Let $a\in\mathbb{C}$ a complex number such that $|a|=1$ and $c$ an irrational real number.
Prove: The set $a^c$ is dense in the unit circle.
The problem is taken from Notes on Complex Function Theory by Donald Sarason.
Would appreciate any help.
Let $a\in\mathbb{C}$ a complex number such that $|a|=1$ and $c$ an irrational real number.
Prove: The set $a^c$ is dense in the unit circle.
The problem is taken from Notes on Complex Function Theory by Donald Sarason.
Would appreciate any help.
Copyright © 2021 JogjaFile Inc.
Let $c$ and $a$ be fixed as in the question. For any complex number $w$ with $|w|=1$ define $$ S_w:=\{w\exp(2\pi i nc):n\in\mathbb{N}\} $$
According to your specifications if $s=Arg(a)$ then the set you are asking about is precisely $S_w$ where we choose $w=\exp(ics)$.
However we can show $S_w$ is dense for any choice of $w$ on the unit circle. There are two steps.
To do step 1, this is the usual "irrational rotation". There are many sources with this proof. Here are two from the network
Dense set in the unit circle- reference needed
Prove that the orbit of an iterated rotation of 0 (by (A)(Pi), A irrational) around a circle centered at the origin is dense in the circle.
Added later: In light of the comments below the question, I can help clarify. While $a$ and $c$ are fixed, the intention seems to be to not fix a branch of logarithm, leading to a set of values for $a^c=\exp(c\log(a))$, in which $\log(a)$ takes any value $i(Arg(a)+2\pi n)$ as $n$ ranges over natural numbers. Since $c$ is irrational we obtain a countable set of values, and this is the set of values that one wants to show is dense. In other words
Source: https://en.wikipedia.org/wiki/Exponentiation#Complex_exponents_with_complex_bases
Note that if $a=1$ then formally one can still obtain a set of values in the same way, which would be the set $S_1$ above. Though this is abusing the notation of complex exponentiation.