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.)
Show that $\frac{3}{5} + i\frac{4}{5}$ isn’t a root of unity
roots-of-unity
Best Answer
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.