How many elements are in the set $$ \left\{ \left( \frac{2+i}{2-i} \right) ^n : n \in \mathbb N\right\}$$
My attempt:
$\left( \frac{2+i}{2-i} \right) ^n = \left( \frac{3}{5} + \frac{4}{5}i \right)^n=e^{\arctan(4/3)ni}$
So if for $n_1 < n_2$ it is true that $e^{\arctan(4/3)n_1 i}=e^{\arctan(4/3)n_2 i}$, then $z = n_2-n_1 = \frac{2\pi}{\arctan(4/3)}k, k\in \mathbb Z$
Because it is not really clear what $\arctan(4/3)$ is, I don't know whether we can find $z,k$ to satisfy the equation above.
One approach:
Let's call $\arctan\left(\frac{4}{3}\right)=\theta$. Some quick algebra gives that $\cos\theta=\frac{3}{5}$. Our approach is going to be to show that $\cos n\theta$ are distinct for $n\ge1$, and this will tell us that the real parts of all the listed numbers are distinct - and hence, that the numbers themselves are.
Using the cosine sum formula and setting $c_n=\cos n\theta$, one can derive the following recurrence:
$$c_{n+2}=\frac{6}{5}c_{n+1}-c_n$$
To get a sense for what this implies, let's use it to pump out the first few terms of this sequence:
$$1,\frac{3}{5},\frac{-7}{25},\frac{-119}{125},...$$
Quickly, we spot that the denominators are increasing powers of $5$ and the numerators are coprime to $5$, and this gives us something to work with - if we can show that $c_n=\frac{\text{something coprime to 5}}{5^n}$, then they're clearly all distinct.
Towards this end, let's call $c_n5^n=a_n$, and note that $a_0=1,a_1=3$. Rearranging our recurrence for $c_n$ tells us that
$$a_{n+2}=6a_{n+1}-25a_n$$
This gives us easily that $a_n$ are always integers (by a simple induction, as expected), but it gives us more: reducing the recurrence $\mod 5$ tells us that $a_{n+2}=a_{n+1} \mod 5$, and so that for all $n$, $a_n$ is coprime to $5$, i.e. $\cos n\theta=\frac{a_n}{5^n}$ in lowest terms, and thus that all of these terms are distinct.
We are thus done.
[Another way of looking at things: $\mathbb{Z}[i]$ is a UFD, and $2\pm i$ are primes which are not associates. This is sufficient to determine that all of the elements described will be distinct - in fact, that $\{\frac{(2+i)^m}{(2-i)^n}:m,n \in \mathbb{Z}\}$ are all distinct.]
I was recently given off line a solution to this problem which to me seems a bit down to earth but probably easier to grasp.
If we have $a+bi$ such that $a = 3 \mod 5$ and $b = 4 \mod 5 $, i.e. $a = 5n + 3, b = 5m + 4$ for $m,n \in \mathbb Z$, then $(a+bi)^2 = x + yi$ and $x = 3 \mod 5$ and $y = 4 \mod 5 $ (can be easily checked by a direct calculation).
By induction $\forall n \in \mathbb N, (a+bi)^n = x + yi$ and $x = 3 \mod 5$ and $y = 4 \mod 5 $.
Now if we have $n_1<n_2$ such that $$\left( \frac{3}{5} + \frac{4}{5}i \right)^{n_1}=\left( \frac{3}{5} + \frac{4}{5}i \right)^{n_2}$$ then there should be $n_0 \in \mathbb N$ such that $$\frac{1}{5^{n_0}}\left( 3 + 4i \right)^{n_0}=\frac{1}{5^{n_0}}\left( x + yi \right)=1$$
So $x=5^{n_0}$, i.e. $x = 0 \mod 5$, but we just showed that $x = 3 \mod 5$. So all elements in the set are different.
**text**), it won't look like SCREAMING – AlexR Apr 17 '16 at 15:13