# How many solutions to $z^{6!}-z^{5!}\in \mathbb{R}$ and $|z|=1$

How many solutions $\Im(z^{720}-z^{120})=0$ where $|z|=1$? (AIME I $2018$ #$6$)

The following is my interpretation of the first solution.

The next step is to see how many solutions to $\sin(720\theta)-\sin(120\theta)=0$. (1)

One way to do this is consider $\sin(6\omega)-\sin(\omega)$, which has $12$ solutions. (2)

For each solution to (2), corresponds $120$ solutions to (1). I understand that.

But how do I know that $sin(6\omega)=sin(\omega)$ has $12$ solutions?

I'd say during a real contest I could come up with everything but the $12$ solutions part. I tried sketching both graphs on the same plane for $0<x<2\pi$ and I only got 9 intersections, while I can see all $12$ clearly on Desmos, so my drawing is unreliable.

$$\sin6w=\sin w\iff 6w=2k\pi+w \text{ or } 6w=2k\pi+\pi-w\iff w=\dfrac{2k\pi}{5} \text{ or } w=\dfrac{2k+1}{7}\pi$$ with $k\in\mathbb{Z}$. then $$w=0, \dfrac{2\pi}{5}, \dfrac{4\pi}{5}, \dfrac{6\pi}{5}, \dfrac{8\pi}{5}, \dfrac{\pi}{7}, \dfrac{3\pi}{7}, \dfrac{5\pi}{7}, \dfrac{7\pi}{7}, \dfrac{9\pi}{7}, \dfrac{11\pi}{7}, \dfrac{13\pi}{7}$$

By the standard sum-to-product formula, $$\sin 6\omega - \sin \omega=2 \cos \frac{7\omega}{2} \sin \frac{5 \omega}{2}$$ The first factor has $7$ roots in $[0,2\pi]$ while the second has $5$, and there is no overlap.

Since \begin{align} \sin(\theta) &=\sin(\pi-\theta)\\ &=\sin(3\pi-\theta)\\ &=\sin(5\pi-\theta)\\ &=\sin(7\pi-\theta)\\ &=\sin(9\pi-\theta)\\ &=\sin(11\pi-\theta)\\ &=\sin(\underbrace{13\pi-\theta}_{6\theta}) \end{align}\tag1 if $6\theta$ equals any of the arguments to $\sin(x)$ on the right, then $\sin(\theta)=\sin(6\theta)$. That is, $$\theta\in\left\{\frac\pi7,\frac{3\pi}7,\frac{5\pi}7,\pi,\frac{9\pi}7,\frac{11\pi}7,\frac{13\pi}7\right\}\tag2$$

Since \begin{align} \sin(\theta) &=\sin(\theta)\\ &=\sin(2\pi+\theta)\\ &=\sin(4\pi+\theta)\\ &=\sin(6\pi+\theta)\\ &=\sin(\underbrace{8\pi+\theta}_{6\theta}) \end{align}\tag3 if $6\theta$ equals any of the arguments to $\sin(x)$ on the right, then $\sin(\theta)=\sin(6\theta)$. That is, $$\theta\in\left\{0,\frac{2\pi}5,\frac{4\pi}5,\frac{6\pi}5,\frac{8\pi}5\right\}\tag4$$

$(2)$ and $(4)$ give $12$ solutions to $\sin(\theta)=\sin(6\theta)$. Now we will show that there are at most $12$ solutions.

$\sin(\theta)=0$ has two solutions, $\theta\in\{0,\pi\}$, both of which are solutions to $\sin(\theta)=\sin(6\theta)$. If $\sin(\theta)\ne0$, then \begin{align} 1 &=\frac{\sin(6\theta)}{\sin(\theta)}\\ &=\frac{e^{i6\theta}-e^{-i6\theta}}{e^{i\theta}-e^{-i\theta}}\\[3pt] &=e^{i5\theta}+e^{i3\theta}+e^{i\theta}+e^{-i\theta}+e^{-i3\theta}+e^{-i5\theta}\tag5 \end{align} which says that $e^{i\theta}$ is a root of $$x^{10}+x^8+x^6-x^5+x^4+x^2+1=0\tag6$$ which has at most $10$ roots.

Thus, the $12$ solutions listed in $(2)$ and $(4)$ are all there are.