How to prove that the complex number $\frac{1+\sqrt{15}i}{4}$ of absolute value $1$ is not a root of unity of any order? Just in case, this number is a root of the polynomial $2x^2 – x +2$.
Prove that a given complex number of modulus 1 is not a root of unity
complex numbersroots-of-unity
Related Solutions
From Vieta's formula, $\omega_1+\omega_2+1 = 0$, where $\omega_1,\omega_2$ are two distinct complex roots of unity.
You already knew $\omega_1 = \omega_2^2$ and $\omega_2=\omega_1^2$. Either way you get $\omega+\omega^2+1 = 0$, and multiply both sides by $\omega^4$ to get the result.
$(3+4i)^2 = -7 + 24i = (3+4i)+5(-2+4i)$
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.
Suppose $(a+ib)/c$ is on the unit circle, with $a$ and $b$ coprime.
First, note that if $c$ is even then so are $a$ and $b$ (because $a^2+b^2=c^2$ and squares are congruent to $0$ or $1$ mod $4$), and we supposed they weren't, so $c$ must be odd.
Also, $a$ and $b$ are coprime with $c$ (if $a$ shared a prime factor with $c$, $b$ would also have it).
Now we look at things modulo $(c)$. Let $j \in \Bbb Z/(c)$ such that $a+jb=0$ (this has a unique solution because $b$ is invertible mod $c$).
Then $j^2 = -1 \pmod c$ ($0 = (a+jb)(a-jb) = a^2-j^2b^2$ and since $a^2=-b^2$ and $-b^2$ is invertible, we must have $j^2+1 = 0$), and we can check that the map $\phi : \Bbb Z[i]/(c) \to (\Bbb Z/(c))^2$ defined by $\phi(x+iy) = (x+jy,x-jy)$ is a ring morphism.
If $\phi(x+iy)=0$ then $x+jy=x-jy=0$ from which you get $2x=2y=0$. But $c$ is odd so $2$ is invertible, and so this implies $x=y=0$. This shows that $\phi$ is injective, and since we're dealing with finite rings, it is an isomorphism of rings.
(note that there can be many isomorphisms between those two rings, but this one in particular is the useful one to study $(a+ib)^n$)
Now we have $\phi(a+ib) = (a+jb,a-jb) = (0,a-jb) = (0,2a)$ : Multiplying by $(a+ib)$ is annihilating the first component and multiplying the second by $2a$. And so $\phi((a+ib)^n) = (0^n, (2a)^n)$.
Since $2a$ is invertible mod $c$, the second component is periodic. The first component stays at $0$ from $n \ge 1$, and so the sequence $(\phi((a+ib)^n))_{n \ge 1}$ is periodic. Since $\phi$ is an isomorphism, the sequence $((a+ib)^n \pmod c)_{n \ge 1}$ is also periodic.
Conjugation in $\Bbb Z[i]/(c)$ is translated into the automorphism $(u,v) \mapsto (v,u)$ of $(\Bbb Z/(c))^2$. An element is "real mod $c$" if and only if it is fixed by conjugation. In the $\Bbb Z[i]/(c)$ world this is when the coefficient of $i$ is $0$ (as usual), in the $(\Bbb Z/(c))^2$ world this is when both components are equal.
In particular, since the first component of $\phi((a+ib)^n)$ for $n \ge 1$ is always $0$, it can only be real mod $c$ if both components are $0$. But (unless $c=\pm 1 $ !!) this is impossible because $2a$ is invertible and so its powers can't be zero. So $(a+ib)^n$ is real mod $c$ only when $n=0$ and never again.
Equivalently, this means that the imaginary part of $(a+ib)^n$ is never again a multiple of $c$, and this implies that $(a+ib)^n$ is never real again.
Best Answer
For simplicity, consider $\alpha=1+\sqrt{-15}$. We will argue that the imaginary part of $\alpha^n$ is never $0$ (for $n≥1$). Clearly, that will suffice. We remark that the minimal polynomial of $\alpha$ is $x^2-2x+16$
If we define $$\alpha^n=a_n+b_n\sqrt {-15}$$
with $a_n, b_n\in \mathbb Z$, we must have $$b_n=2b_{n-1}-16b_{n-2}$$
with $b_1=1, b_2=2$.
A routine induction tells us that the order of $2$ in $b_n$ increases by exactly $1$ as $n$ increases by $1$, hence $b_n$ never vanishes, and we are done.