I want to show that if $z\in\mathbb{C}$ such that $|z|=1$ and $1+z^{k_1}+z^{k_2}=0$ for integers $k_1<k_2$, than $z$ is a root of unity.
Here is my approach:
Suppose $z=\cos(\theta)+i\sin(\theta)$. Then we have $$1+\cos(k_1\theta)+\cos(k_2\theta)+i(\sin(k_1\theta)+\sin(k_2\theta))=0.$$
This would require $$\sin(k_1\theta)+\sin(k_2\theta)=2\sin\left(\frac{(k_1+k_2)\theta}{2}\right)\cos\left(\frac{(k_2-k_1)\theta}{2}\right)=0.$$
This would mean either
$$\frac{(k_1+k_2)\theta}{2}=n\pi$$
or
$$\frac{(k_2-k_1)\theta}{2}=\frac{(2n+1)\pi}{2}$$
so either
$$\theta=\frac{2n\pi}{k_1+k_2}, \frac{(2n+1)\pi}{k_2-k_1}$$
Now note that
$$1+\cos(k_1\theta)+\cos(k_2\theta)=1+2\cos\left(\frac{(k_1+k_2)\theta}{2}\right)\cos\left(\frac{(k_2-k_1)\theta}{2}\right)$$
But now I do not know how to continue, because I would need the real part to be $0$. Any hints on how to continue? My guess is that it would be a $k_1+k_2$-root of unity. Also, is there an algebraic way/trick of doing this without resorting to using polar forms and trig identities? Thank you.
Best Answer
The key point is that $z^{k_1}$ and $z^{k_2}$ are complex conjugate.
Indeed, since $|z|=1$, let $z=e^{i\theta}$ then
$$1+e^{ik_1\theta}+e^{ik_2\theta}=0$$
requires that $e^{ik_1\theta}+e^{ik_2\theta}$ is real that is $e^{ik_2\theta}=e^{-ik_1\theta}$ with
$$e^{ik_1\theta}+e^{-ik_1\theta}=2\cos (k_1 \theta)=2\Re(z^{k_1})=-1$$
therefore
Following your way, we have that the condition
$$1+\cos(k_1\theta)+\cos(k_2\theta)+i(\sin(k_1\theta)+\sin(k_2\theta))=0$$
requires
$$\sin(k_1\theta)+\sin(k_2\theta) =0 \iff k_2\theta =-k_1\theta +2n\pi \; \lor \; k_2\theta =\pi +k_1\theta +2n\pi$$
then for $k_2\theta =-k_1\theta +2n\pi$ we obtain
$$1+\cos(k_1\theta)+\cos(k_2\theta) =0 \iff 1+2\cos(k_1\theta) =0$$
that is
or
which are equivalent, while for $k_2\theta =\pi +k_1\theta +2n\pi$ we obtain
$$1+\cos(k_1\theta)+\cos(k_2\theta) =0 \iff 1=0$$