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Distinct powers of z=cosa*pi+isina*pi 
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#1
Jul1703, 11:20 AM

P: 364

How many distinct powers of z=cosa[pi]+isina[pi] are there if a is rational? Irrational?
Concerning the rational part of the question, it was easy to show that if a=p/q, then there are q distinct powers of z if p is even, and 2q if p is odd. (q+n)p[pi]/q=p[pi]+np[pi]/q (p is even) (2q+n)p[pi]/q=2p[pi]+np[pi]/q (p is odd) I was wondering for this instance if there is a more direct way to get the answer (as opposed to just chosing q+n for n). For example, I at first attempted to write an equation such that the difference between two angles is a multiple of 2[pi]. n_{2}p[pi]/q  n_{1}p[pi]/q=k2[pi] k is some positive integer (p/q)[pi](n_{2}  n_{1})=k2[pi] But this method has poor results. n_{2}  n_{1}=2kq/p ...... As for the second instance, I haven't the slightest clue how to go about showing that there is an infinite number of distinct powers. My equation above certainly would have no application here because there should be no two angles that have a difference that is a multiple of 2[pi]. And as for just arbitrarily selecting values of n, I don't think that will get me very far since I will have to select an infinite number of n's to show there is no repeat. 


#2
Jul1703, 02:50 PM

P: 513

na[pi]= k*2[pi], or na=2k Now, let a = p/q, then np=2qk p and q have no common factor larger than 1. So, If p is odd, then p=k, so n=2q. If p is even, then p=2k, so n=q. As you already stated correctly. Now, what solutions does na=2k have if a is irrational? 


#3
Jul1703, 10:24 PM

P: 364

I need to know how you determined p=k .... before I can answer this: 


#4
Jul1803, 01:36 PM

P: 513

Distinct powers of z=cosa*pi+isina*pi
60 = 2*2*3*5 85 = 5*19 ... Low let a be rational, i.e. a = p/q. If p and q have any common prime factors, then we can divide them out. Let's assume we have already done this, so p and q have no common prime factors. Now, if np = 2qk, then the prime factors on both sides are the same. Now, let p be odd. This means, n is even. Let n = 2n'. Then, 2n'p = 2qk, or n'p = qk. Any prime factor of p cannot be in q, so it must be in k. And vice versa, no prime factor of q can be in p, so it must be in n'. This yields p = k and q = n', so n = 2q. Now, let p be even. Then let p = 2p', so 2np' = 2qk, or np' = qk. The same argument as above yields p' = k and n = q. 


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