our lecture notes are very thin on complex numbers, and a lot of us have been asked a question that we havent been taught in a workshop. So does anyone have an idea to:
Recall that a primitive nth root of unity for any integer n>/1 is a complex number z simultaneously satisfying the conditions:
z^n=1 and z^m does not equal 1 for all m<n
Let hence k,n are in Z whith n>/1.
Prove that z=cos((2*k*pi)/n)+i*sin((2*k*pi)/n) is a primitive nth root of unity if and only if k is relatively prime to n. That is the only common divisors of k and n in Z are +1 and -1
Thanks for any help, i really dont understand.
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- Thread Starter
- 18-01-2010 17:50
Offline14ReputationRep:Wiki Support Team
- Wiki Support Team
- 18-01-2010 18:10
An nth root of unity (remember "unity" just means "1" ) is a number z such that z^n = 1. A primitive nth root of unity is a number such that z^n = 1, but z, z^2, ..., z^(n-1) aren't equal to 1. So, for example, i is a 4th root of unity (because i^4 = 1) and an 8th root of unity (because i^8 = 1), but it's only a primitive 4th root of unity.
So, using de Moivre's theorem and noting that 1 = cos(2pi) = cos(4pi) = ..., you should be able to prove that, if k is relatively prime to n, we never have z^i = 1 unless i is a multiple of n (and so the smallest is i = n, exactly what is needed to be a primitive nth root of 1). Conversely, if z^n = 1, you should be able to write it as cos(2k.pi/n) + i.sin(2k.pi/n) for some k; suppose k shares a non-trivial factor with n, and show that z^i = 1 for some i smaller than 1, contradicting the claim that z is a primitive root of 1.