Finding the number of points on unit circle satisfying a criteria.

Question

Find the number of numbers$(z)$(Or the number of solution for $z$) on the unit circle such that :-
$z^{6!}-z^{5!}$ is a real number.

Answer 1

For any point on the unit circle we have $z = e^{i\theta}$, whose imaginary part is $\sin(\theta)$. Now for $z^{6!} - z^{5!}$ to be real we must have that both $z^{6!}$ and $z^{5!}$ to have the same imaginary part. This corresponds to:

$$\sin(6!\theta) = \sin(5!\theta)$$

Using a well-known trigonometric identity this corresponds to:

$$\sin\left(\frac{6!-5!}{2}\theta \right)\cos\left(\frac{6!+5!}{2}\theta \right) = 0$$

So we either have $\sin(3\theta) = 0$ or $\cos(420\theta) = 0$. Now finding all values for $\theta$ shouldn't be hard. Can you continue from here?

Answer 2

Hint: $w$ is real iff $w-\bar{w}=0$ this gives the condition $$z^{6!}-\bar{z}^{6!}=z^{5!}-\bar{z}^{5!}$$ where $z=e^{it}$ will be $$\sin6!t=\sin5!t$$ this leads to $6!t=5!t+2k\pi$ or $6!t=-5!t+2k\pi+\pi$.