than the original. So, bn of z, if any general polynomial of the n's degree in the field of complex numbers is always represented, can be represented as a polynomial of the smaller by one degree and z minus c where z is the root. Okay. So, this is the first corollary, the first very simple, well it's a little lengthy but it's really simple consequence of the fundamental theorem of algebra. Because the fundamental theorem says that c always exists for any polynomial including this one of course. And that's why it is possible to do it this way. Now, let's do a different theorem. It's another very simple consequence of corollary. Since I know that this is true, then exactly the same theorem can be applied to this polynomial. It also has some root according to the fundamental theorem of algebra, right? So, if I will call this root c1 which means my first root. Now, this can be represented as this polynomial of the n minus 2 degree and c2 would be a root of this polynomial, et cetera. Obviously, it's true for this as well. And finally, it would be z minus c1 times z minus c2 times et cetera times z minus. Now, for any polynomial real polynomial including the polynomial of the first degree I will have some kind of root. So, the polynomial of the first degree would also have some root. Now, what is the polynomial of the first degree? Let's go to this. Well, obviously this is something like this. This is the first degree, right? And it has a root. I mean, obviously it has a solution to the equation az plus b is equal to zero. And we can represent it as this particular solution times some kind of a constant, right? Actually, this constant is supposed to be equal to a where cm is the solution. So, there is a constant at the very end. This is the result of the very last representation of the polynomial. The first degree has z minus its root and some kind of constant, which is no longer a polynomial. So, no longer our fundamental theorem is applicable anymore. So, that's basically the final form. Any polynomial of the nth degree can be represented at. Well, incidentally, a is not just a. For obvious reason, this is a zero, the coefficient at z to the nth degree. Because you can only get z to the nth degree from the multiplication of these if everywhere you take z. z, z, z, z, and you will have one actually coefficient at z to the nth degree, right? There are only n multipliers. So, to get z to the nth degree, if you open all the parentheses, you have to multiply z by z by z n times. And the coefficient will be one, so you need a zero to convert it into a zero z to the nth degree. Basically, that's it. This is the end of this second corollary to the fundamental theorem of algebra. Any polynomial can be represented in this particular form. And by the way, incidentally, it means that it has exactly n roots. Because every time I have a root, I can always represent it in this way. So, I can always find n roots not necessarily different, because sometimes c1 and c2 or whatever, they can actually be exactly the same value by accident. But anyway, there are n of them, and it's exactly n of them. Because every time I have a new root, I'm reducing the power from n to n minus 1, from n minus 1 to n minus 2, et cetera, down to zero. So, there is no other solutions. It's always n, and you can always represent your original polynomial in the field of complex numbers again and repeat it again and again as the product of these z minus roots. And as I said before, some of them might actually be exactly the same values. Like, for instance, if you have z squared is equal to 1, well, I'm sorry, if you have z minus 1 squared equals to zero, this. You have root 1, and then you have another root 1. So, there are two roots 1, c1 is equal to 1, and c2 is equal to 1, right? So, the roots can be multiple, how should I say it? Multiple roots can have exactly the same value. Let's put it this way. It can be double roots, triple roots, et cetera. But in any case, the theorem is always formulated as a polynomial of the nth degree in the field of complex numbers has exactly n roots and can be represented in this particular form where a0 is the coefficient at z to the nth degree and c1, c2, et cetera, cn are roots. Okay. Basically, this statement, it seems to be a little bit stronger than the original fundamental theorem of algebra because fundamental theorem says that any polynomial has at least one root. This is a slightly stronger statement. It has exactly n roots. However, these two statements are basically, I mean, this is kind of consequential and the derivation is so simple that in many cases I've actually heard about the fundamental theorem of algebra stating that any polynomial has exactly n roots. Well, it doesn't really matter. It's basically, it has some maybe historical value, but no more than that. That's it for the fundamental theorem of algebra. Try to go through this material again, maybe just from the notes and try to follow the proof, whatever I have presented here. There is one in the notes. And again, don't forget to register and go through the exams. Thank you very much. That's it for today.