Root of a quadratic equation that has modulus $1$

Question

Let us suppose $\alpha \in \mathbb C$ and $|\alpha|=1$ and $\alpha$ satisfies a monic quadratic equation. Then prove that $\alpha^{12} =1$.

Show me the right way to solve this. Thanks in advance.

Answer 1

Suppose $\alpha\in \bf C$ is a root of some quadratic polynomial $x^2+ax+b\in{\bf Q}[x]$ and $|\alpha|=1$.

Then $\displaystyle\alpha=\frac{-a\pm\sqrt{a^2-4b}}{2}$. If $\alpha\ne\pm1$ then $a^2<4b$, and

$$1=\alpha\overline{\alpha}=\frac{a^2-(a^2-4b)}{4}=b,\quad -2<a<+2.$$

It is false that $\alpha$ must be a $12$th root of unity; check either root of e.g. $x^2+\frac{1}{2}x+1$. However if we restrict $a,b\in\bf Z$ to be integers, then $-2<a<2\iff a\in\{-1,0,+1\}$, and it can be checked that

• the values $\alpha=\pm1$ are second roots of unity,
• roots of $x^2+1$ are primitive fourth roots of unity, $\pm i$,
• roots of $x^2-x+1$ are primitive sixth roots of unity,
• roots of $x^2+x+1$ are primitive cube roots of unity,

and all of the above are twelfth roots of unity. However, observe that none of these is a primitive twelfth root; the possible values for $\alpha$ are precisely the nonprimitive twelfth roots of unity.

The last two bullet points above are covered in the theory of cyclotomic polynomials, as the two polynomials listed are in fact $\Phi_6(x)$ and $\Phi_3(x)$.

• If $x^2+x+1=0$ then multiply by $x-1$ to obtain the relation $x^3-1=0$, but $x=1$ is impossible so $x$ must be a primitive cube root.
• If $x^2-x+1=0$ then multiply by $(x^2+x+1)(x^2-1)$ to obtain $x^6-1=0$, but $x$ cannot be a root of $x^2-1$ or of $x^2+x+1$ at the same time, so it is not a second or third root of unity while it is a sixth root, hence it is a primitive sixth root of unity.