SOLUTION: find all the solutions in the complex numbers, of z^4 + z^3 + z^2 + z + 1 = 0. use: (1-r)(1+r+r^2+...+r^n)=1-r^(n+1) which of the solutions are conjugates?

According to the formula for the sum of a geometric progression,


     = 


It is the same as 

     = .    (*)


        (and the formula, which is the hint to your problem, says the same).


From the formula (*), every zero of the polynomial p(z) = z^4 + z^3 + z^2 + z + 1  is the root of the polynomial q(z) = .


In other words, if   = 0,  then   = 0,  or, which is the same,   = 1.


Thus every root of the polynomial    is the complex root of the degree 5 of 1.


And vice versa, every root of the degree 5 of 1, different from 1, is the root of the polynomial  .   It follows from the formula (*).


Thus the four roots of the polynomial    are  all complex root of degree 5 of 1, different of 1.


OK, very good.  


From complex number theory, you can conclude then that these roots have the form

     = ,  where k = 1, 2, 3, 4.


Geometrically, these four complex numbers are vertices of the regular pentagon of the radius 1 centered at (0,0), the origin of the complex plane.


These four roots are vertices of the regular polygon, different of its vertex (1,0).


Of these roots,   is conjugate to   and   is conjugate to .