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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.

BulGali
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  • What is $a^c$? For instance, what is $i^\pi$? – José Carlos Santos Jun 28 '20 at 14:43
  • Can assuming $a = \cos \theta +i \sin \theta = e^{i \theta}$ help? – UmbQbify Jun 28 '20 at 14:48
  • @JoséCarlosSantos $a^c=e^{clog(a)}$ – BulGali Jun 28 '20 at 14:57
  • And what is $\log(a)$? – José Carlos Santos Jun 28 '20 at 14:58
  • This is equivalent to the statement in real numbers: If $x$ is irrational and $a,b\in [0,1]$ be any two real numbers, $nx-[nx]$ (where [y] is the greatest integer less than or equal to $y$) lies between $a$ and $b$ for some $n\in \mathbb{N} \cup {0}$ ...

    Can you prove this statement? The equivalence carries forward quite easily after that.

     – Fawkes4494d3 Jun 28 '20 at 14:58
  • You've written this like $a$ and $c$ are fixed. Please clarify the set which is to be dense. – halrankard Jun 28 '20 at 14:59
  • @halrankard $a$ and and $c$ are indeed fixed. – BulGali Jun 28 '20 at 15:00
  • @JoséCarlosSantos $log(a)=ln|a|+i(Arg(a)+2\pi n)=i(Arg(a)+2\pi n)$ – BulGali Jun 28 '20 at 15:02
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    @BulGali you still need to specify the set that you want to be dense in the unit circle. Maybe you mean the set ${a^{nc} \mid n\in \mathbb{N}, a,c \text{ given as above}}$? – Fawkes4494d3 Jun 28 '20 at 15:03
  • @Fawkes4494d3 $a^c$ is defined as $a^c=e^{clog(a)}$ and $log(a)= ln|a|+i(Arg(a)+2\pi n) , n\in\mathbb{N}$ is a set. – BulGali Jun 28 '20 at 15:05
  • Um.. what if $a = 1$? Also the wording is screwed up: "Let a∈C a complex number such that |a|=1 and c an irrational real number". Then the set ${a^c}$ contains one single element and is not dense in anything but itself. – fleablood Jun 28 '20 at 15:42
  • @fleablood Going by similar questions asked in the previous comments, the question looks to be about the set I put in my answer. But I agree that the original post needs these clarifications. – halrankard Jun 28 '20 at 15:45
  • "a and and c are indeed fixed" Then the statement is utterly false. There is only one point in the set ${a^c}$ and if you take any other point and let a radius be anything less than the distance between $a^c$ and the point, the neighborhood of that point will not contain any point ${a^c}$ and and so no other points can belong to a set that is dense in ${a^c}$. – fleablood Jun 28 '20 at 15:47
  • @fleablood The point seems to be to not fix a choice of log when defining $a^c$ which produces a set. I updated my answer with these remarks. (Though yes...I agree this should be made clear in the initial question.) – halrankard Jun 28 '20 at 16:55

1 Answers1

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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.

  1. Show $S_1$ is dense (details below).
  2. Given arbitrary $w$, write $S_w=w\cdot S_1$. So $S_w$ is a uniform rotation of the dense set $S_1$ and hence is dense itself.

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

If $z$ is an irrational number, then the countably infinitely many choices of $\log w$ lead to infinitely many distinct values for $w^z$.

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.

halrankard
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