Prove that $\frac{3}{5} + \frac{4}{5}i$ is not a root of unity

Question

I want to prove that $z=\frac{3}{5} + \frac{4}{5}i$ is not a root of unity, although its absolute value is 1.

When transformed to the geometric representation: $$z=\cos{\left(\arctan{\frac{4}{3}}\right)} + i\sin{\left(\arctan{\frac{4}{3}}\right)}$$ According to De Moivre's theorem, we get: $$z^n= \cos{\left(n\arctan{\frac{4}{3}}\right)} + i\sin{\left(n\arctan{\frac{4}{3}}\right)}$$

Now, if for $n\in \mathbb{N}: z^n=1$, then the imaginary part of the expression above must be zero, therefore:

$$\sin{\left(n\arctan{\frac{4}{3}}\right)}=0 \iff n\arctan{\frac{4}{3}} = k\pi, \ \ \ k \in \mathbb{N}$$

And we get that for $z$ to be a root of unity for some natural number $n$, $n$ must be in the form: $$n = \frac{k\pi}{\arctan{\frac{4}{3}}}, \ \ \ k \in \mathbb{N}$$

On the other hand, for $z^n=1$ it must be that:

$$\cos{\left(n\arctan{\frac{4}{3}}\right)} = 1 \iff n\arctan{\frac{4}{3}} = 2l\pi, \ \ \ l \in \mathbb{N}$$

and thus

$$n = \frac{2l\pi}{\arctan{\frac{4}{3}}}, \ \ \ l \in \mathbb{N}$$

By comparing those two forms of $n$, it must be the case that $k=2l$ and for $n$ to satisfy $z^n = 1$. What follows is that $n$ should be in the form

$$n = \frac{2l\pi}{\arctan{\frac{4}{3}}}, \ \ \ l \in \mathbb{N}$$

But, at the same time, $n$ must be a natural number. Should I prove now that such $n$ cannot even be a rational number, let alone a natural one? Or how should I approach finishing this proof?

Answer 1

As is shown in this answer, Niven's theorem says that $\sin(\pi p/q)$ is rational only when $\sin(\pi p/q)\in\left\{-1,-\frac12,0,\frac12,1\right\}$. However, $\sin\left(\arg\left(\frac35+\frac45i\right)\right)=\frac45$, so we know that $\arg\left(\frac35+\frac45i\right)$ is not a rational multiple of $\pi$.

Answer 2

You've reduced your problem to showing that $\arctan(4/3)$ is not a rational multiple of $\pi$. Take a look at Niven's theorem. (This is pretty much equivalent to your problem in difficulty - as far as I know the bulk of any proof that $\arctan(x)$ is not a rational multiple of $\pi$ for some specific rational $x$ can be generalized to all rational $x$ for which it is true.)

If you haven't seen this theorem before, and want to try to come up with a proof, I'll give you a hint: the key ingredients are Chebyshev polynomials and the rational root theorem.

Answer 3

$$z=\frac{3+4i}5=\frac{(1+2i)^2}5=\frac{1+2i}{1-2i}.$$ The ring $\Bbb Z[i]$ is a UFD and $1+2i$ and $1-2i$ are non-associate primes therein. So in $\Bbb Z[i]$, $(1+2i)^n$ is never divisible by $1-2i$ so that $$z^n=\frac{(1+2i)^n}{(1-2i)^n}$$ is always in lowest terms and cannot cancel to equal $1$.

Answer 4

From here, we know that

If $x$ is a rational multiple of $\pi$, then $2\cos(x)$ is an algebraic integer.

So if $x=\arctan(4/3)$ is a rational multiple of $\pi$, then $2\cos(x)=6/5$ is an algebraic integer, which is also a rational number, and hence an integer. This is a contradiction. Therefore $\frac{3+4i}5$ is not a root of unity.

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Hope this helps.

Answer 5

Since $\mathbf Z[i]$ is a UFD it is integrally closed (the rational roots theorem holds in $\mathbf Z[i][x]$). Therefore any root of some $x^n-1$ in $\mathbf Q(i)$ is in $\mathbf Z[i]$, so no number in $\mathbf Q(i)$ that is not in $\mathbf Z[i]$ could be a root of unity (integral over $\mathbf Z$ and thus also over $\mathbf Z[i]$).

Answer 6

Hint: Look at the real and imaginary parts of $(3+4i)^n$, mod 5.

If $(3+4i)^n$ never has real and imaginary parts a multiple of 5, then $\left(\dfrac{3+4i}5\right)\!^{\Large n}$ never has real and imaginary parts an integer. (For $n\ge2$, anyway.) In particular, it never equals $1$, and $\dfrac{3+4i}5$ is not a root of unity.

Answer 7

Here's a simple proof using only divisibility in Gaussian integers $\,\Bbb G = \Bbb Z[i] = \{ j\! +\! k\, i\, :\, j,k\in\Bbb Z\}.\,$
Here $\ b\mid a\ $ means $\,b\,$ divides $\,a\,$ in $\Bbb G,\,$ i.e. $\, bx=a\,$ for some $\,x\in \Bbb G,\,$ and $\,\overline{j\!+\!k\,i} = j\! -\! k\, i\,$

Let $\,p = 2\!-\!i\,$ and $\,a = \dfrac{3\!+\!4i}5 = \dfrac{\bar p\bar p}{p\bar p} = \dfrac{\bar p}p\,.\ $ Then $\ a^n = 1\,\Rightarrow\, p^n = \bar p^n\Rightarrow\,p\mid \bar p^n\,$ by $\,n>0.$

But $\ p\mid \color{#c00}{\bar px}\,\Rightarrow\, p\mid x.\ $ Proof $\,\ p\mid(p\!+\!\color{#c00}{\bar p})\color{#c00}x = 4x,\,\ p\mid p{\bar px} = 5x,\,\\ $ so $\,\ p\mid 5x\!-\!4x= x$

So $\ \ p\mid \bar p \bar p^{n-1}\Rightarrow\, p\mid \bar p^{n-1}\cdots\Rightarrow\, p\mid \bar p\cdot 1\,\Rightarrow\, p\mid 1,\,$ contra $\,\dfrac{1}p = \dfrac{\bar p}{\bar p p} = \dfrac{2\!+\!i}5 \not\in \Bbb G =\Bbb Z[i]$

Remark $\ $ More generally. RRT = Rational Root Test extends the above to show that $a$ is not the root of any monic polynomial $\in \Bbb G[x].\,$ In fact - just as for $\Bbb Z\,$ - RRT holds for all fractions over $\,\Bbb G\,$ because it enjoys a Euclidean division algorithm, so gcds exist, so the RRT proof still works.