Explore Irrational Exponents: {e^(2ki)|k=intiger}

[eczeno]

I am playing around with the set {e^(2ki)|k=intiger}. all of these numbers when raised to the pi power give one (at least as one possible value). In other words, they can all be thought of as values of 1^(1/pi). There are a countable infinity of them, and I believe that these numbers are dense on the complex unit circle, but I am having a hard time thinking of how, even to begin, to show that. any thoughts on that, or this set in general?

 

[tiny-tim]

welcome to pf!

hi eczeno! welcome to pf!

(have a pi: π and try using the X² icon just above the Reply box )

one way would be to assume that they aren't dense, so there's an interval which has none of them, and to prove that that's impossible

 

[eczeno]

hi tim,

thanks for the reply.

that is an idea i have considered. it is probably a good way to go, but i guess my real problem is trying to state an assumption like that in a rigorous way. the way i have presented the set above is a bit of a cheat. for all values of k where |k|>1 the argument lies outside of the interval (-pi,pi]. any attempt to restrict the domain introduces a discontinuity somewhere on the circle and i don't know how to talk about density on an interval that contains a discontinuity.

well, if nothing else, the problem has become a littler clearer to me.

 

[Landau]

Hint: $$\mathbb{Z}+2\pi \mathbb{Z}$$ is dense in $$\mathbb{R}$$.

 

[eczeno]

thanks landau. if i am understanding you correctly my conjecture is equivalent to:

2Z+2piZ is dense in the reals and so it is dense in any subset of the reals, such as (-pi,pi].

now all i have to do is prove your hint is true .

 

[Landau]

Proving my hint is the most work. Once you've done that, use the homeomorphism
$$\mathbb{R}/2\pi \mathbb{Z}\to S^1$$
and the fact that a surjective continuous map sends a dense set to a dense set.

To prove my hint, prove the generalized statement:
$$\mathbb{Z}+r\mathbb{Z}$$ is dense in $$\mathbb{R}$$,
where r is any irrational number.

 

[eczeno]

thanks again landau. this is exactly the kind of help i was looking for.