To prove that element $\frac{3}{n}+i\frac{4}{5}$ has an infinite order in $\mathbb{C}$ for any $n\in\mathbb{Z}\backslash \{0\}$ 
The problem is to prove that element $z=\frac{3}{n}+i\frac{4}{5}$ has an infinite order in the group $(\mathbb{C},\, \cdot\, )$ for any non-zero integer $n$.

Let's consider the case $|n|\neq 5$. The absolute value of $z$ then is not equal $1$ and so there's no such $m\in\mathbb{N}$ that $|z|^m=1$ and hence $z$ has an infinite order.
Now let's consider the case $n = 5$. In this case $|z|=1$ and $\arg z = \arctan\frac{4}{3}$. So $z^m = 1 \iff \frac{m\cdot\arg z}{2\pi}\in\mathbb{Z}$. In other words, $z$ has a finite order iff there exists such $m,k\in\mathbb{Z}, m>0$ that $m\cdot \arctan\frac{4}{3} = 2\pi k$. Rephrasing it one more time we have $Ord(z)<\infty \iff \exists q\in\mathbb{Q}: \arctan\frac{4}{3}=q\pi$. And that's where I'm stuck.
So the question is why there's no such rational number $q$ that $\arctan(4/3)=q\cdot\pi$?
 A: An element $z$ with finite order is a zero of the polynomial $t^r-1$ for some positive integer $r$, which makes $z$ an algebraic integer. But $(1/5)(3+4i)$ is not an algebraic integer. Its minimal polynomial is $5t^2-6t+5$, which is not monic. 
A: I found more straightforward but less elegant way than the one of Gerry Myerson. Also this proof requires less background knowledge of abstract and linear algebra, hence it's easier to understand (for a student). 
So, we consider the case $n=5$, $z=\frac{3}{5}+\frac{4}{5}i$, and try to prove that $Ord(z)<\infty$. We observe that
$$
\left(\frac{3}{5}+\frac{4}{5}i\right)^2 = -\frac{7}{25}+\frac{24}{25}i
$$
Suppose that
$$
\left(\frac{3}{5}+\frac{4}{5}i\right)^n=\frac{a(n)}{5^n}+\frac{b(n)}{5^n}i
$$
then
$$
\left(\frac{3}{5}+\frac{4}{5}i\right)^{n+1}
=
\left(\frac{a(n)}{5^n}+\frac{b(n)}{5^n}i\right)\left(\frac{3}{5}+\frac{4}{5}i\right)
=
\left(\frac{3a(n)-4b(n)}{5^{n+1}}+\frac{3b(n)+4a(n)}{5^{n+1}}i\right)
$$
so
$$
a(1) = 3, \qquad a(n+1)=3a(n)-4b(n),\\
b(1) = 4, \qquad b(n+1)=3b(n)+4a(n).
$$
Here we note that $b(n)\equiv 4 (\mathrm{mod}\ 5)$ for all $n\in\mathbb{N}$, since $b(n)\equiv 4 (\mathrm{mod}\ 5)$ implies that $b(n+1)\equiv 3\cdot4+4\cdot 3 =24 \equiv 4 (\mathrm{mod}\ 5)$ and $b(1)\equiv 4 (\mathrm{mod}\ 5)$. That means that $b(n) \neq 0$ for all $n\in\mathbb{N}$ and so $\left(\frac{3}{5}+\frac{4}{5}i\right)^n \neq 1$ for all $n\in\mathbb{N}$ which was to be proved. For the case $n=-5$ argumentation is simular.
As a corollary, we have this non trivial statement about linear independency of numbers $\pi$ and $\arctan(\frac{4}{3})$ over $\mathbb{Q}$ proven.
