Date: 03/11/2003 at 06:41:57 From: Shawn Yapp Subject: Complex Numbers z^4 + z^3 + z^2 + z + 1 = 0 This is a question the class got asked to factorise. Not even the teacher could get it! Please show me how to get the solution.
Date: 03/11/2003 at 12:23:27 From: Doctor Douglas Subject: Re: Complex Numbers Hi, Shawn, Thanks for submitting your question to the Math Forum. This is a very interesting problem. Here is one way to do it, guided by the fact that the problem has something to do with complex numbers. Consider the number z0 = exp(2*pi*i/5). This is one of the five fifth-roots of unity. These five roots lie on the unit circle in the complex plane, equally spaced, and have the form z0, z0^2, z0^3, z0^4, 1 It is the similarity of these roots to the terms in the z-polynomial that gives us a clue as to how to proceed. We also know from geometry that the vector sum of the five fifth-roots is zero (this is the key realization that saves us a huge amount of algebraic work). Let's just plug the number z0 into your original z-polynomial and use the fact that the sum of the five fifth roots is zero. We see that 1 + z0 + z0^2 + z0^3 + z0^4 = 0 (vector sum is zero) is equal to zero, so that (z-z0) = (z - exp(2*pi*i/5)) is one of the factors of the original equation. We could divide this factor into the original quartic z-polynomial and work with the cubic that remains, but we can continue with the same reasoning (and therefore have to perform less algebra). We consider the next fifth-root of unity: z1 = z0^2. If we plug this into the z-polynomial, we get 1 + z1 + z1^2 + z1^3 + z1^4 = 1 + z0^2 + z0^4 + z0^6 + z0^8 = 1 + z0^2 + z0^4 + z0^1 + z0^3 because z0^5 = 1. Again, we see that this must be zero, and so we've found another factor of the original z-polynomial: (z-z1)=(z-z0^2) = (z-exp(4*pi*i/5)). Similarly, you can verify that z0^3 and z0^4 are also roots of the original polynomial, and we obtain the final factorization as z^4 + z^3 + z^2 + z + 1 = (z-z0)(z-z0^2)(z-z0^3)(z-z0^4) where z0 = exp(2*pi*i/5). You can take this number and plug it back in and multiply everything out to check that this equation is true. I hope that helps! - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/
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