How many elements are in the set $ \{ \left( \frac{2+i}{2-i} \right) ^n : n \in \mathbb N\}$ 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.

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. 
