How to prove that a complex number is not a root of unity?

$(3+4i)^2 = -7 + 24i = (3+4i)+5(-2+4i)$

With this you can easily prove by induction that for every positive integer $n \ge 1$ there are integers $a_n,b_n$ such that $(3+4i)^n = (3+4i) + 5(a_n+b_n i)$.

In particular, since $4$ is not a multiple of $5$, its imaginary part cannot vanish, so $(3+4i)^n$ can't be a real number when $n \ge 1$.

Here we were lucky that $(3+4i)^n$ mod $5$ is $1$-periodic (that $3+4i = 1 \pmod {2-i}$), in general you can get a cyclic behaviour with larger periods, so it can be a bit longer to write down, but the method works just as well.

The minimal polynomial of $\frac35+\frac45i$ is quadratic. If it were a primitive $n$th root of unity, the degree should be $\phi(n)$. Hence only few $n$ are possible ($n=3$, $n=4$), but of course this is neither a third nor a fourth root of unity.

Write $e^{i\theta} = \frac{3}{5} + i\frac{4}{5}$. Then $e^{ni\theta} = 1$ only if $\cos(n\theta) = 1$, that is, if $$\cos(n\theta) = T_n(\cos\theta) = T_n\left(\frac{3}{5}\right) = 1$$ where $T_n$ is the $n$-th Chebyshev polynomial.

But the leading term of $T_n$ for $n > 0$ is $2^{n-1}$, thus the minimum denominator of any rational root of $T_n - 1$ must divide $2^{n-1}$. Since 5 does not, $T_n\left(\frac{3}{5}\right) - 1 \neq 0$ for all $n > 0$.

In $Z_5$ : $(3+4i)(3+4i) = (3+4i)$

$\forall n \ge 1$ : $(3+4i)^n \notin \Re$

$(3+4i)/5$ is not a root of unity.

If $(3+4i)^n=5^n$ then, in $\mathbb Z[i]$, we have $(2+i)^{2n}=(2+i)^n(2-i)^n$ hence $(2+i)^{n}=(2-i)^n$.

But $2+i$ (and $2-i$, but we don't need it) is a prime element in $\mathbb Z[i]$ (why?), so we must have $2+i\mid 2-i$, impossible. (If $2-i=(2+i)(m+ni)$, then $2=2m-n$ and $-1=m+2n$, therefore $m=\frac35\notin\mathbb Z$.)

We use the formula for $\tan n t$ in terms of $\tan t$ (when $n\in\mathbb N$).

(1) For odd prime $p$, let $ q=(p-1)/2.$ Let $x=\tan \pi/p.$ Then $0=\tan \pi=\tan p(\pi/p) = A(x)/B(x)$ where, for any $z,$ $$A(z)=\sum_{j=0}^q(-1)^j z^{2 j+1}\binom {p}{2 j + 1}$$ $$\text {and }\quad B(z)=\sum_{j=0}^q(-1)^j z^{ j}\binom {p}{2 j}.$$ By Eisenstein's Criterion, the polynomial $A(z)$ is irreducible over $\mathbb Q$, and $\deg(A(z))=p>1$, so any solution of $A(z)=0$ is irrational. Since $A(\tan \pi/p)=A(x)=0,$ we have $$\tan \pi/p\not \in\mathbb Q.$$

(2) Assume $\tan \pi m/n \in\mathbb Q,$ where $m, n\in\mathbb Z^+$, with $\gcd (m,n)=1$, and $n$ is divisible by an odd prime $p.$ Let $r\in\mathbb Z$ such that $m r\equiv 1 \pmod n.$ Then $m r\pi/n=\pi (k+1/n)$ where $k\in Z$, so $\tan m r \pi/n=\pm \tan \pi/n.$

By the formula for $\tan (\pi m/n)r$ in terms of $\tan \pi m/n,$ we have $\tan \pi/n\in\mathbb Q.$ Let $n=p s .$ By the formula for $\tan (\pi/n)s$ in terms of $\tan \pi/s,$ we have $\tan \pi/p\in\mathbb Q.$ This contradicts (1). Therefore $\tan \pi m/n\not \in\mathbb Q$.

(3) We can also show that $\tan m 2^{-m}\pi\not \in\mathbb Q$ when $m,n \in\mathbb N$ with odd $m$ and $n\geq 3$.

(4) Therefore if $z=\cos T+i \sin T$ is an $n$th root of $1$, for some $n\in \mathbb N$, and $z\not \in \{\exp (\pi i j/4) :j\in N\}$ then $\tan T\not \in\mathbb Q.$

In particular, if $\tan T=4/3,$ then $T\ne q\pi$ for any $q\in\mathbb Q,$ so $(3+4 i)/5$ is not an $n$th root of $1.$