Number of complex numbers such that $z^{80} = 1$ and other properties

Question

Let $$A = \left \{ z \in \mathbb{C} : \Re z > 0, \Im z < 0, z^{80} = 1 \right \}$$

Then, the number of elements in $A$ is $19$, $20$, $21$, or $22$?

I just started studying complex numbers and I've come across this question. I really don't know how to approach it. Do you think it's solvable with basic techniques?

I think we state the problem as finding the number of roots $r$ of the polynomial $z^{80}-1$ such that $r$ is in the fourth quadrant of the complex plane. I also recall that if $z$ is a root, then $\overline{z}$ is one too? That restricts the answers to the even ones, so $20$ or $22$, right?

Answer 1

Hint: To find all the solutions of $z^{80}=1$ write $z$ is polar form $$ z=rcis(\theta) $$

then we have $$ r^{80}cis(80\theta)=1=1cis(0) $$

and so $r=\sqrt[80]{1}=1$ and $80\theta=2\pi k$ with $k\in\mathbb{Z}$. The solutions are $$ \theta=\frac{\pi k}{40};\, k=0,1,...,79 $$

Now find which of them satisfies the rest of the conditions

Answer 2

Another approach: Note that the roots of unity of order n form a regular n-sided polygon inscribed in the unit circle.

Answer 3

$$A= \{ e^{\frac{2 k\pi i}{80}} : k \in \mathbb{Z}, 60 < k < 80\}$$ hence the answer is 19.