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My attempt:
If $z^{5040} - z^{720}$ is real, that means that their imaginary parts are equal.
$e^{5040i\theta} - e^{720i\theta} = k$ , where $k$ is a real number
$\sin{5040\theta} = \sin{720\theta} $

Let $u=720\theta$
$\sin 7u = \sin u$
By graphing $y=\sin 7u$ and $y=\sin u$, there are $16$ intersections between $0$ and $2\pi$ (including $0$ but not including $2\pi$)
Because the substitution $u=720\theta$ scaled the functions by a factor of $720$ parallel to the x-axis,
there are $11520$ intersections for $\sin{5040\theta} = \sin{720\theta}$ from $\theta=0$ to $\theta= 2\pi$ (not including $2\pi$ because $0+2k\pi=2\pi$ when $k=1$.

Is $11520$ complex numbers correct? There was no answer provided.

helpme
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3 Answers3

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Note that $\sin(u)=\sin(7u)$ has only $14$ solutions on $[0,2 \pi)$. But here is a slightly different approach following the same lines.

Let’s ask the same question for $z^7-z$ first. If $\lvert z\rvert = 1$ then $\overline{z} = z^{-1}$. So then $ z^7-z$ is real if and only if $$z^7-z=z^{-7}-z^{-1}$$ or equivalently $$z^{14}-z^8+z^6-1=(z^8+1)(z^6-1)=0.$$ This equation has $14$ solutions on the unit circle. Now the substitution $z\leftarrow z^{720}$ increases this amount by a factor of $720$ (since the mapping is $720$ to one). So $720\cdot 14 = 10080$ possible values for $z$.

WimC
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Your work is good up to now, but let’s go back to $\sin(5040\theta) = \sin(720\theta)$ so it’s simpler. Call the sines arguments $x$ and $y$. Using the trig circle: either $x = y \pmod{2\pi}$ or $x = \pi-y \pmod{2\pi}$.

In the first case, $5040\theta = 720\theta + 2k\pi$ so $\theta = \frac{k}{2160}\pi$. Analogously, in the second case $\theta = \left(\frac1{5760} + \frac k {2880}\right) \pi$.

Consider those sets of solutions $\mathrm{mod}\ 2\pi$ and the number of elements of $S_1\cup S_2$, which is $|S_1| + |S_2| - |S_1\cap S_2|$, and you’re done.

Lucas Henrique
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Solve $\sin{7u} = \sin{u} $ as follows,

$$ \sin{7u} - \sin{u} = 2\cos 4u \sin 3u=0$$

$\sin4u = 0$ yields $ u = \frac {\pi n}4$ and $\cos3u=0$ yields $u=\frac\pi6+ \frac {\pi n}3$. Then,

$$z= e^{ \frac { i2\pi n}{8\cdot 720}}, \>\>\>\>\> z= e^{ \frac{i\pi}{6\cdot 720 }+i\frac { i2\pi n}{6\cdot 720}}$$

Thus, number of distinct solutions are $8\cdot 720 + 6\cdot 720= 10080$.

Quanto
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