Show that $\frac{3}{5} + i\frac{4}{5}$ isn't a root of unity

First of all, we have to convert this complex number to trigonometric form. In particular: $$\theta=\arctan\left(\frac{4}{3}\right)$$ Now, we use De-Moivre formula for rising $z$ to the $n$ power, and we have: $$z^n=1^n(\cos(n\theta)+i\sin(n\theta))$$ Substituing, we have: $$z^n=\cos\left(n\arctan\left(\frac{4}{3}\right)\right)+i\sin\left(n\arctan\left(\frac{4}{3}\right)\right)=1=\cos\left(\frac{\pi}{2}\right)+i\sin\left(\frac{\pi}{2}\right)$$ This, leads to: $$n\arctan\left(\frac{4}{3}\right)=\frac{\pi}{2}$$ Now, we have to show that $\pi$ and $\arctan\left(\frac{4}{3}\right)$ are incommensurable.

If $\alpha=\arctan\frac{4}{3}$ and $\pi$ were commensurable, then it would be true that $p\alpha=q\pi$ for some integers $p,q\ne 0$. Take $z=\frac{3}{5}+\frac{4}{5}i$: then $z=\cos\alpha+i\sin\alpha$. Thus $z^p=\cos(p\alpha)+i\sin(p\alpha)=\cos(q\pi)+i\sin(q\pi)=\pm 1$.

Now this means that $(5z)^p=(3+4i)^p=\pm 5^q$. Note that $3+4i=i(2+i)^2$, so:

$$i^p(2+i)^{2p}=\pm(2+i)^p(2-i)^p$$

Cancelling $(2+i)^p$, we get:

$$i^p(2+i)^p=\pm(2-i)^p$$

which is impossible because, in the ring $\mathbb Z[i]$ of Gaussian integers, $2-i$ and $2+i$ are distinct prime elements, so the unique factorisation would be violated.

Here is another kind of proof. There is perhaps room to explore the hinterland of the concept of an algebraic integer - this is a concept, though, which emerges naturally from this kind of question.

Let $5x=3+4i$ then $(5x-3-4i)(5x-3+4i)=25x^2-30x+25=0$ so that $5x^2-6x+5=0$ is the minimal polynomial for $x$ over the integers. Since it is not monic, $x$ is not an algebraic integer and can't therefore be a root of unity (roots of unity are algebraic integers).