Sum of squares of elements of $A=\left\{\left|z^{n}+\frac{1}{z^{n}}\right| \mid n \in \mathbb{N}, z \in \mathbb{C}, z^{4}+z^{3}+z^{2}+z+1=0\right\}$ $A=\left\{\left|z^{n}+\frac{1}{z^{n}}\right| \mid n \in \mathbb{N}, z \in \mathbb{C}, z^{4}+z^{3}+z^{2}+z+1=0\right\}$
Sum of the squares of the elements of A = ?
I have no idea how to approach this. All I could figure out was that ${z}^5=1$ and that I should do something with Moivre formula. I actually have a solution to this in a textbook, but I simply could not understand it, so please explain in detail if possible.
 A: It looks like the set $A$ might be big and complicated, but in fact it isn't.  It only has three elements in it! The first part of this problem is going to be to whittle down those infinitely many elements by figuring out which are duplicates.  It's hard to say how to approach that without knowing your background, but here's the idea:

*

*There are four options for $z$,  the roots of the polynomial $T^4+T^3+T^2+T+1$ (and you're right that they'll all satisfy $z^5=1$ - what are the complex numbers that satisfy $z^5=1$?) What are they?


*For each of those options we get a set $\left\{\left|z^n+\tfrac1{z^n}\right||n\in\mathbb N\right\}$, and $A$ is the union of those four sets.  Each of them will turn out to be finite, so use the fact that $z^5=1$ to figure out which elements are in there.


*Finally, figure out the which elements between the four sets from the previous step are duplicates.
Feel free to ask questions if there are parts of this outline you don't understand or can't figure out!
After doing all of that, you should wind up with just a finite list of explicit elements that make up the set $A$.  And from there, you can compute the sum of their squares explicitly.
