Show that $\frac{3}{5} + i\frac{4}{5}$ isn't a root of unity Intuitively, I see why $\frac{3}{5} + i\frac{4}{5}$ is not a root of unity because $\frac{2\pi}{\arctan(4/3)}$ appears to be irrational when I plug into my calculator. But how to I show this rigorously?I think contradiction should work, but still I wasnt able to show this rigorously. Any help will be appreciated. (Also: I first tried using the idea that roots of unity had angle $\frac{2\pi}{n}$ for n in the natural numbers but then I realized that n should be replaced by rational number because $e^{\frac{i4\pi}{7}}$ is a root of unity too.) 
 A: 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.
A: 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).
A: This can be a very nice elementary proof, in my opinion, using an infinite descent argument.
Suppose there is a $p$ so that $z^p=1$. Than we start with:
$(3+4i)^p=5^p$ [1]
We can get this lemma:

Lemma: given $a,b$ integers so that $(a+bi)^q=5^p$. If 5 does not divide $a$, $a>1$ and $p,q \ge 1$, than $q$ is even.
Proof: if we expand the right left term with the binomial expansion:
$(a+bi)^q=\sum_{k=0}^q {q \choose k} a^{n-k} b^k i^{k}=5^p$
we see that only terms with $k$ even contribute to the real part. If $q$ were odd than $a$ would divide the real part. This is acontradiction because $(a,5)=1$ and $a>1$: only powers of $5$ divide instead the right member.

So let's start with $a_0=3$ and $b_0=4$. Since $(3,5)=1$ we can apply the lemma to [1] and conclude that $p$ is even. Therefore $p=2p',p' \ge 1$ and $(-7+24i)^{p'}=5^{p}$. Now we apply the lemma again with $a_1=-7$,$b_1=24$, $q=p'$. Again $(-7,5)=1$ so that $p'$ is even by the lemma. Note that we are generating a series of $(a_i,b_i)$ and the process stops when $5$ divides $a_i$ or $a_i$ gets equal to $1$.
It is easy to check that we are generating the recurrent sequence $a_{i+1}=a_i^2-b_i^2$ and $b_{i+1}=2a_i b_i$. Starting with $(a_0,b_0)=(3,4)$ this gives mod 5 the constant series $(3,4)$. Therefore $5$ will never divide $a_i$ nor will it become equal to $1$ and the process can go on for ever. This means that $p$ shall be infinitly divisible by 2, a contradiction.
A: Hint: An $n$th root of unity has the form $e^{2\pi ik/n}$ with $0\leq k\leq n-1$.
By Euler's formula, $r(\cos \phi + i\sin\phi) = re^{i\phi}$.
