Show that $\frac{3}{5} + \frac{4}{5}i$ number in multiplicative complex numbers field(apart from $0)$ has infinite order Show that $\frac{3}{5} + \frac{4}{5}i$ number in multiplicative complex numbers field(apart from $0)$ has infinite order and prove that $\frac{1}{\pi}\arctan(\frac{4}{3})$ is irrational.
by contradiction $\exists n$ s.t. $(\frac{3}{5}+\frac{4}{5}i)^n=1$
$(3+4i)^n=5^n$
when $n=2$, $(3+4i)^2=3+4i\pmod5$
stuck how prove that $(3+4i)^n$=$3+4i\pmod5$ use induction? if yes how in this case?
for second part of the question again by contradiction.
$\frac{1}{\pi}\arctan(\frac{4}{3})=\frac{m}{n}$
$\phi=\arctan(\frac{4}{3})=\frac{\pi m}{n}$ how continue from here?

For the second part of the question: O.P. assumed that $\frac{1}{\pi}\arctan(\frac{4}{3})=\frac{m}{n}$ is rational. Then, $\cos(\frac{m}{n}\pi)=\frac{3}{5}$.
Let $U_{n-1}(x)$ be the Chebyshev polynomial of second kind of order $n-1$. Then $U_{n-1}(\frac{3}{5})=\frac{\sin m\pi)}{\sin(\frac{m}{n}\pi)}=0.$ We will show that this is impossible giving us a contradiction.
From https://en.wikipedia.org/wiki/Chebyshev_polynomials, I got
$$U_{n}(x)=\frac{(x+\sqrt{x^2-1})^n-(x-\sqrt{x^2-1})^n}{2\sqrt{x^2-1}}.$$
I am not sure if this is for all $x$. I hope so. Then, when we put $x=\frac{3}{5}$ we get
$$(3+4i)^{2n}=5^{2n}$$
which is not possible due to the answer of S. Eberhard.
