How many distinct powers of z=cosa[pi]+isina[pi] are there if a is rational? Irrational?(adsbygoogle = window.adsbygoogle || []).push({});

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.

**Physics Forums - The Fusion of Science and Community**

# Distinct powers of z=cosa*pi+isina*pi

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: Distinct powers of z=cosa*pi+isina*pi

Loading...

**Physics Forums - The Fusion of Science and Community**