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?
 A: As you observed, $(3+4i)^2 = -7 + 24i \equiv 3 + 4i \pmod 5$. Therefore $$(3+4i)^n =(3+4i)^{n-2} (3+4i)^2 \equiv (3+4i)^{n-2} (3+4i) = (3+4i)^{n-1} \pmod 5$$
for all $n > 1$, so $(3+4i)^n \equiv 3+4i \pmod 5$ by induction. Hence $(3+4i)^n \neq 5^n$ for $n > 0$.
If you know some algebraic number theory, you can say the following. We can write $z = (3+4i)/5 = (2+i)/(2-i)$, and $2+i$ and $2-i$ are primes in $\mathbf{Z}[i]$, which is a UFD, so $z^n = 1$ is impossible by unique factorization.
I will leave the second part to you. Hint: write $(3+4i)/5$ in polar form.
A: 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.
