How to factor complex polynomials

Free ebook http://bookboon.com/en/introduction-t... Every polynomial p(z) of degree n has at least one root in the set of complex numbers. That is, there is at least one number α such that p( α)=0. That is what we mean when we say that α is a root of a polynomial, p( α)=0. Ok, now, if the coefficients are all real so that a, n’s, etc are all real, then, the roots of complex numbers appear in conjugate pairs. And we have seen examples of this in previous videos, where I solved certain polynomial type equations with real coefficients. Now there is a factor theorem as well. If α is a root of the polynomial p, then, z is a factor of p(z). So, this is like an extension of the factor theorem that you see at school. Ok. 00:23 - 02:29. Every polynomial of degree n can be factored into n linear parts. That is, I can write my polynomial as the product of z – each of the roots. Now the a sub n out here is the coefficient of the leading term if you like. Okay, alright, so there are 4 important points. Let us talk about the following example. Here, we have got a polynomial, p(z) defined to be z^6+64. Okay, so the polynomials there; it is a polynomial of degree 6, and the coefficients are all real. So, we know that the roots of this polynomial will appear in conjugate pairs. and lastly, the second thing were asked to do there is to factorise p into linear parts. Ok, so we want to find the roots, essentially, and write them like this. Ok, so, how do we do that. Well, let us solve this equation for z. Find the zeroes of the roots of the polynomial. Ok, once you know that, the factor theorem says we can write p(z) like this. Ok, so we have that other questions like the first part. So let us just refresh our memory. For this type of problem, we will convert it to polar exponential form. 02:32-04:34. Ok, so, we have the following: z^6 = - 64. Let me just adjust the focus on this. Ok, so, we write the right hand side in polar exponential form. Ok, so, if I plot -64 in the complex plane, its going to lie over here somewhere. It is just a real number. Ok, now, so I go around half a revolution and I go out say 64 units. Now, if I want to write that in terms of the polar exponential form, then I can do the following. You have a length 64 units from that point to the origin and you have gone around half a revolution. Now, the first part can be written as 2^6 and we have the 6 there and that is going to simplify. Ok, alright, so, now we have got this, what we can do is take the power of 6 to the other side and try to simplify. So if I take power of 6 of both sides, that will disappear, that will disappear, and I divide this exponent by 6, ok, and this is , again, k = 0, +-1,+-2 etc. Ok, so here then, is like a sequence of solutions to our problem. Ok, some of these values of k are going to give identical solutions. Alright, so what we are going to do, we are going to start with the simplest value of k and just write out 6 solutions to our problem. Ok, so, let us start with k = 0 in here. So, by the subscript, I am taking the value k = 0, so that would disappear. I would get 2e^π*i/6. So that is one solution. Ok, let us take another simple value of k, k = 1. Ok, that will give me 2e^π… I will get 3 in there and then ’i’ over 6, that would reduce down to the following. Alright, let us keep on going. let us put in say, sorry k = +2. So I will get, again, something like this, and I will get 5 in there. Ok, now, let us choose another simple value of k, say k = -1, so, I am going to get -1 in here. Ok, let us try k = -2. Ok, so up here, I am going to get -3 in there. So that would simplify down to the following and lastly, k = -3. Ok. So now I have 6 solutions. I can stop. If I kept going with other values of k, I will keep getting back to one of these solutions. 06:30-09:31. Ok, so here are the 6 solutions to my problem. Ok, so, um, notice that they do appear in conjugate pairs just like we have hoped them to appear from this part here and I know from the factor theorem, I can write this, the p(z) as z – that times z – that times z- than etc. Ok, I can simplify a little but more in here. e^(π*i/2), we go around a ¼ of a revolution and out 2 units. that is lying on the imaginary axis. So that is just 2i and same here, you go the other way, a ¼ of a revolution, and that would be -2i. So now, I can write the following… it is just the following here… z – this, z – that, times a z – 2i times z + 2i times z – that, times z – that. So it is a bit of a mess but we got there. Ok, now if you were to plot these in a complex plane, they would lie on a circle centred at the origin with radius 1(2?). So let me just give you a basic freehand drawing. Right, so you have got 1 there, 1 there, 1 there 1 there and 1 there. Ok, so those points are your roots. So, very rough sketch there. Ok, so that is it, just applying the factor theorem, finding the roots etc etc. 09:31-12:53.